EA311 2022 TUTORIAL QUESTION AND SOLUTION
(1) An electrical system consists of four components as illustrated in figure 1. The system works
if components A and B work and either of the components C or D work. The reliability
(probability of working) of each component is also shown in Figure 1. Find the probability that
(a) The entire system works, and
(b) The component C does not work, given that the entire system works. Assume that four
components work independently.
Solution
(A)The probability that the entire system works;
Note: For the entire system to work, it must pass through either C
or D as they are parallel to each other, so let’s first find the
probability that either C or D works
P (C or D) = P(C ꓴ D) = P(C) + P(D) – P(C ꓵ D) = P(C) + P(D) – P(C) P( D)
From the question, we got P(C) = 0.9 and P(D) = 0.9
P(C ꓴ D) = 0.9 + 0.9 – (0.9×0.9) = 0.99
So let the probability that C or D works be P(E) = 0.99
Therefore the probability that the entire systems works is P(W) ;
P(W)=(A and B and E works) = P(A ꓵ B ꓵ E) = P(A) × P(B) × P(E)
P(W) = 0.9 × 0.9 × 0.99
P(W) = 0.8019
Therefore the probability that the entire systems works is 0.8019
(B) Probability that C does not work but the entire system works.
Probability that C will work is P(C) = 0.9
Probability that C will not work is P(C’) = 1-P(C) = 1-0.9 = 0.1
P=
(
)
(
)
=
( ) ( ) (
(
) ( )
)
=
. × . × . × .
.
= 0.09
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(2). In a certain region of the country it is known from (last, experience that the: probability of
selecting an adult over 40 years of age: with cancer is 0.05, If the probability of a doctor correctly
diagnosing a person with cancer as having the disease is 0.78 and the: probability of incorrectly
diagnosing a person without cancer as having the disease is 0.06, what is the probability that, a
person is diagnosed as having cancer?
SOLUTION
Let P(C) be person with cancer that is P (C) = 0.05
Therefore person without cancer is P(C’) = 1-P(C) = 1-0.05=0.95
Let P(D/C) be correct diagnosis given the person has cancer that is P(D/C) = 0.78
Let P(D/C’) be incorrect diagnosis given the person does not have cancer that is
P(D/C’) = 0.06
The probability that a person is diagnosed of having cancer therefore is;
The probability that the person has cancer and was correctly
diagnosed. That is P(C and D/C) = P(C ꓵ D/C ) = P(C) × P(D/C)
PLUS
The probability that the person does not have cancer and was
incorrectly diagnosed. That is,
P (C’ and D/C’) = P (C’ ꓵ D/C’) = P (C’) × P (D/C’)
To compute we have;
P = P(C and D/C) + P(C’ and D/C’)
P = P(C ꓵ D/C ) + P(C’ ꓵ D/C’ )
P = P(C) × P (D/C) + P (C’) × P (D/C’)
P = (0.05) (0.78) + (0.95) (0.06)
P = 0.096
Therefore probability that a person is diagnosed of having the disease is 0.096
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(3) Police plan to enforce speed limits by using radar traps at 4 different locations within the city
limits. The radar traps at each of the locations L1, L2, L3 and L4 are operated 40%, 30%, 20%, and
30% of the time, and if a person who is speeding on his way to work has probabilities of 0.2, 0.1,
0.5, and 0.2, respectively, of passing through these locations, what is the probability that he will
receive a speeding ticket?
SOLUTION
Let P (L1) be the probability that he will pass through location L1, that is P (L1) = 0.2 × 0.4
Let P (L2) be the probability that he will pass through location L2, that is P (L1) = 0.1 × 0.3
Let P (L3) be the probability that he will pass through location L3, that is P (L1) = 0.5 × 0.2
Let P (L4) be the probability that he will pass through location L4, that is P (L1) = 0.2 × 0.3
Therefore the probability that he will receive a speeding ticket is the probability that he will
pass through either of the four locations;
P (L1 ꓵ L2 ꓵ L3 ꓵ L4) = P (L1) + P (L2) + P (L3) + P (L4)
P = (0.2 × 0.4) + (0.1 × 0.3) + (0.5 × 0.2) + (0.2 × 0.3)
P = 0.08 + 0.03 + 0.1 + 0.06
P = 0.27
Therefore the probability that he will receive a speeding ticket is 0.27
(4) The probabilities that a service station will pump gas into 0, 1, 2, 3, 4, or 5 or more cars during
a certain 30-minute period are 0.03, 0.18, 0.24, 0.28, 0.10, and 0.17, respectively. Find the
probability that in this 30-minute period
(a) More than 2 cars receive gas;
(b) At most 4 cars receive gas;
(c) 4 or more cars receive gas.
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SOLUTION
Let P(C = 0) denote zero cars, that is P(C = 0) = 0.03
Let P(C = 1) denote one car, that is P(C = 1) = 0.18
Let P(C = 2) denote two cars, that is P(C = 2) = 0.24
Let P(C = 3) denote three cars, that is P(C = 3) = 0.28
Let P(C = 4) denote four cars, that is P(C = 4) = 0.10
Let P(C ≥5) denote five cars, that is P(C ≥5) = 0.17
(A) More than 2cars received gas that is from three cars above
We have P(C = 3) or P(C = 4) or P(C ≥5) (range above 2)
P = P(C = 3) + P(C = 4) + P(C ≥5)
P = 0.28 + 0.10 + 0.17
P = 0.55
(B) At most 4 cars received gas that is from 0 to the 4 th car
We have P(C = 0) or P(C = 1) or P(C = 2) or P(C = 3) or P(C =4)
P = P(C = 0) + P(C = 1) + P(C = 2) + P(C = 3) + P(C = 4)
P = 0.03 + 0.18 + 0.24 + 0.28 + 0.10
P = 0.83
(C) 4 or more cars received gas that is from 4 to ≥5
We have P(C =4) or P(C ≥5)
P = P(C = 4) + P(C ≥5)
P = 0.10 + 0.17
P = 0.27
(5) A producer of a certain type of electronic component ships to suppliers in lots of twenty.
Suppose that 60% of all such lots contain no defective components, 30% contain one defective
component, and 10% contain two defective components. A lot is selected and two components
from the lot are randomly selected and tested, and neither is defective.
(a) What is the probability that zero defective components exist in the lot?
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(b) What is the probability that one defective exists in the lot?
(c) What is the probability that two defectives exist in the lot?
SOLUTION
Let’s first bring out the given data:
Let P(C) be two non-defective components are selected
Let P (G) be lot contains non-defective components, that is P (G) = 60% = 0.6
Let P (H) be lot contains one defective component, that is P (H) = 30% = 0.3
Let P (I) be lot contains two defective components, that is P (I) = 10% = 0.1
(a) What is the probability that zero defective components exist in the lot?
Note: The event where two components are selected out of twenty given that none is
defective is:
From combination and permutation, C(n,r)
n is the number of lot (n=20)
r is the total number of selection made
C(n,r)=
(
!
)! !
Therefore the probability of two non-defective components are selected P(C), given lot
contains non-defective components P (G) is P(C/G) =
(
(
!
)!× !
!
)!× !
=
=1
Tip: the highlighted portion of the fraction shows that
the selection P (G) contains no defective component
and the un-highlighted portion simply shows that two
selection out of twenty was made. Check next selection
for further knowledge
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The probability of two non-defective components are selected P(C), given lot
contains one defective components P (H) is P(C/H) =
!
)!× !
!
)!× !
(
(
=
=
Tip: In this case the lot contains one defective thus we have n = 20-1 defective = 19
The probability of two non-defective components are selected P(C), given lot contains two
defective components P (I) is P(C/I) =
!
)!× !
!
)!× !
(
(
=
=
Tip: In this case the lot contains two defective thus we have n = 20-2 defective = 18
To find P (G/C) that is the probability that zero defective is selected given that the two nondefective components are selected, we use Bayes’ theorem
P (G/C) =
( / )×
( )
( )
by the law of total possibility;
P(C) = P(C/G) ×P(G) + P(C/H) × P(H) + P(C/I) × P(I)
P(C) = 1 × 0.6 +
× 0.3 +
×0.1
P(C) = 0.9505
P (G/C) =
( / )× ( )
( )
× .
=
= 0.631
.
Therefore the probability that zero defective exist in the lot is 0.631
(b) What is the probability that one defective exist in the lot
P (H/C) =
( / )× ( )
( )
=
× .
.
= 0.284
The probability that one defective exist in the lot is 0.284
(C) What is the probability that two defectives exist in the lot?
P (I/C) =
( / )× ( )
( )
=
× .
.
= 0.084
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(6) A cereal manufacturer is aware that the weight of the product in the box varies slightly
from box to box. In fact, considerable historical data has allowed the determination of the
density function that describes the probability structure for the weight ounces). In fact, letting X
be the random variable weight, in ounces, the density function can be described as;
(a) Verify that this is a valid density function.
(b) Determine the probability that the weight is smaller than 24 ounces.
(c) The company desires that the weight exceeding 26 ounces is an extremely rare occurrence.
What is the probability that this "rare occurrence" does actually occur?
SOLUTION
(a) Verify that this is a valid density function.
The condition for probability density function is
∫ 𝑓(𝑥)𝑑𝑥 = 1
For the function 𝑓(𝑥) =
, 23.75 ≤ 𝑥 ≤ 26.25
0,
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
to be a probability density function, 𝑓(𝑥) =
1
2
5
.
𝑑𝑥 =
.
2(26.25 − 23.75)
5
= =1
5
5
This shows that 𝑓(𝑥) is a probability density function
(b) Determine the probability that the weight is smaller than 24 ounces.
The weight smaller than 24 gives the upper limit of 24 (that is the weight
should not exceed 24) and lower limit of 23.75 and let X be the random variable
P(X < 24) =
∫
.
𝑑𝑥 =
(
.
)
=
.
= 𝟎. 𝟏
Therefore the probability that the weight is smaller than 24 ounces is 0.1
(c)The company desires that the weight exceeding 26 ounces is an extremely rare
occurrence. What is the probability that this "rare occurrence" does actually occur?
The weight exceeding 26 gives the lower limit of 26 (that is the weight should
not be less than 26) and upper limit of 26.25 and let X be the random variable
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P(X > 26) =
∫
.
𝑑𝑥 =
(
.
)
=
.
= 𝟎. 𝟏
Therefore the probability that the weight exceeding 26 ounces that is the rare
occurrence is 0.1
(7) According to Chemical Engineering Progress (Nov. 1990), approximately 30% of all
pipework failures in chemical plants are caused by operator error.
(a) What is the probability that out of the next 20 pipework failures at least 10 are due to
operator error?
(b) What is the probability that no more than 4 out of 20 such failures are due to operator
error?
(c) Suppose for a particular plant that, out of the random sample of 20 such failures, exactly 5
are operational errors. Do you feel that the 30% figure stated above applies to this plant?
Comment.
SOLUTION
Related topic is binomial and random variable.
(a) What is the probability that out of the next 20 pipework failures at least 10 are due to
operator error
Let X be the random variable denoting number of pipework failure among 20.
Probability of failure is 30% = 0.3
From binomial, X~Binomial (n,r)
X = random variable
n is the number of trial = 20,
r is the probability of failure = 0.3
The probability that at least 10 are due to operator error is 1 minus the
summation from X=0 to X=9
P(X < 10) =
(𝑛𝐶𝑥 )
(𝑟) × (1 − 𝑟)
Now with the expression generated for the 10 pipework
we move from X=0 to X=9
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P(X = 0) = 20𝐶 × (0.3) × (1 − 0.3)
=(
!
)!× !
× (0.3)
× (0.7)
20!=20×19×18×17×16×15×14×13×12×
11×10×9×8×7×6×5×4×3×2×1
P(X = 0) = 0.00079
Same procedure applies for X=1 up to X=9
P(X = 1) = 0.00683
P(X = 2) = 0.028
P(X = 3) = 0.072
P(X = 4) = 0.1304
P(X = 5) = 0.1789
P(X = 6) = 0.1916
P(X = 7) = 0.1642
P(X = 8) = 0.1144
P(X = 9) = 0.065
P(X < 10) = 0.00079 + 0.00683 + 0.028 + 0.072 + 0.1304 + 0.1789 + 0.1916 + 0.1642 + 0.1144 +
0.065 = 0.95212
P(X ≥ 10) = 1 − P(X < 10)
P(X ≥ 10) = 1 − 0.95212
P(X ≥ 10) = 0.479
Therefore the probability that out of the next 20 pipework failures, at least 10 are due to
operator error is 0.479
(b) What is the probability that no more than 4 out of 20 such failures are due to operator
error?
This will be from X=0 to X=4 because 4 is included
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P(X ≤ 4) = 0.00079 + 0.00683 + 0.028 + 0.072 + 0.1304
P(X ≤ 4) = 0.238
The probability that no more than 4 out of 20 such failures are due to operator error is 0.238
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(c) Exactly 5 are operational errors
That is X=5
P(X = 5) = 0.1789
TEST 01 EA311
1. Find the probability of the following events,
a. The sum 8 appears in a single toss of two fair four faced dice
Solution
The total possible outcome when two four faced dice are tossed once is 4 × 4 = 16
The chances of getting the sum of 8 include (4+4) which shows an expected outcome
of 1 therefore, the probability of the sum of 8 appearing is P(sum of 8) =
𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒐𝒖𝒕𝒄𝒐𝒎𝒆
𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒐𝒖𝒕𝒄𝒐𝒎𝒆
=
𝟏
𝟏𝟔
b. A non-defective bolt will be found next if out of 600 bolts examined, 12 were
defective
Solution
The probability of finding 12 defective bolts out of 600 is P(D) =
=
The probability of finding a non-defective bolt is P (N) = 1-P (D) = 1 −
= 0.98
c. At least 1 head appears in two tosses of a coin
Solution
Using binomial probability, the probability that at-most no head is P (at-most no
head) = 1 – P (head)
First: toss for head
Second: toss for head
The odds that both tosses are head is the P (at-most no head) =
× =
P (at least a head) = 1-P (at-most no head) = 1 − = 0.75
d. A student will be from chemical or petroleum engineering from a group of nine students
with each from a department from faculty of engineering
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Solution
Given number of students in each department is 9 and the probability of a
student from a department from faculty of engineering be P(chemical
engineering) = , P(petroleum engineering) = ,
Therefore the probability that a student will be from chemical or petroleum is
𝟏
𝟏
𝟏
𝟏
𝟗
𝟗
𝟗
𝟗
P ( 𝒐𝒓 ) = +
= 𝟎. 𝟐𝟐
5. For what value of n is 3 × (𝑛 + 1 𝐶 ) = 7 × (𝑛𝐶 )
SOLUTION
)!
×(
(
)!× !
)!
×(
(
)!×
(
(
)!
)!×
× !
=(
=(
=(
)!× !
× !
)!×
× !
)!×
Note: (1-3=-2, 3! = 6 and 2! = 2)
Note: (3/6 = 2)
Multiplying both side by (𝑛 − 2)! × 2
(𝑛 + 1)! = 7𝑛!
(n+1)n! = 7𝑛!
Dividing both side by n!
n+1=7
n = 7-1
n=6
More solution to come soon.
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