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Importance of Maths in the Field of Computing
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Unit – 11
Maths for Computing
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Maths for Computing
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Unit 11 :Maths for Computing
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LO1 Use applied number theory in practical computing scenarios.
Pass, Merit & Distinction P1
P2
M1
D1
Descripts
LO2 Analyse events using probability theory and probability distributions.
Pass, Merit & Distinction P3
Descripts
P4
M2
D2
LO3 Determine solutions of graphical examples using geometry and vector methods.
Pass, Merit & Distinction P5
P6
M3
D3
Descripts
LO4 Evaluate problems concerning differential and integral calculus.
Pass, Merit & Distinction P7
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P8
M4
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Unit – 11
3
Maths for Computing
Assignment 01
General Guidelines
1. A Cover page or title page – You should always attach a title page to your assignment. Use
previous page as your cover sheet and make sure all the details are accurately filled.
2. Attach this brief as the first section of your assignment.
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Word Processing Rules
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9. Non-submission of work without valid reasons will lead to an automatic RE FERRAL. You
will then be asked to complete an alternative assignment.
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citation and a reference list.
11. If you are proven to be guilty of plagiarism or any academic misconduct, your grade could
be reduced to A REFERRAL or at worst you could be expelled from the course.
Unit – 11
4
Maths for Computing
Assignment 01
Student Declaration
I hereby, declare that I know what plagiarism entails, namely, to use another’s work and to present
it as my own without attributing the sources in the correct way. I further understand what it means
to copy another’s work.
1. I know that plagiarism is a punishable offence because it constitutes theft.
2. I understand the plagiarism and copying policy of the Edexcel UK.
3. I know what the consequences will be if I plagiaries or copy another’s work in any of the
assignments for this program.
4. I declare therefore that all work presented by me for every aspects of my program, will be
my own, and where I have made use of another’s work, I will attribute the source in the
correct way.
5. I acknowledge that the attachment of this document signed or not, constitutes a binding
agreement between myself and Edexcel UK.
6. I understand that my assignment will not be considered as submitted if this document is not
attached to the attached.
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Unit – 11
5
Maths for Computing
Date:
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Assignment 01
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Unit – 11
6
Date
Maths for Computing
Assignment 01
Assignment Brief
Student Name /ID Number
Unit Number and Title
Unit 11 :Maths for Computing
Academic Year
2021/2022
Unit Tutor
Importance of Maths in the Field of Computing
Assignment Title
Issue Date
Submission Date
IV Name & Date
Submission Format:
This assignment should be submitted at the end of your lesson, on the week stated at the front of this brief. The
assignment can either be word-processed or completed in legible handwriting.
If the tasks are completed over multiple pages, ensure that your name and student number are present on each
sheet of paper.
Unit Learning Outcomes:
LO1
Use applied number theory in practical computing scenarios.
LO2
Analyse events using probability theory and probability distributions.
LO3
LO4
Determine solutions of graphical examples using geometry and vector methods.
Evaluate problems concerning differential and integral calculus.
Unit – 11
7
Maths for Computing
Assignment 01
Assignment Brief and Guidance:
Activity 01
Part 1
1. A tailor wants to make square shaped towels. The required squared pieces of cloth will be cut from
a ream of cloth which is 20 meters in length and 16 meters in width.
a) Find the minimum number of squared pieces that can be cut from the ream of cloth without
wasting any cloth.
b) Briefly explain the technique you used to solve (a).
2. On the first day of the month, 4 customers come to a restaurant. Afterwards, those 4 customers
come to the same restaurant once in 2,4,6 and 8 days respectively.
a) On which day of the month, will all the four customers come back to the restaurant together?
b) Briefly explain the technique you used to solve (a).
Part 2
3. Logs are stacked in a pile with 24 logs on the bottom row and 10 on the top row. There are 15 rows
in all with each row having one more log than the one above it.
a) How many logs are in the stack?
b) Briefly explain the technique you used to solve (a).
4. A company is offering a job with a salary of Rs. 50,000.00 for the first year and a 4% raise each
year after that. If that 4% raise continues every year,
a) Find the total amount of money an employee would earn in a 10-years career.
b) Briefly explain the technique you used to solve (a).
Part 3
5. Define the multiplicative inverse in modular arithmetic and identify the multiplicative inverse of 6
mod 13 while explaining the algorithm used.
6. Prime numbers are important to many fields. In the computing field also prime numbers are
applied. Provide examples and in detail explain how prime numbers are important in the field of
computing.
Activity 02
Part 1
1. Define ‘Conditional Probability’ with a suitable example.
2. The manager of a supermarket collected the data of 25 customers on a certain date. Out of them 5
purchased Biscuits, 10 purchased Milk, 8 purchased Fruits, 6 purchased both Milk and Fruits.
Let B represents the randomly selected customer purchased Biscuits, M represents the
randomly selected customer purchased Milk and F represents the randomly selected customer
purchased Fruits.
Represent the given information in a Venn diagram. Use that Venn diagram to answer the
following questions.
Unit – 11
8
Maths for Computing
Assignment 01
a)
b)
Find the probability that a randomly selected customer either purchased Biscuits or Milk.
Show that the events “The randomly selected customer purchased Milk” and “The randomly
selected customer purchased Fruits” are independent.
3. Suppose a voter poll is taken in three states. Of the total population of the three states, 45% live in
state A, 20% live in state B, and 35% live in state C. In state A, 40% of voters support the liberal
candidate, in state B, 30% of the voters support the liberal candidate, and in state C, 60% of the
voters support the liberal candidate.
Let A represents the event that voter is from state A, B represents the event that voter is from
state B and C represents the event that voter is from state C. Let L represents the event that a
voter supports the liberal candidate.
a) Find the probability that a randomly selected voter does not support the liberal candidate and
lives in state A.
b) Find the probability that a randomly selected voter supports the liberal candidate.
c) Given that a randomly selected voter supports the liberal candidate, find the probability that
the selected voter is from state B.
4. In a box, there are 4 types [Hearts, Clubs, Diamonds, Scorpions] of cards. There are 6 Hearts
cards, 7 Clubs cards, 8 Diamonds cards and 5 Scorpions cards in the box. Two cards are selected
randomly without replacement.
a)
Find the probability that the both selected cards are Hearts.
b)
Find the probability that one card is Clubs and the other card is Diamonds.
c)
Find the probability that the both selected cards are from the same type
Part 2
5. Differentiate between ‘Discrete Random Variable’ and ‘Continuous Random Variable”.
6. Two fair cubes are rolled. The random variable X represents the difference between the values of
the two cubes.
a) Find the mean of this probability distribution. (i.e. Find E[X] )
b) Find the variance and standard deviation of this probability distribution.
(i.e. Find V[X] and SD[X])
The random variables A and B are defined as follows:
A = X-10 and B = [(1/2)X]-5
c) Show that E[A] and E[B].
d) Find V[A] and V[B].
e) Arnold and Brian play a game using two fair cubes. The cubes are rolled, and Arnold records
his score using the random variable A and Brian uses the random variable B. They repeat this
for a large number of times and compare their scores. Comment on any likely differences or
similarities of their scores.
7. A discrete random variable Y has the following probability distribution.
Y=y
1
P(Y=y)
1/3
where k is a constant.
a) Find the value of k.
b) Find P(Y≤3).
c) Find P(Y>2).
Unit – 11
9
2
1/6
Maths for Computing
3
1/4
Assignment 01
4
k
5
1/6
Part 3
10. The “Titans” cricket team has a winning rate of 75%. The team is planning to play 10 matches in
the next season.
a) Let X be the number of matches that will be won by the team. What are the possible values of
X?
b) What is the probability that the team will win exactly 6 matches?
c) What is the probability that the team will lose 2 or less matches?
d) What is the mean number of matches that the team will win?
e) What are the variance and the standard deviation of the number of matches that the team will
win?
11. In a boys’ school, there are 45 students in grade 10. The height of the students was measured. The
mean height of the students was 154 cm and the standard deviation was 2 cm. Alex’s height was
163 cm. Would his height be considered an outlier, if the height of the students were normally
distributed? Explain your answer.
12. The battery life of a certain battery is normally distributed with a mean of 90 days and a standard
deviation of 3 days.
For each of the following questions, construct a normal distribution curve and provide the answer.
a) About what percent of the products last between 87 and 93 days?
b) About what percent of the products last 84 or less days?
For each of the following questions, use the standard normal table and provide the answer.
c) About what percent of the products last between 89 and 94 days?
d) About what percent of the products last 95 or more days?
13. In the computing field, there are many applications of Probability theories. Hashing and Load
Balancing are also included to those. Provide an example for an application of Probability in
Hashing and an example for an application of Probability in Load Balancing. Then, evaluate in
detail how Probability is used for each application while assessing the importance of using
Probability to those applications.
Unit – 11
10
Maths for Computing
Assignment 01
Activity 03
Part 1
1. Find the equation (formula) of a circle with radius r and center C(h,k) and if the Center of a circle
is at (3,-1) and a point on the circle is (-2,1) find the formula of the circle.
2. Find the equation (formula) of a sphere with radius r and center C(h, k, l) and show that x2 + y2 + z2
- 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius.
3. Following figure shows a Parallelogram.
If a=(i+3j-k) ,b=(7i-2j+4k), find the area of the Parallelogram.
Part 2
4. If 2x - 4y =3, 5y = (-3)x + 10 are two functions. Evaluate the x, y values using graphical method.
5. Evaluate the surfaces in ℝ3 that are represented by the following equations.
i. y = 4
ii. z = 5
6. Following figure shows a Tetrahedron.
Construct an equation to find the volume of the given Tetrahedron using vector methods and if the
vectors of the Tetrahedron are a=(i+4j-2k) ,b=(3i-5j+k) and c=(-4i+3j+6k), find the volume of the
Tetrahedron using the above constructed equation..
Unit – 11
11
Maths for Computing
Assignment 01
Activity 04
Part 1
1. Determine the slope of the following functions.
i. f(x) = 2x – 3x4 + 5x + 8
ii. f(x) = cos(2x) + 4x2 – 3
2. Let the displacement function of a moving object is S(t) = 5t3 – 3t2 + 6t. What is the function for the
velocity of the object at time t.
Part 2
3. Find the area between the two curves f(x) = 2x2 + 1 and g(x) = 8 – 2x on the interval
(-2) ≤ x ≤ 1 .
4. It is estimated that t years from now the tree plantation of a certain forest will be increasing at the
rate of 3t 2 + 5t + 6 hundred trees per year. Environmentalists have found that the level of Oxygen
in the forest increases at the rate of approximately 4 units per 100 trees. By how much will the
Oxygen level in the forest increase during the next 3 years?
Part 3
5. Sketch the graph of f(x) = x5- 6x3 + 3 by applying differentiation methods for analyzing where the
graph is increasing/decreasing, local maximum/minimum points [Using the second derivative test],
concave up/down intervals with inflection points.
6. Identify the maximum and minimum points of the function f(x)= 2x3 - 4x4 + 5x2 by further
differentiation. [i.e Justify your answer using both first derivative test and second derivative test.]
Unit – 11
12
Maths for Computing
Assignment 01
Grading Rubric
Grading Criteria
Achievement
(Yes/No)
Feedback
LO1 : Use applied number theory in practical computing
scenarios.
P1 : Calculate the greatest common divisor and least common
multiple of a given pair of numbers.
P2 :Use relevant theory to sum arithmetic and geometric
progressions.
M1: Identify multiplicative inverses in modular arithmetic.
D1 : Produce a detailed written explanation of the importance of
prime numbers within the field of computing.
LO2 :Analyse events using probability theory and
probability distributions.
P3 :Deduce the conditional probability of different events occurring
within independent trials.
P4 : Identify the expectation of an event occurring from a discrete,
random variable.
M2 : Calculate probabilities within both binomially distributed and
normally distributed random variables.
D2 : Evaluate probability theory to an example involving hashing
and load balancing.
Unit – 11
Maths for Computing
Assignment 01
13
LO3 : Determine solutions of graphical examples using
geometry and vector methods.
P5 : Identify simple shapes using co-ordinate geometry.
P6 :Determine shape parameters using appropriate vector methods.
M3 : Evaluate the coordinate system used in programming a simple
output device.
D3 :Construct the scaling of simple shapes that are described by
vector coordinates.
LO4 : Evaluate problems concerning differential and
integral calculus.
P7 : Determine the rate of change within an algebraic function.
P8 :Use integral calculus to solve practical problems involving area.
M4 :Analyse maxima and minima of increasing and decreasing
functions using higher order derivatives.
D4 :Justify, by further differentiation, that a value is a minimum.
Unit – 11
Maths for Computing
Assignment 01
14
Acknowledgement
Unit – 11
Maths for Computing
Assignment 01
15
Table of Contents
Task 01 ................................................................................................................................................... 22
Part 01 .................................................................................................................................................... 22
1) ............................................................................................................................................................ 22
a) Find the minimum number of squared pieces that can be cut from the ream of cloth without wasting
any cloth................................................................................................................................................. 22
b) Briefly explain the technique you used to solve (a). ........................................................................ 23
2) ............................................................................................................................................................ 23
a) On which day of the month, will all the four customers come back to the restaurant together? ....... 23
B) Briefly explain the technique you used to solve (a). ......................................................................... 24
Part 2 ...................................................................................................................................................... 24
3) ............................................................................................................................................................ 24
B) Briefly explain the technique you used to solve (a). ......................................................................... 25
4) ............................................................................................................................................................ 26
a) Find the total amount of money an employee would earn in a 10-years career. ............................... 26
The above method shows the equation of the geometric progression alongside the complete sum in 10
years. The representative would get a sum of Rs. 600,305 toward the finish of 10 years. ................... 26
b) Briefly explain the technique you used to solve (a). ......................................................................... 27
Part 3 ...................................................................................................................................................... 27
5) Modular arithmetic ............................................................................................................................ 27
Multiplicative inverse ............................................................................................................................ 28
6) ............................................................................................................................................................ 29
Usage of prime numbers for computing ................................................................................................ 29
Task 02 ................................................................................................................................................... 30
Part 1 ...................................................................................................................................................... 30
1) ............................................................................................................................................................ 30
Conditional probability .......................................................................................................................... 30
a) Find the probability that a randomly selected customer either purchased Biscuits or Milk .............. 32
a) Probability of a random customer buying either biscuits or milk. .................................................... 33
b) Showing that the events “randomly selected customer purchased milk” and “the randomly selected
............................................................................................................................................................... 34
3) ............................................................................................................................................................ 34
Tree diagram .......................................................................................................................................... 35
a) Randomly selected voter does not support liberal candidate and lives in State A. ........................... 35
b) Probability of a randomly selected voter supports liberal candidate. ................................................ 35
Unit – 11
Maths for Computing
Assignment 01
16
c) Given that the random selected voter supports liberal candidate, find the probability that the
selected .................................................................................................................................................. 36
4) ............................................................................................................................................................ 36
Complete Diagram ................................................................................................................................. 37
Hearts ..................................................................................................................................................... 38
Clubs ...................................................................................................................................................... 39
Diamonds ............................................................................................................................................... 40
Scorpions ............................................................................................................................................... 41
a) Find the probability that the both selected cards are Hearts. ............................................................. 42
b) Find the probability that one card is Clubs and the other card is Diamonds. .................................... 42
c) Find the probability that the both selected cards is from the same type. ........................................... 43
Part 2 ...................................................................................................................................................... 44
5) ............................................................................................................................................................ 44
Random variables .................................................................................................................................. 44
Types of Random Variables .................................................................................................................. 44
Key Differences between Discrete and Continuous Variable................................................................ 45
6) ............................................................................................................................................................ 46
Discrete Random Variables ................................................................................................................... 46
Here I utilized a basic equation of X = C1 – C2 to get the final output values of the X variable.
Presently that we have all the values ready to turn to our questions ..................................................... 46
a) Find the mean of this probability distribution. (I.e. Find E[X]) ........................................................ 47
b) Find the variance and standard deviation of this probability distribution. ........................................ 49
(I.e. Find V[X] and SD[X]) ................................................................................................................... 49
c) ............................................................................................................................................................ 53
d) ............................................................................................................................................................ 53
Variance for A ....................................................................................................................................... 53
Variance for B ........................................................................................................................................ 54
e) Differences or similarities of scores .................................................................................................. 54
7) ............................................................................................................................................................ 55
a) Find the value of K ............................................................................................................................ 55
b) Find P (Y≤3) ...................................................................................................................................... 55
c) Find P(Y>2) ....................................................................................................................................... 56
Part 3 ...................................................................................................................................................... 56
10) .......................................................................................................................................................... 56
Binomial Distribution ............................................................................................................................ 56
a) Let X is the number of matches that will be won by the team. What are the possible values of X? . 56
Unit – 11
Maths for Computing
Assignment 01
17
b) What is the probability that the team will win exactly 6 matches? ................................................... 56
c) What is the probability that the team will lose 2 or fewer matches? ................................................. 58
d) What is the mean number of matches that the team will win? .......................................................... 59
e) What are the variance and the standard deviation of the number of matches that the team will win?
............................................................................................................................................................... 59
11) .......................................................................................................................................................... 60
Normal Probability Distribution ............................................................................................................ 60
12) .......................................................................................................................................................... 62
a) About what percent of the products last between 87 and 93 days? ................................................... 62
b) About what percent of the products last 84 or fewer days?............................................................... 64
c) About what percent of the products last between 89 and 94 days? ................................................... 66
d) About what percent of the products last 95 or more days? ............................................................... 68
13) .......................................................................................................................................................... 70
Hashing .................................................................................................................................................. 70
Load balancing ....................................................................................................................................... 71
Task 03 ................................................................................................................................................... 73
Part 01 .................................................................................................................................................... 73
Circle ...................................................................................................................................................... 73
1) Find the equation (formula) of a circle with radius r and center C (h, k) and if the Center of a circle
is at (3,-1) and a point on the circle is (-2,1) find the formula of the circle. ......................................... 73
2) Find the equation (formula) of a sphere with radius r and center C (h, k, l) and show that x2 + y2 +
z2 - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius ............................ 74
Area of the Sphere ................................................................................................................................. 74
3) If a= (i+3j-k), b= (7i-2j+4k), find the area of the Parallelogram. ...................................................... 76
Area of the Parallelogram ...................................................................................................................... 76
Part 2 ...................................................................................................................................................... 77
4) If 2x - 4y =3, 5y = (-3)x + 10 are two functions. Evaluate the x, y values using graphical method . 77
Functions in graph ................................................................................................................................. 77
Graph ..................................................................................................................................................... 79
5) ............................................................................................................................................................ 81
Evaluating the surfaces in R3 as the following...................................................................................... 81
6) ............................................................................................................................................................ 82
Volume of Tetrahedron .......................................................................................................................... 82
Task 04 .......................................................................................................................................................... 84
Part 1 ...................................................................................................................................................... 84
Differentiation ........................................................................................................................................ 84
Unit – 11
Maths for Computing
Assignment 01
18
1) Determine the slope of the following functions. ............................................................................... 84
i.
f(x) = 2x – 3x4 + 5x + 8 .................................................................................................................. 84
ii.
f(x) = cos(2x) + 4x2 – 3 .............................................................................................................. 84
2) Let the displacement function of a moving object is S (t) = 5t3 – 3t2 + 6t. What is the function for
the velocity of the object at time t? ........................................................................................................ 85
Part 2 ...................................................................................................................................................... 86
3) Find the area between the two curves f(x) = 2x2 + 1 and g(x) = 8 – 2x on the interval (-2) ≤ x ≤ 1 86
Area Between two curves ...................................................................................................................... 86
4) ............................................................................................................................................................ 90
Finding units through integration .......................................................................................................... 90
Part 3 ...................................................................................................................................................... 91
5) ............................................................................................................................................................ 91
Differentiation with graphs .................................................................................................................... 91
Concavity Intervals ................................................................................................................................ 95
6) ............................................................................................................................................................ 98
Differentiation with graphs .................................................................................................................... 98
1st differentiation Method ...................................................................................................................... 98
2nd Derivative method .......................................................................................................................... 101
References ............................................................................................................................................ 102
Unit – 11
Maths for Computing
Assignment 01
19
Table of Figures
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5
Figure 6
Figure 7
Figure 8
Figure 9
Figure 10
Figure 11
Figure 12
Figure 13
Figure 14
Figure 15
Figure 16
Figure 17
Figure 18
Figure 19
Figure 20
Figure 21
Figure 22
Figure 23
Figure 24
Figure 25
Figure 26
Figure 27
Figure 28
Figure 29
Figure 30
Figure 31
Figure 32
Figure 33
Figure 34
Figure 35
Figure 36
Figure 37
Figure 38
Figure 39
Figure 40
Figure 41
Figure 42
Figure 43
Figure 44
Figure 45
Method ......................................................................................................................................... 22
Extended Proof ............................................................................................................................. 22
Method ......................................................................................................................................... 23
Methods for the sum of logs in the pile ....................................................................................... 25
Formula ........................................................................................................................................ 26
Equation of the geometric progression ....................................................................................... 26
Clock method ............................................................................................................................... 27
Multiplicative inverse of a modular arithmetic term ................................................................... 28
Conditional probability formula ................................................................................................... 30
Venn diagram 1 ............................................................................................................................ 32
Venn diagram 2 ............................................................................................................................ 33
Tree diagram ................................................................................................................................ 35
Complete Diagram ....................................................................................................................... 37
Hearts ........................................................................................................................................... 38
Clubs ............................................................................................................................................. 39
Diamonds ..................................................................................................................................... 40
Scorpions ...................................................................................................................................... 41
Sample space of the difference of values in two cubes............................................................... 46
Probability table ........................................................................................................................... 47
Method......................................................................................................................................... 48
After solving the equation ........................................................................................................... 48
Step by step procedure table 1 & 2 ............................................................................................. 49
Step by step procedure table 3 .................................................................................................... 50
Answer after adding all the new values of X................................................................................ 51
Standard deviation ....................................................................................................................... 52
Mean values of A and B (E (A) and E (B ........................................................................................ 53
Variance of A ................................................................................................................................ 53
Variance of B ................................................................................................................................ 54
Final result of the question .......................................................................................................... 57
Final probability ........................................................................................................................... 58
Extreme Values ............................................................................................................................ 61
Normal distribution curve between 87 and 93 days ................................................................... 62
after finding the answer ............................................................................................................... 63
Normal distribution curve for 84 or less days .............................................................................. 64
after finding the answer ............................................................................................................... 65
Normal distribution curve between 89 and 94 days ................................................................... 66
after finding the answer ............................................................................................................... 67
Normal distribution curve for 95 or more days ........................................................................... 68
after finding the answer ............................................................................................................... 69
Equation to recognize the two fundamental factors ................................................................... 73
after finding the equation ............................................................................................................ 74
Equation of sphere ....................................................................................................................... 76
Method of getting the suitable equation for the circle ............................................................... 76
Area of the Parallelogram ............................................................................................................ 77
Finding the X and Y values for our graph (pic 1) .......................................................................... 78
Unit – 11
Maths for Computing
Assignment 01
20
Figure 46
Figure 47
Figure 48
Figure 49
Figure 50
Figure 51
Figure 52
Figure 53
Figure 54
Figure 55
Figure 56
Figure 57
Figure 58
Figure 59
Figure 60
Figure 61
Figure 62
Figure 63
Figure 64
Figure 65
Figure 66
Figure 67
Figure 68
Finding the X and Y values for our graph (pic 2) .......................................................................... 78
Graph ............................................................................................................................................ 79
3D model of R3 space .................................................................................................................. 81
Volume of Tetrahedron................................................................................................................ 82
after effectively executing the derived equation with the given values (pic 1) .......................... 83
after effectively executing the derived equation with the given values (pic 2) .......................... 83
Answer for the function 1 ............................................................................................................ 84
Answer for the function 2 ............................................................................................................ 84
Velocity Function for the moving object...................................................................................... 85
Values of the integrated values of the functions (pic 1) .............................................................. 86
Values of the integrated values of the functions (pic 2) .............................................................. 87
Area between the curves ............................................................................................................. 88
Area between the curves in a graph ............................................................................................ 89
No. of oxygen units increased ...................................................................................................... 90
Method......................................................................................................................................... 91
discovering the plots where the graph is curved ......................................................................... 92
identifying the second derivative to find the concave upwards/ downwards ............................ 93
Graph 1 ......................................................................................................................................... 95
Graph 2 ......................................................................................................................................... 96
1st differentiation Method .......................................................................................................... 98
The values of X and Y ................................................................................................................... 99
Graph 3 ....................................................................................................................................... 100
2nd Derivative method .............................................................................................................. 101
Unit – 11
Maths for Computing
Assignment 01
21
Task 01
Part 01
1)
a) Find the minimum number of squared pieces that can be cut from the ream of cloth without
wasting any cloth.
Have to discover the least number of square shaped towels without wasting any cloth.
Area of cloth ream = 20*16
In arrange to discover the least number; I would be utilizing the HCF method.
Therefore the answer would be 4.
Minimum number of square shaped towels = (20*16) / (4*4)
=320/16
=20
20 square shaped towels could be made by the cloth ream without any wastage.
Method
Figure 1 Method
Extended Proof
Figure 2 Extended Proof
Unit – 11
Maths for Computing
Assignment 01
22
b) Briefly explain the technique you used to solve (a).
The procedure I used to settle the above sum was HCF. The highest common factor (HCF) is utilized to
introduce the best factor between at least two numbers. This is likewise called as Greatest Common
Factor (GCF). This can be one of the foremost fundamental and imperative strategies utilized in
mathematics. HCF is utilized to track down the best common divisor between the introduced numbers
with no leftovers. This is the way we tracked down the base number of square towels which can be made
through the area 20*16) of the cloth ream with no remnants. The graphical form of the square towels
demonstrates the answer.
2)
In this question we are given four individuals visiting a restaurant on the first day of the month. They
again visit the restaurant once in a 2,4,6,8 days respectively.
a) On which day of the month, will all the four customers come back to the restaurant together?
To discover the day where every one of the four customers will meet up I would utilize the LCM strategy.
As indicated by LCM we get the appropriate response 24.
This implies the four customers will meet in the restaurant 24 days after the first day of the month.
The answer would be that the four customers will meet on the 25th (1st + 24 days) day of the month.
Method
Figure 3 Method
Unit – 11
Maths for Computing
Assignment 01
23
Extended Proof
Customer A = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25
Customer B = 1, 5, 9, 13, 17, 21, 25
Customer C = 1, 7, 13, 19, 25
Customer D = 1, 9, 17, 25
B) Briefly explain the technique you used to solve (a).
The procedure I used to tackle the above issue was LCM. Comparable to HCF however rather than the best
common divisor, LCM gives the slightest common numerous of at least two numbers. LCM is the smallest
or least positive integer that is separable by a progression of numbers. The extended proof approves the
appropriate response received by the LCM technique.
Part 2
3)
This question depends on the subject of arithmetic progression. As indicated by the question, logs are
piled in a sequence. There are 24 at the bottom and 10at the top. There are 15 rows of logs and each line
has one log more than the one above it. First I might want to make an arithmetic progression equation to
approve the progression. I would track down the last log (15) value and it should be 24 as in the question.
Working for arithmetic progression
1st term = 10 (a)
Common difference = 1
Row No = 15
An = a1+ (n-1) d
a15 = 10 + (15-1)1
a15=10+14
a15= 24
Now that we received the same value as expected, we could find the total value of logs in the pile.
Unit – 11
Maths for Computing
Assignment 01
24
Methods for the sum of logs in the pile
Figure 4 Methods for the sum of logs in the pile
There are 255 logs in the pile.
B) Briefly explain the technique you used to solve (a).
Since the above question depended on arithmetic progression I utilized the method of discovering the sum
of the nth terms. Arithmetic progression is an arrangement of numbers that has a consistent normal
distinction between the terms. Factors like the initial term, last term, normal difference and the nth term
are utilized to unravel the sums. In the above question, I found both the sum and an nth term for
demonstrating. The equation I used to get the sum is utilized when the initial term and the all outnumber
of terms are known
Unit – 11
Maths for Computing
Assignment 01
25
4)
This question is based on geometric progression.
a) Find the total amount of money an employee would earn in a 10-years career.
1st term – Rs. 50,000
4 % increase of salary each year.
2nd year – 50000*104/100 = 52000
Common ratio – 52000/50000 = 26/25
Total sum of money earned by the employee in 10 years
Figure 5 Formula
Formula to be used
Method
Figure 6 Equation of the geometric progression
The above method shows the equation of the geometric progression alongside the complete sum in 10
years. The representative would get a sum of Rs. 600,305 toward the finish of 10 years.
Unit – 11
Maths for Computing
Assignment 01
26
b) Briefly explain the technique you used to solve (a).
The method I used to unravel the above issue was a geometric progression. Geometric progression is
comparative to arithmetic progression but rather than including terms, the geometric progression has
multiplying terms. There is a typical ratio between each term. This ratio is generally utilized while
tackling issues. Within the assignment, I settled the equation by discovering the sum of the geometric
progressions by the particular equation.
Part 3
5) Modular arithmetic
A/B = Q remainder R
In this equation we can see that: A – Dividend
B – Divisor
Q – Quotient
R – Remainder
In some circumstances it is important for us to look for the remainder in these cases we use the term
called modulo operator (MOD).
To find the mod in this equation we can change it as: - A MOD B = R
Using the clock method to find the MOD is widely used in the present Ex: - 8 MOD 4
Figure 7 Clock method
From the above clock method we can see that since the last term ended in 0. 8 MOD 4 will be 0.
So it’s clear that using the clock method is helpful in solving questions based on MODs
Unit – 11
Maths for Computing
Assignment 01
27
Multiplicative inverse
Multiplicative inverse is another chapter included in modular arithmetic. The multiplicative inverse of
any number is simply 1/n. this is also called reciprocal of a number similar to opposite. When this comes
into modular arithmetic it’s a bit complicated. There are many methods to solve this. We use the MOD
function to find the multiplicative inverse in Modular arithmetic instead of division.
Modular inverse of A (MOD C) is A^-1
(A* A^-1) ≡ 1 (MOD C) or equal to (A* A^-1) MOD C =1
With the above two equation it’s clear that multiplicative inverse of modular is different than the normal
multiplicative inverse.
In our question we are about to find the multiplicative inverse of 6MOD13
We are using a different method to approach this question. The equation which I am using to solve this is:
- A*B MOD C = 1.
Method
Figure 8 Multiplicative inverse of a modular arithmetic term
By using the above method we can easily figure the multiplicative inverse of a modular arithmetic term.
6 MOD 13 = 11
Unit – 11
Maths for Computing
Assignment 01
28
6)
Usage of prime numbers for computing
This rule, called the prime factorization rule, is called something as well: the fundamental the borem of
Arithmetic. It makes sense when we think about what primes are, numbers that can’t be pulled apart any
further. So as we try to pull apart any number into two numbers, and then pull those apart into two
numbers if possible, and so on, we will eventually be left only with primes.
This all might seem like nothing more than a cool mathematical oddity. But it becomes important thanks
to one simple additional fact: As far as the best mathematician and computer scientist have been able to
determine, It is totally impossible to come up with a truly efficient formula for factoring large numbers
into primes.
That is to say, we have ways of factoring large numbers into crimes, but if we try to do it with a 200 Digit number, or a 500 - digit number, using the same algorithms we would use to factor a 7 - digit
number, the world’s most advanced super computers still take absurd amount of time to finish. Like
timescales longer than the formation of the planet and, for extremely large numbers, longer than the age
of the universe itself.
So, there is a functional limit to the size of the number we can factor into prime, and this fact is absolutely
essential to modern computer security. Pretty much anything that computers can easily do without being
able to easily undo will be of interest to computer security. Modern encryption algorithms exploit the fact
that we can easily take two large primes and multiply them together to get a new, super large number, but
that no computer yet created can take that super large number and quickly figure out which to primes
went into making it.
This math level security allows what’s called public key cryptography or encryption where we don’t have
to worry about publishing a key to use in encrypting transmissions, because simply having that key (A
very large number) won’t help anyone to undo the encryption it created. In order to undo the encryption,
and read the message, you need the prime factors of the key used for encryption and as we have been
saying, that’s not something you can just figure out on your own.
(Esoft 2021, 2021)
Unit – 11
Maths for Computing
Assignment 01
29
Task 02
Part 1
1)
Conditional probability
Conditional probability is one of the foremost imperative concepts in probability theory. This theory can be
essentially applied to any movement when there is extra data. Condition probability is the probability of an
event happening when another event has already happened. It is additionally important to be beyond any
doubt that conditional probability doesn't generally have a causal relationship and it moreover doesn't say
that both events take place simultaneously. One event must happen before the other.
Figure 9 Conditional probability formula
The idea of the conditional probability formula is one of the quintessential ideas in probability theory. The
probability conditional formula gives the proportion of the probability of an event, say B given that another
event, say A has occurred.
The Bayes' theorem is utilized to decide the conditional probability of event A, considering that event B
has occurred, by knowing the conditional probability of event B, considering that event A has occurred,
additionally the individual probabilities of events A and B.
The formula for conditional probability is:
P (A | B) = P (A and B)/P (B)
It can also be written as,
P (A|B) = P (A∩B) P (B)
(cuemath)
Unit – 11
Maths for Computing
Assignment 01
30
Example for a conditional probability event
The scenario is as follows, a teacher gives two tests to students namely Math and IT. 25% students passed
both tests while 40% of the students passed the math test. What percent of those who passed the math test
also passed the IT test?
Solution
Students who passed Math test = 40%
Students who passed both test = 25%
I will take event A as students who passed Math and event B as students who passed IT.
P (A) = 40% = 0.40
P (A ⋂ B) = 25% = 0.25
To find the students who passed math test also passed the science test. I would apply the formula.
Condition of B given that A
= P (B | A) = P (A ⋂ B) / P (B)
= 0.25 / 0.40
= 0.625
= 62.5%
The above arrangement shows us the essential idea of conditional probability and how it functions.
Understanding the information and applying them likewise is the key point in this theory. Rather than this
theory the tree diagram, Bayes theorem, and so on can be utilized to discover the answer.
Unit – 11
Maths for Computing
Assignment 01
31
a) Find the probability that a randomly selected customer either purchased Biscuits or Milk
Customers = 25
Purchased biscuits = 05
Represented by B
Purchased milk = 10
Represented by M
Purchased fruits = 08
Represented by F
Purchased both milk and fruits = 06
Representing the information in a Venn diagram
Figure 10 Venn diagram 1
This is the first Venn diagram I made with every one of the raw data. In the wake of addressing the data
in the Venn diagram, I had the option to make it in a more valid and clear manner as the underneath
diagram.
Unit – 11
Maths for Computing
Assignment 01
32
Figure 11 Venn diagram 2
The second Venn diagram appears a bit clear view with legitimate data representation. Since Biscuits
didn’t have any connection with milk or fruits I separated it away as an individual event. Besides, ready
to see that 8 customers didn’t purchase anything as well. Presently able to utilize this Venn diagram to
unravel the questions.
a) Probability of a random customer buying either biscuits or milk.
Probability of Milk = 10/25
Probability of Biscuits = 5/25
Probability of a customer buying either biscuit or milk (B ∪ M)
10/25 + 5/25 = 15/25 = 3/5
Probability of a customer buying either biscuit or milk is 15/25 or 3/5.
Unit – 11
Maths for Computing
Assignment 01
33
b) Showing that the events “randomly selected customer purchased milk” and “the randomly
selected customer purchased Fruits” are independent.
Formula to find whether the events are independent = P (A ∩ B) = P (A). P (B)
When connected with our scenario A and B changes into Milk and Fruits, therefore able to utilize this
situation to discover the solution.
P (M ∩ F) = P (M). P (F)
6/25
= 10/25 * 8/25
6/25
= 16/125
Since they are not equal, it demonstrates that the events are not independent
3)
Total population
A
45%
B
20%
C
35%
Support for Liberal Candidate
A
40%
B
30%
C
60%
To move on with the questions I would like to draw my tree diagram for the scenario. I would be using
the tree diagram technique to solve the following questions with evidence.
Unit – 11
Maths for Computing
Assignment 01
34
Tree diagram
Figure 12 Tree diagram
The above tree diagram shows a summary of the probabilities of liberal and non-liberal candidates
winning and losing in each state according to their population. With these data we can solve our
questions.
a) Randomly selected voter does not support liberal candidate and lives in State A.
45/100 * 60/100 = 27 /100
The probability of a voter does not support liberal and lives in state A is 27/100.
b) Probability of a randomly selected voter supports liberal candidate.
To get this answer we should add up all the fractions of people who support liberal candidates.
18/100 + 6/100 + 21/100 = 45/100
The probability of a voter who supports liberal in all 3 states is 45/100.
Unit – 11
Maths for Computing
Assignment 01
35
c) Given that the random selected voter supports liberal candidate, find the probability that the
selected voter is from state B.
P (voter lives in state B |voter supports liberal candidate) = P (B ∩ L) / P (L)
= 20/100 (population of B) * 30/100 (liberal percentage in B) / 45/100 (total B)
= 1/5 * 3/10 / 45/100
=3/50 / 45/100
=3/50 * 100/45
= 6/45 or 2/15
The probability of a selected voter supporting liberal candidate and that voter is from state B is
6/45
4)
To solve this arrangement of questions I would utilize the tree diagram procedure. The tree diagram will
include 4 branches for the very first set and afterward grow from that point. I might want to display the
tree diagram first to comprehend the information all the more clearly.
Hearts
6 Cards
Clubs
7 Cards
Diamonds
8 Cards
Scorpions
5 Cards
Since there are 4 variables in the tree diagrams, it is too big when showing the probability of the events.
In this manner, I would separate my tree diagram into more modest parts for a clear arrangement.
Unit – 11
Maths for Computing
Assignment 01
36
Complete Diagram
Figure 13 Complete Diagram
Unit – 11
Maths for Computing
Assignment 01
37
Hearts
Figure 14 Hearts
In the above diagram, we can see the Probability of each event of the hearts in the box.
Unit – 11
Maths for Computing
Assignment 01
38
Clubs
Figure 15 Clubs
In the above diagram, we can see the probability of each event of the Clubs in the box.
Unit – 11
Maths for Computing
Assignment 01
39
Diamonds
Figure 16 Diamonds
In the above diagram, we can see the probability of each event of the diamonds in the box.
Unit – 11
Maths for Computing
Assignment 01
40
Scorpions
Figure 17 Scorpions
In the above diagram, we can see the probability of each event of the scorpions in the box
From these diagrams, we have seen the tree diagram of each variable and event we can get a clear thought
of how the probability is divided among all the possible results.
Unit – 11
Maths for Computing
Assignment 01
41
a) Find the probability that the both selected cards are Hearts.
In this scenario, we need to consider that the card picked on the two events is hearts. In this manner, as
indicated by that we need to discover its probability. By referring to the tree diagram we can get the
probability of all the sets needed for the sum.
Method
6/26 * 5/25 = 3/65
Within the first sum, I multiplied the full value of hearts with the 1st card and after that result with the
probability of the 2nd card.
So the final probability of receiving both cards as heart is 3/65
b) Find the probability that one card is Clubs and the other card is Diamonds.
In this scenario, we need to discover all the possibilities of getting one card as club and another card as
diamond. First of all, I would like to list out those possibilities.
P (Club, Diamond), P (Diamond, Club)
Method
P (Club, Diamond)
7/26 * 8/25 = 28/325
P (Diamond, Club)
8/26 * 7/25 = 28/325
28/325 + 28/325 = 56/325
The probability of getting diamond and club is 56/325
Unit – 11
Maths for Computing
Assignment 01
42
c) Find the probability that the both selected cards is from the same type.
For this question, we have to find a series of probabilities and after that add them to urge the collective
probability. I will utilize the same method I utilized within the first question to discover the same sort of
cards on both events.
Method
Hearts
6/26 * 5/25 = 3/65
3/65 * 4/24 = 1/130
Clubs
7/26 * 6/25 = 21/325
21/325 * 5/24 = 7/520
Diamonds
8/26 * 7/25 = 56/650
56/650 * 6/24 = 7/325
Scorpions
5/26 * 4/25 = 2/65
2/65 * 3/24 = 1/260
Probability of getting the same cards on both instances
1/130 + 7/250 + 7/325 + 1/260 = 121/2600
Unit – 11
Maths for Computing
Assignment 01
43
Part 2
5)
Random variables
A random variable (stochastic variable) may be a sort of variable in statistics whose possible values
depend on the results of a certain random phenomenon. Since a random variable can take on distinctive
values, it is commonly labeled with a letter (e.g., variable “X”). Each variable has a particular probability
distribution function (a mathematical function that represents the probabilities of events of all conceivable
results).
Types of Random Variables
Random variables are classified into discrete and continuous variables. The most contrast between the
two categories is the sort of possible values that each variable can take. In addition, the sort of (random)
variable suggests the specific strategy of finding a probability distribution function.
Discrete Random Variables
A discrete random variable is a (random) variable whose values take just a limited number of values. The
best illustration of a discrete variable is a dice. Throwing dice is a simple random event. At the same
time, the dice can take just a limited number of results {1, 2, 3, 4, 5, and 6}.
Every result of a discrete random variable contains a specific probability. For instance, the probability of
each dice result is 1/6 on the grounds that the results are of equal probabilities. Note that the entire
probability result of a discrete variable is equal to 1.
Continuous Random Variables
Not at all like discrete variables, can continuous random variables take on a boundless number of possible
values. One of the cases of a continuous variable is the returns of stocks. The returns can take a boundless
number of possible values (as percentages).
Due to the over reason, the probability of a certain result for the continuous random variable is zero. In
any case, there's continuously a non-negative probability that a certain result will lie inside the interval
between two values.
(corporatefinanceinstitute)
Unit – 11
Maths for Computing
Assignment 01
44
Key Differences between Discrete and Continuous Variable






The statistical variable that assumes a limited set of data and a countable number of values, at that
point it is called a discrete variable. As against this, the quantitative variable which takes on an
infinite set of data and an uncountable number of values is known as a continuous variable.
For non-overlapping or something else known as mutually inclusive classification, wherein both
the class constrain are included, is appropriate for the discrete variable. On the contrary, for
overlapping or say commonly exclusive classification, wherein the upper class limit is excluded, is
appropriate for a continuous variable.
In a discrete variable, the range of specified numbers is complete, which isn't within the case of a
continuous variable.
Discrete variables are the variables, wherein the values can be obtained by counting. On the other
hand, Continuous variables are the random variables that measure something.
Discrete variables accept independent values whereas continuous variables assume any value in a
given range or continuum.
A discrete variable can be graphically represented by isolated points. Not at all like a continuous
variable which can be shown on the chart with the assistance of associated points.
Discrete
Continuous
Discrete variable alludes to the variable that Continuous variable implies the variable
expects a limited number of isolated values. which assumes an interminable number of
diverse values.
Complete
Incomplete
Values are obtained by counting.
Values are obtained by measuring.
Non-overlapping
Overlapping
Distinct or separate values.
Any value between the two values.
Isolated points
Connected points
(keydifferences)
Unit – 11
Maths for Computing
Assignment 01
45
6)
Discrete Random Variables
Two fair cubes are rolled and x represents the difference between the values of the two cubes.
Ex: - cube1 – cube2
=1–5
= -4
Such as this we have to discover out all the possible results of the difference between the values of two
cubes. To solve this part I would like to utilize a sample space to show all the possible results of the
scenario.
Sample space of the difference of values in two cubes
Figure 18 Sample space of the difference of values in two cubes
Here I utilized a basic equation of X = C1 – C2 to get the final output values of the X variable. Presently
that we have all the values ready to turn to our questions
Unit – 11
Maths for Computing
Assignment 01
46
a) Find the mean of this probability distribution. (I.e. Find E[X])
The mean of the probability distribution or in other terms the expected value is the weighted average of all
possible values that the random variable can take on. To find the expected value of the random variable the
sum of the probability of each possible result has to be multiplied with the value itself. The equation for
finding the mean is
E[X] = Ex . P (x)
To move forward I will add the probability table with its values.
Figure 19 Probability table
Unit – 11
Maths for Computing
Assignment 01
47
Next I will replace the values from the table to the mean equation.
Method
Figure 20 Method
Within the above picture, I replaced the equations with x variables and the probability values. Since the
denominator is the same (36) I cut it off within the below picture and utilized only the numerator to solve
the question.
Figure 21 After solving the equation
After solving the equation the result of the mean value of variable X is
E [X] = 0/36
Unit – 11
Maths for Computing
Assignment 01
48
b) Find the variance and standard deviation of this probability distribution.
(I.e. Find V[X] and SD[X])
Next I have to find the variance and the standard deviation of the discrete random variable.
Variance – Var (X) = E (X2) – [E (X)] 2
Standard Deviation – √Var (X)
By utilizing the over equation I got to discover the variance and the standard deviation of the probability
distribution.
As the equation states I had to take the values of E (X2) first to minus it with the E (X)2. The picture
portrays the step by step procedure of how I solved the new equation E (X2).
Figure 22 Step by step procedure table 1 & 2
Unit – 11
Maths for Computing
Assignment 01
49
Figure 23 Step by step procedure table 3
Unit – 11
Maths for Computing
Assignment 01
50
Figure 24 Answer after adding all the new values of X
After adding all the new values of X I got the answer as 5.83333 as a recurring value. With this, I moved
forward to urge the square of the earlier anticipated value of our probability. Since it was 0 the value of E
(X) 2 will be 0 as well. In this manner at last when I subtract both the values I will ultimately pick up the
value of 5.83333 (recurring) value as our variance for our discrete random variable.
Variance = 5.83333 (recurring)
Unit – 11
Maths for Computing
Assignment 01
51
In order to find the standard deviation I have to find the square root of the Var (X).
Figure 25 Standard deviation
Since our variance is 5.83333 (recurring) I had to find the square root of it which brings us a value of
2.4152.
So our standard deviation is = 2.4152
The following half of the question consists of a replacement equation. Here we need to replace the values
of X with the given equation. The following are the conditions given.
A = X-10 and B = [(1/2) X]-5
In this manner, again I have to create a new probability table after replacing these values.
Replaced values for X appear on the following page.
Both the following pictures appear, how X is replaced with new values according to the given equations.
Unit – 11
Maths for Computing
Assignment 01
52
c)
Figure 26 Mean values of A and B (E (A) and E (B
E (A) = -10
E (B) = -5
From the above image I would like to say that the mean values of A and B (E (A) and E (B)) were
successfully found.
d)
Variance for A
To find the variance of both A, I would be using the below equation.
Figure 27 Variance of A
As the above images depicts the
Variance of A is = 5.83 (Recurring)
Unit – 11
Maths for Computing
Assignment 01
53
Variance for B
To find the variance of B, I will use the below equation.
Figure 28 Variance of B
As the above picture depicts the
Variance of B is = 1.4583
e) Differences or similarities of scores
Arnold and Brian roll two fair cubes. Arnold records his score using random variable A. Brian records his
score using random variable B. They repeat this process for a large number of times. Compare their scores
and comment any likely differences or similarities of their scores.
o The similarity between the scores of Arnold and Brian will be that their total scores will be roughly
zero.
o The contrast will be that Arnold’s score should be more different than Brian's.
Unit – 11
Maths for Computing
Assignment 01
54
7)
Within the following question, we are given the probability distribution of the variable Y which is a
constant variable. Here the value K is constant.
Y=y
1
2
3
4
5
P(Y=y)
1/3
1/6
1/4
k
1/6
a) Find the value of K
In arrange to discover the value of K, we have to get it one of the fundamental theorems in a probability
distribution. That's the sum of the probabilities is equal to 1. So This means the sum of the probability P(Y)
will be equal to 1. In this manner in case we implement this in our issue we would be able to urge the value
of K.
Method
1/3 + 1/6 + 1/4 + K + 1/6 = 1
If we take 12 as the common factor then,
4/12 + 2/12 + 3/12 + 2/12 + K = 1
11/12 + K = 1
1 – 11/12 = K
1 / 12 = K
K = 1 / 12
b) Find P (Y≤3)
Value of probability when P (Y < 3)
Method
P (Y < 3) = P (3) + P (2) + P (1)
= 1/4 + 1/6 + 1/3
= 9 /12 = 3 / 4
P (Y < 3) = 3 / 4
Unit – 11
Maths for Computing
Assignment 01
55
c) Find P(Y>2)
Value of probability when P (Y >2)
Method
P (Y > 2) = P (3) + P (4) + P (5)
= 1/4 + 1/12 + 1/6
= 6 /12 = 1 / 2
P (Y > 2) = 1 / 2
Part 3
10)
Binomial Distribution
Titans Cricket team has a winning rate of 75 %. The team is playing 10 matches in the next season.
a) Let X is the number of matches that will be won by the team. What are the possible values of X?
X = Number of matches won by the team
Possible values of X – 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10
b) What is the probability that the team will win exactly 6 matches?
Finding the probability that the team will win exactly 6 matches
Binomial dispersion has as it were two results success and failure. With this theory, we seem to recognize
the probability of certain scenarios. In this question, we got to discover the probability of the team winning
exactly six matches. Before solving the question I would like to point out the vital factors within the
equation.
P – Success
Q – Failure
n – Number of trials
x – 0, 1, 2…10
Success rate = P = 0.75 or 75 %
Failure rate = q = 0.25 or 25 %
I will use these factors to solve the questions.
Unit – 11
Maths for Computing
Assignment 01
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Method
Figure 29 Final result of the question
The over picture portrays the workings and the final result of the question. As the question states, I had to
discover the probability of the team winning exactly 6 matches. Hence I replaced the factors with the given
values and after that solved the question concurring to the equation (appeared within the picture).
So the probability of the team winning exactly 6 matches is 0.1457001
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c) What is the probability that the team will lose 2 or fewer matches?
Under this scenario, I have to replace the values of X with the given conditional clause which is “two or
less”. This implies that the team is progressing to lose 2 or fewer matches which implies at that point they
are planning to win 8 or more matches. The working for this question is as takes after.
Figure 30 Final probability
After extricating the values of each value of X, These have to be included together to urge the final result.
0.281559 + 0.1877 + 0.056314 = 0.525573
So the final probability of the team losing 2 or less matches is 0.526
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d) What is the mean number of matches that the team will win?
Finding the mean number of matches the team will win.
The equation to find the mean of a binomial distribution is,
E (X) = n * p
So based on this equation we can discover the mean value of the team winning matches.
E (X) = n * p
E (X) = 10 * 0.75
= 7.5
Mean of the number of matches the team will win is 7.5
e) What are the variance and the standard deviation of the number of matches that the team will
win?
Variance and the standard deviation of the number of matches the team will win.
Formula for Variance and Standard Deviation are,
Var (X) = n * p * q
S.D (X) = – √Var (X)
When we apply this equation to the scenario we will get the variance and the standard deviation.
Var (X) = n * p * q
= 10 * 0.75 * 0.25
= 1.875
Variance of X is 1.875
S.D (X) = √Var (X)
= √ 1.875
= 1.3693
Standard deviation of X is 1.3693
Unit – 11
Maths for Computing
Assignment 01
59
11)
Normal Probability Distribution
Normal probability distribution includes continuous variables.




45 Students in grade 10
Mean height of the students – 154cm
Standard deviation – 2cm
Height of Alex – 163cm
Would Alex’s height be an outlier in the event that the height of the students was ordinarily conveyed?
So with the over factors, we need to discover whether Alex’s height is an exception or not. In ordinary
probability distribution, we consider three fundamental factors they are,
M – Mean
σ – Standard deviation
σ2 - Variance
Other than this a bell-formed chart can be drawn to represent the data.
Next I will replace all the variables with the factors

M – 154

σ – 2cm

Alex’s height – 163
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Next I will be utilizing an equation to discover the extraordinary values of the instance. In case Alex’s
height resides above or underneath the value limit at that point it’s considered as an outlier.
Extreme Values are under and above
Figure 31 Extreme Values
The values should be in between
148 < X < 160
So according to our theorem, since Alex’s height (163) lies above 160 is considered as an extreme
value or in other terms an outlier.
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12)
Battery life is normally distributed with a mean of 90 days and a standard deviation of 3 days.
a) About what percent of the products last between 87 and 93 days?
Percent of the products last between 87 and 93 days. Giving the normal distribution curve
Method
Figure 32 Normal distribution curve between 87 and 93 days
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Mean = 90
SD = 3
Figure 33 after finding the answer
To obtain the percentage I have to multiply the final value by 100 which leave us with the answer
0.68268 * 100 = 68.3%.
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b) About what percent of the products last 84 or fewer days?
Battery life last 84 or less days
Method
Figure 34 Normal distribution curve for 84 or less days
Unit – 11
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Assignment 01
64
Figure 35 after finding the answer
To find the percentage I multiply the Z score with 100.
The percentage of X being 84 or less is 0.2275 * 100 = 22.75%
Unit – 11
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Assignment 01
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c) About what percent of the products last between 89 and 94 days?
Percentage of products last between 89 and 94 days
Method
Figure 36 Normal distribution curve between 89 and 94 days
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Figure 37 after finding the answer
The percentage of products lasting between 89 to 94 days is 0.53754 * 100 = 53.8%
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d) About what percent of the products last 95 or more days?
Percentage of products lasting 95 days or more
Method
Figure 38 Normal distribution curve for 95 or more days
Unit – 11
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Figure 39 after finding the answer
The percentage of products lasting more than 95 days is
0.04746 * 100 = 47.5%
Unit – 11
Maths for Computing
Assignment 01
69
13)
In computer science, probability theories are employed more frequently. Huge support for these notions
has resulted in numerous new innovations and creations. Most natural causes are also challenged by the
theories covered under likelihood. People want to believe in luck and chance, but when they consider
these possibilities in terms of probability, they see the tremendous potential of this topic. From throwing a
coin to gambling, probability has come a long way. It has also played an important part in computer
science up to this point. Probability is employed in a variety of computer settings, from calculating a
computer's lifespan to machine learning.As can be seen, probability theory has played an important role
in nearly every element of computing. Those creations include hashing and load balancing.
Hashing
Hashing is the most frequent method for creating hash tables. A hash table is a list that maintains
key/value pairs and allows you to access any entry by its index. We may use a hash function to map the
keys to the size of the table because there is no limit to the number of key/value pairs; the hash value
becomes the index for a given element. A simple hashing method would be to compare the key's modular
(if it's numerical) to the table's size:
Index = key MOD table size
Data encryption also employs hashing. Passwords can be saved as hashes, preventing plaintext passwords
from being accessed even if a database is hacked. Popular cryptographic hashes include MD5, SHA-1,
and SHA-2.
The birthday paradox is a probability application. This has an odd idea behind it, but it is mathematically
provable. Consider that did at least two people with the same birthday appear in nearly half of the
groupings of 23 or more people. When comparing birthday probabilities, it's sometimes easier to look at
the probability that persons don't share a birthday. The birthdate of a person is one of 365 possibilities
(excluding February 29 birthdays). Because there are 364 days that are not a person's birthday, the
likelihood of having the same birthday as another person is 364 divided by 365. This means that there's a
364/365, or 99.726027 percent, probability that no two people's birthdays are the same.
As previously stated, there are 253 possible comparisons or combinations in a group of 23 people. So
we're looking at 253 different comparisons instead of just one. Every one of the 253 combinations has the
same chance of not being a match: 99.726027 percent. If you multiply 99.726027 percent by 253, or
calculate (364/365)253, you'll get that there's a 49.952 percent chance that none of the 253 comparisons
Unit – 11
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70
are identical. As a result, the chances of finding a birthday match in those 253 comparisons are 1 – 49.952
percent = 50.048 percent, or slightly more than half! The closer the actual chance approaches 50%, the
more trials you conduct.
This is where the birthday paradox and hashing come into play. We can find the rate of hash collisions
using this birthday paradox now that we know what a hash collision is. If we consider the buckets in a
hash table to be the days in a calendar, there is a good possibility that two values will fall into the same
index. This is when the birthday paradox comes into play in determining the hash table's likelihood rate
of collision.Double hashing is a better developed solution that normal hashing. It generates probe
sequences that depend on the key instead of being the same for every key. It also hashes the key for the
second time using a different hash function, to avoid collisions between keys and encodings.
Load balancing
Load balancing is a strategy that is employed to prevent this from happening. The word itself explains the
technique's nature, which divides the work evenly across the servers. The division of labor is influenced
by a number of factors.Load balancing can be explained in two ways: concretely and analytically. In the
concrete form, a public-facing server accepts user requests and forwards them to a back-end server, which
processes them and responds to the user. When you have a billion users and a million servers, you want to
route requests so that no single server receives an excessive number of requests, or the users would
experience delays. Furthermore, you're concerned that the League of Tanzanian Hackers is attempting to
bring your website down by sending you requests in a specific order in order to disrupt your load
balancing mechanism.This is mostly employed in the networking business due to its ability to efficiently
distribute job load. These concerns affect even the largest data centers of significant corporations on a
daily basis.Let's consider a case in which we have a finite number of servers. Node services must be
available 24 hours a day, seven days a week.
Consistent hashing is an example of a technique for dividing the workforce. Another technique that can
be used to connect users and servers is the equation server = node MOD no.of servers.Other algorithm
techniques include: First Come First Serve (FCFS) – this refers to the moment a user connects to the
server.
Round robin method - In a group brainstorming context, the "Round robin" option is a technique for
generating and developing ideas. It is based on an iterative process that builds on the contributions of
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each participant in a series of written or verbal variations. This uses a simple formula that sends nodes to
servers in the same order. Server 1 is served by nodes 1, 3,5,7,9, whereas server 2 is served by nodes
2,4,6,8. If there is a performance disparity between the servers, this solution will not work. If that's the
case, we can utilize the weight round robin method to assign one server *more nodes than the other.In our
example we took the MOD values for differentiating. By this we can see applications of probability
playing a helpful role in load balancing. In order to protect the uptime of the server we use the hashing
technique to connect the servers with the nodes. This is another instance where we use probability to help
maintain the stability of the system.
Finally, I'd like to point out that hashing and load balancing are two major innovations in two critical
areas of IT. And probability applications have played a critical role in successfully elevating these
functions.
(stanford)
(educative)
(scientificamerican)
(mindtools)
(jeremykun)
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Task 03
Part 01
Circle
1) Find the equation (formula) of a circle with radius r and center C (h, k) and if the Center of a
circle is at (3,-1) and a point on the circle is (-2,1) find the formula of the circle.
Method
Figure 40 Equation to recognize the two fundamental factors
The above picture portrays the equation to recognize the two fundamental factors, the point in a circle and
the center. Here we utilized the equation (x-h)2 + (y-k)2 = r2 to recognize the radius of the circle first. At
that point, we utilized the same equation once again with the radius to derive the new equation of the
circle. The below picture appears the workings of this process.
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The formula of the
circle is:(X - 3)2 + (Y + 1)2= 29
Figure 41 after finding the equation
OR
X2 – 6X + Y2 + 2Y -19 = 0
2) Find the equation (formula) of a sphere with radius r and center C (h, k, l) and show that x2 + y2 +
z2 - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius
Area of the Sphere
The 3D shape of the circle is the sphere. Same as there's an area to 2D shapes there's a way to discover the
area to 3D shapes as well. In this question, I have to discover the equation of a sphere together with its
center and radius.
Sphere consists of:Radius – r
Center – C (h, k, l)
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Prove that x2 + y2 + z2 - 6x + 2y + 8z - 4 = 0 is an equation of the sphere.
Find the center and the radius.
Method
Unit – 11
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Assignment 01
75
Figure 42 Equation of sphere
The over picture
Figure 43 Method of getting the suitable equation for the circle
appears as the method
of getting the suitable
equation for the circle. Here I made the given equation to squares so that I can get the center through the
second digit of the brackets. Also, I found the radius by the square root. Subsequently, as the over picture
appears, I have effectively found the equation for the circle along with the radius and the center of the
circle.
Center – (3, -1, -4)
Radius - √30 – 5.477
3) If a= (i+3j-k), b= (7i-2j+4k), find the area of the Parallelogram.
Area of the Parallelogram
Find the area of the parallelogram when, a= (i+3j-k), b = (7i-2j+4k)
Method
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Figure 44 Area of the Parallelogram
The final area of the parallelogram for the given variable is 27.386cm2
Part 2
4) If 2x - 4y =3, 5y = (-3)x + 10 are two functions. Evaluate the x, y values using graphical method
Functions in graph
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Figure 45 Finding the X and Y values for our graph (pic 1)
Unit – 11
Figure 46
Maths for Computing
Finding the X and Y values for our graph (pic 2)
Assignment 01
78
I have found the X and Y values for our graph. Next step is to plot it in a Cartesian plan.
Graph
Unit – 11
Figure 47 Graph
Maths for Computing
Assignment 01
79
In the above graph, the black dots represent the points of our X and Y axis. Other than that we seem to see
the intersection of both these straight lines are (2.5,0.5).
Values of X and Y in the equation 4y =2x -3
X
-2
-1
0
1
2
Y
-1.75
-1.25
-0.75
-0.25
0.25
Values of X and Y in the equation 5y = (-3) x +10
X
-2
-1
0
1
2
Y
3.2
2.6
2
1.4
0.8
So the final X and Y values of the intersection will be (2.5,0.5).
Unit – 11
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5)
Evaluating the surfaces in R3 as the following
y=4
z=5
Figure 48 3D model of R3 space
The above picture is a 3D model of R3 space. As we know this represents
{(x, y, z), z = 5} as of all points in R3 whose coordinate is equal to 5. Moreover, we are able to show that
the Z plane is horizontal (red) and parallel to the XY plane (Red and blue) by 5 units.
So I could say that Z plane is parallel to both XY plane by 5 units. Whereas the Y Plane is too
parallel to ZX by 4 units
Unit – 11
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6)
Volume of Tetrahedron
Method
Figure 49 Volume of Tetrahedron
As appeared within the above picture the volume of a substance or a shape is found by the area of base
multiplied by its perpendicular height. To discover the volume of the tetrahedron I am using the vector
strategies. The first equation shown within the picture incorporates the cos value but to solve the condition
I expect that cos is 0. So that I can get an amount of 1 (Cos 0).
So as the final equation we will be getting 1/6 | (a * b). h |
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Figure 50 after effectively executing the derived equation with the given values (pic 1)
Figure 51 after effectively executing the derived equation with the given values (pic 2)
As the over pictures portray, after effectively executing the derived equation with the given values. I
received the final volume of the tetrahedron as 16.5 cm3
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Task 04
Part 1
Differentiation
1) Determine the slope of the following functions.
i.
f(x) = 2x – 3x4 + 5x + 8
ii.
f(x) = cos(2x) + 4x2 – 3
The slope of the function can be solved through the following method.
I.
F (x) = 2x – 3x4 + 5x + 8
Figure 52 Answer for the function 1
The slope of the above function is F’ (x) = – 12 X3 + 7
II.
F (x) = cos (2x) + 4x2 – 3
Figure 53 Answer for the function 2
The slope of the above function is F’ (x) = -2Sin (2X) + 8X
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2) Let the displacement function of a moving object is S (t) = 5t3 – 3t2 + 6t. What is the function for
the velocity of the object at time t?
I have given a displacement function S (t) = 5t3 – 3t2 + 6t and I have to velocity of the object as a
function.
Figure 54 Velocity Function for the moving object
Velocity Function for the moving object is
V (t) = 15t2 – 6t + 6
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Part 2
3) Find the area between the two curves f(x) = 2x2 + 1 and g(x) = 8 – 2x on the interval (-2) ≤ x ≤ 1
Area Between two curves
Finding the area between two curves;
F(x) = 2x2 + 1 and g(x) = 8 – 2x on the interval (-2) ≤ x ≤ 1
First I have to figure out the bigger function out of f(x) and g(x).
Method
Figure 55 Values of the integrated values of the functions (pic 1)
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Figure 56 Values of the integrated values of the functions (pic 2)
The answers got from the over two tables are the values of the integrated values of the functions. By
solving the two functions I had the option to get two values and through this, I could sort out the greater
value to discover the area.
F (X) = 9<G (X) = 27
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The area between the curves will be;
Figure 57 Area between the curves
Area between the two curves are = 18
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Assignment 01
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Graph
Figure 58 Area between the curves in a graph
Above picture shows the area of the two curves. (Purple Color)
Unit – 11
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89
4)
Finding units through integration
In this question, I need to discover the increment of trees in a specific forest through a rate of 3t 2 + 5t + 6
every year. This must be trailed by the level of oxygen that must be increased regarding the trees. This
oxygen increment will be 4 units for every 100 trees. Moreover, the measure of oxygen should be found for
the following three years.
Method
Figure 59 No. of oxygen units increased
No. of oxygen units increased =270 units
Unit – 11
Maths for Computing
Assignment 01
90
Part 3
5)
Differentiation with graphs
In this question we are asked to sketch a graph for f(x) = x5- 6x3 + 3by applying differentiation methods
to find:

Increasing and decreasing

Maximum and minimum

Concave up and concave down intervals

Inflection points
To draw the chart, I need to recognize the fixed points in the equation. X and Y values must be found
before drawing the chart. In this way, I utilized the first derivative of the equation to recognize the values.
Method
Figure 60 Method
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Figure 61 discovering the plots where the graph is curved
In the above pictures, I found the X values for plotting the diagram and I replaced these X values into the
main to recognize the Y values. From this method, we had the option to discover the plots where the graph
is curved.
Points are:
(0, 3)
(-1.9, 19.39)
(1.9, -13.39)
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The next step is to identify the second derivative to find the concave upwards/ downwards.
Method
Figure 62 identifying the second derivative to find the concave upwards/ downwards
Here the 2nd derivative will be - 20x3 -36x
The values found by this equation will be used to identify the concave upwards and downwards.
Unit – 11
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X
Increase/ Decrease
-∞ , -1.9
Increase
-1.9 , 0
Decrease
0 , 1.9
Decrease
1.9, ∞
Increase
To find the minimum and maximum value, the increasing and decreasing intervals can be used. From the
first derivative test the intervals obtained were as follows (from the graph):(-1.9, 19.39) - Maximum
(0, 3) – Inflection
(1.9, 13.39) – Minimum
This can be proved by taking the second derivative test. As shown in the above figure when the x values
(critical points are applied to the 2nd derivative) the answers obtained can be tested as follows.
F (0) = 0 = 0 (Inflection point)
F (-1.9) = -68.7 < 0 (Maximum)
F (1.9) = 68.7 > 0 (Minimum)
Therefore by using both the test, the student has proved the minimum and maximum values.
Unit – 11
Value
Minimum/Maximum
-1.9
-68.7 < 0 in 2nd derivative so Maximum
0
0 = 0 Inflection Point
1.9
68.7 > 0 in 2nd derivative so Minimum
Maths for Computing
Assignment 01
94
Graph
Figure 63 Graph 1
Concavity Intervals
First I would be finding the inflection points for concavity intervals. For this task I would be using the
second derivative to get the stationary points for x.
20x3 – 36x
4x (5x - 9)
X = 1.342 or -1.342 or 0
Next the student will be replacing the x values for the default equation which is x5 -6x3+3
According to that then;
X (0) - 05-6*03+3 = 3
X (1.34) – 1.345-6*1.343+3 = -7.12
X (-1.34) - -1.345-6*1.343+3 = 13.12
Unit – 11
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95
x
Values
0
3
-1.34
13.12
1.34
-7.12
Figure 64 Graph 2
The red line represents Concave down, the green line represents Concave up and the blue straight
line represents the Inflation points
Unit – 11
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To discover the concavity changes, second derivative test will be used.
Our f’’(x) = 20x3 – 36x
X=0
= 20*03 – 36*0
= 0 is the inflection point (place where the curve changes)
X = 1.9
= 20*1.93 – 36*1.9
= 68.78 (Positive value therefore concave upward)
X = -1.9
= 20*-1.93 – 36*-1.9
= -68.78 (negative value therefore concave downward)
Unit – 11
M
Concave Up/ Down
-∞ , -1.9
Concave Down
-1.34, 13.12
Inflection point
-1.34, 0
Concave Up
0, 3
Inflection Point
0, 1.34
Concave down
1.34, -7.12
Inflection Point
1.34, ∞
Concave Upward
Maths for Computing
Assignment 01
97
6)
Differentiation with graphs
Function of the graph is f (x) = 2x3 - 4x4 + 5x2. I have to find the minimum and maximum points by
differentiation.
1st differentiation Method
Figure 65 1st differentiation Method
The above image shows the method used to find the X values from the 1st deviation. According to that our
stationary values will be
X=0
X = -5/8
X=1
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Following this, I would be finding adjacent Y values for the X by replacing the X values for the main
function.
Figure 66 The values of X and Y
The values of X and Y will be as follows:
(0, 0)
(-5/8, 0.855)
(1, 3)
Unit – 11
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99
Increasing and decreasing intervals
X
Increase/ Decrease
-∞ , -0.625
Increase
-0.625, 0
Decrease
0,1
Increase
1, ∞
Decrease
To discover the minimum and maximum value, the increasing and decreasing intervals can be utilized.
From the first derivative test the intervals obtained were as follows (from the graph):
(-5/8, 0.855) - Maximum
(0,0) – Minimum
(1, 3) –Maximum
Figure 67 Graph 3
Unit – 11
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This can be proved by taking the second derivative test. As shown in the below figure when the x values
(critical points are applied to the 2nd derivative) the answers obtained can be tested as follows.
F (0) = 10 > 0 (Minimum)
F (-1.9) = -16.25< 0 (Maximum)
F (1.9) = -26< 0 (Maximum)
2nd Derivative method
Figure 68 2nd Derivative method
Unit – 11
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101
References
corporatefinanceinstitute. [online]. [Accessed 02/12/2022]. Available from World Wide Web:
<https://corporatefinanceinstitute.com/resources/knowledge/other/random-variable/>
cuemath. [online]. [Accessed 05/12/2022]. Available from World Wide Web:
<https://www.cuemath.com/conditional-probability-formula/>
educative. [online]. [Accessed 09/12/2022]. Available from World Wide Web:
<https://www.educative.io/edpresso/what-is-hashing>
Esoft 2021. 2021.
jeremykun. [online]. [Accessed10/12/2022]. Available from World Wide Web:
<https://jeremykun.com/2015/12/28/load-balancing-and-the-power-of-hashing/>
keydifferences. [online]. [Accessed 11/12/2022]. Available from World Wide Web:
<https://keydifferences.com/difference-between-discrete-and-continuous-variable.html>
mindtools. [online]. [Accessed 13/07/2022]. Available from World Wide Web:
<https://www.mindtools.com/pages/article/round-robin-brainstorming.htm>
scientificamerican. [online]. [Accessed 12/07/2022]. Available from World Wide Web:
<https://www.scientificamerican.com/article/bring-science-home-probability-birthday-paradox/>
stanford. [online]. Available from World Wide Web:
<http://stanford.edu/~dntse/classes/cs70_fall09/n12.pdf>
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