AL-IMRAN HOSH ACADEMY
EXERCISE 3.1
1. A coin is tossed 3 times. Each toss results in head or tail. Find the number of
possible outcomes (result) for the 3 tosses. Illustrate these outcomes using a
tree diagram.
SOLUTION
The possible outcomes of the three tosses are 2𝑛 = 23 = 8 ways
2. There are 3 feeder bus service playing to and from the nearby town center.
Ahmed can ride on one of these services or walk to the town center.
However, he decides that on this way back, he will ride on one of these
services. In how many ways can he go to the town center and back?
Solution
He can go to the town in 4 ways and then he can return in 3 waysSo the
total possible ways are 4 × 3 = 12 𝑤𝑎𝑦𝑠
3. Nasra has to do the following during her lunch break: take lunch,
post latter,go to the bank and buy the afternoon papers. In how
many ways can she do all these?
Solution
She can do all of these in 4! = 4 × 3 × 2 × 1 = 24 ways
4. Six men and five women are available to form a mixed double
pair for atennis match. How many pairs are possible?
Solution
Number of men= 6
Number of women= 5
So, the number of possible pairs= 6 × 5 = 30 𝑤𝑎𝑦𝑠
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AL-IMRAN HOSH ACADEMY
5. Three friends decided to have dinner together and then go
shopping. Five restaurants are proposed for the dinner and four
nearby shopping centers aresuggested. How many possibilities are
there?
Solution
Number of restaurants= 5 and number of shopping centers= 4
Number of possibilities = 5 × 4 = 20 𝑤𝑎𝑦𝑠
6. Eight people have been shortlisted for an interview. In how many ways can
the interviewer see them one after another?
Solution
The possibility that the interviewer can see one after another are
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320 Ways
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AL-IMRAN HOSH ACADEMY
EXERCISE 3.2
1. Find the total number of difference permutations of all the letters of the
word:
a) SIMPLE
b) SECONDARY
Solution
The total number of permutation=6!
=6×5×4×3×2×1
= 720 Ways
The total number of permutation=9!
=9×8×7×6×5×4×3×2×1
= 362880 Ways
2. Find the total number of different 4-digit numbers using all the digits in
the number 4129.
Solution:
We have 4 different digit number (4,1,2,9)
∴ The total number of permutation=4!
=4×3×2×1
= 24 Ways
3. The following duties need to be carried out: clean the board, arrange the
tables, sweep the floor and the clear the waste paper basket. Find the
numberof ways of assigning these duties to 4 students, given that each
student will perform only one duty.
Solution:
The total number of permutation=4!
=4×3×2×1
= 24 Ways
4. In a particular division of a soccer league, there are 9 teams. How many
different end-of- season rankings are possible? Assume that there are no
ties.
Solution
The total number of permutation=9!
=9×8×7×6×5×4×3×
2×1
= 362880 Ways
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
5. There are five finalists at an oratorical competition. In how many ways
canthey be arranged to give their speeches?
Solution
The number of ways=5!
=5×4×3×2×1
= 120 Ways
6. Four students are to go for dinner and order a hamburger, a fish burger, a
cheeseburger and beef burger (one burger for each). When the waiters
returns with the food, she forgets which student orders which items and
simply places a burger before each students. In how many ways can the
waiters do this?
Solution
The number of possible ways is = 4𝑝3
=
=
4!
1!
𝑛!
4!
= (4−3)!
(𝒏−𝒓)!
= 4 × 3 × 2 × 1 = 24 𝑤𝑎𝑦𝑠
7. With out using calculator evaluate
a)
4!
b)
5!
a)
4!
5!
c)
=
10!
6!4!
4!
5×4!
=
=
8!
6!
c)
𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
1
b)
5
10×9×8×7×6!
6!4!
8!
6!
=
=
10!
6!4!
8×7×6
6!
10×9×8×7
4!
=
8 × 7 = 56
= 210
8. Simplify
a)
(𝑛−1)!
b)
𝑛!
(𝑛−3)!
c)
𝑛!
(𝑛!)2
(𝑛−1)!(𝑛−2)!
Solution
a)
b)
(𝑛−1)!
𝑛!
(𝑛−3)!
𝑛!
=
(𝑛−1)(𝑛−2)(𝑛−3)!
𝑛(𝑛−1)(𝑛−2)(𝑛−3)!
=
=
(𝑛−3)!
𝑛(𝑛−1)(𝑛−2)(𝑛−3)!
=
1
𝑛
1
𝑛(𝑛−1)(𝑛−2)
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AL-IMRAN HOSH ACADEMY
(𝑛!)2
c)
(𝑛−1)!(𝑛−2)!
(𝑛)!(𝑛)!
𝑛−1)!(𝑛−2)!
2
=(
=
𝑛(𝑛−1)!𝑛(𝑛−1)(𝑛−2)!
(𝑛−1)!(𝑛−2)!
= 𝑛 (𝑛 − 1)!
9. Without calculator, evaluate
a) 7𝑝3
b) 10𝑝2
Solution
a) 7𝑝3 =
𝑛!
(𝑛−𝑟)!
𝑛!
b) 10𝑝2 = (𝑛−𝑟)!
=
=
7!
=
(7−3)!
10!
(10−2)!
7!
c) 5𝑝5
=
4!
7!
= 8!
=
7×6×5×4!
4!
10×9×8!
8!
= 210
= 90
𝑛!
5!
5!
c) 5𝑝5 = (𝑛−𝑟)!
= (5−5)! = 0! = 5 × 4 × 3 × 2 × 1 = 120
10. In a Mathematics class with 30 students, the teacher wants 2 different
students to present solutions to problem 3 and 5 on the board. In how
manyways can the teacher assign the problems?
Solution
The number of possible ways is = 30𝑝2
𝑛!
30!
=
=
(𝑛 − 𝑟)! (30 − 2)!
30! 30 × 29 × 28!
=
=
= 870
28!
28!
11. A shelf will hold only seven books. Given that 11 different books are
available, find the number of different arrangements that can be made to
fillthe shelf.
Solution
The number of possible arrangements is = 11𝑝7
𝑛!
11!
=
(𝑛 − 𝑟)! (11 − 7)!
11! 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!
=
=
4!
4!
= 11 × 10 × 9 × 8 × 7 × 6 × 5
=
= 1663200
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AL-IMRAN HOSH ACADEMY
12. In how many ways can a judge award first, second and third prizes in a
contest with 9 participants?
Solution
The number of possible arrangement is = 9 × 8 × 7 = 504 𝑤𝑎𝑦𝑠
13. In a survey 10 characteristics of a teacher are listed. You are asked to
indicate in order of importance which 4 of these characteristics a good
teacher. How many response are there?
Solution
The total number of response is = 10𝑝4
𝑛!
10!
=
(𝑛 − 𝑟)! (10 − 4)!
10! 10 × 9 × 8 × 7 × 6!
=
=
6!
6!
=
= 5040
14. How many 5-digit numbers can be formed from the digits 2, 3, 5, 7, 8
and 9if no digit is repeated?
Solution
The number of possible permutations =6𝑝
5
=
=
𝑛!
6!
=
(𝑛 − 𝑟)! (6 − 5)!
6!
= 6×5×4×3×2×1
1!
= 720
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AL-IMRAN HOSH ACADEMY
15. 5 actors and 8 actresses are available for a play which required 3 male
rolesand 4 female roles. Find the number of different possible cast lists.
Solution
The number of possible male is = 5𝑝
3
The number of possible female is = 8𝑝
4
Total number of male and female are= 5𝑝
3
=
5!
8!
×
(5 − 3)! (8 − 5)!
=
=
× 8𝑝4
5!
2!
=
8!
3!
5 × 4 × 3 × 2! 8 × 7 × 6 × 5 × 4 × 3!
×
2!
3!
= (5 × 4 × 3 ×) × (8 × 7 × 6 × 5 × 4)
= 60 × 6720
= 403200
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
EXERCISE 3.3
1. Without using a calculator, evaluate 5𝑐4 and 9𝑐3
Solution
𝑐 (𝑛, 𝑟) =
=
𝑛!
𝑟!(𝑛−𝑟)!
5!
=5
4! (5 − 4)!
𝑐 (𝑛, 𝑟) =
𝑛!
𝑟!(𝑛−𝑟)!
9!
=
= 84
3! (9 − 3)!
2. In a soccer league consisting of the twelve sides, each team plays every
other team once. How many matches are there?
Solution
Number of teams 𝑛 = 12 teams
Number of teams in one match 𝑟 = 2 teams
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐(𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
=
=
=
=
12!
2!(12−2)!
12!
2!10!
12×11×10!
2!10!
12×11
2!
= 66 matches
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
If each team plays every other team once, then they can
play 66 matches
3. A club has 14 members. It has to send a delegation of 5 members to
represent team at particular event. Find the number of possible
delegations.
Solution
Number of members n=14
Number of delegation r=5
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑛!
𝑟!(𝑛−𝑟 )!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
𝑛!
𝑟!(𝑛−𝑟)!
=
5!(14−5)!
12!
=
5!9!
=
=
14!
14×13×12×11×10×9!
5×4×3×2×1×9!
240240
120
= 2002
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AL-IMRAN HOSH ACADEMY
4. At the library, peter found 6 books of interest but he can only borrow 4
books. How many possible selections can he make?
Solution
Number of interest books 𝑛 = 6
Number of borrow books 𝑟 = 4
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐(𝑛, 𝑟 ) =
𝑛!
=
𝑟!(𝑛−𝑟 )!
𝑛!
=
𝑟!(𝑛−𝑟 )!
𝑛!
=
𝑟!(𝑛−𝑟 )!
𝑛!
𝑟!(𝑛−𝑟 )!
=
6!
4!(6−4)!
6!
4!2!
6×5×4!
4!2!
6×5
2×1
= 15
5. The second section of mathematics paper contains 7 questions and a
candidate must answer only 4 questions. In how many ways can the 4
questions be chosen (without regard to order)?
Solution
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑐 (𝑛, 𝑟 ) =
𝑛!
𝑟!(𝑛−𝑟 )!
𝑛!
𝑟!(𝑛−𝑟 )!
𝑛!
𝑟!(𝑛−𝑟 )!
𝑛!
𝑟!(𝑛−𝑟 )!
=
=
=
=
7!
4!(7−4)!
7!
4!3!
7×6×5×4!
4!3!
7×6×5
3×2×1
= 35
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AL-IMRAN HOSH ACADEMY
6. One a piece of paper, 8 points are marked such that no 3 points lie on the
same straight line. How many lines can be drawn passing through any 2
ofthese points?
Solution
To form a line, all we need to do is select any 2 points
The number of ways to select any points (out of 8 distinct points)
will be 8𝑐2
Therefore, the number of the line will be 8𝑐2
=
8 × 7 × 6!
2! 6!
56
=
= 28 𝑙𝑖𝑛𝑒𝑠
2
8!
2!(8−2)!
=
7. Seven points lie on a circle. How many triangles can be drawn using 3
these points as vertices?
Solution
Total number of triangles = 7𝑐3 =
=
7!
3!(7−3)!
7 × 6 × 5 × 4!
3! 4!
=
7×6×5
3×2×1
= 35 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
8. A group of 4 adults and 3 children are to be formed form 8 adults and 5
children. How many possible groups are there?
Solution
𝑐 (𝑛, 𝑟 ) × 𝑐(𝑛, 𝑟 ) =
=
𝑛!
𝑟!(𝑛−𝑟 )!
8!
4!(8−4)!
×
×
𝑛!
𝑟!(𝑛−𝑟 )!
5!
3!(5−3)!
= 70 × 10 = 700
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
9. To promote reading, a teacher decides to feature 3 classics, 4
contemporarynovels and 2 non-fiction books on a notice board. How
many selections canshe make from 5 classics, 6 contemporary novels and
4 non-fiction books?
Solution
The possible number of classics is 𝑛𝑐𝑟
The possible number of novels is 𝑛𝑐𝑟
= 5𝑐3
= 6𝑐4
The possible number of non-fiction books is 𝑛𝑐𝑟
The total number of selections is =
5!
3!(5−3)!
×
6!
4!(6−4)!
×
= 4𝑐2
5𝑐3 × 6𝑐4 × 4𝑐2
4!
2!(4−2)!
= 10 × 15 × 6 = 900 𝑤𝑎𝑦𝑠
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
EXERCISE 3.4
1. An unbiased die is tossed. Find the probability that the number obtained is
a) An even number
b) less than 4
c) Greater
than 6
Solution
a) The probability of an even number is
3
6
b) The probability which is less than 4 is
=
3
6
1
2
=
1
2
c) The probability that is greater than 6 is zero because when
we toss a dieonly 6 outcomes are possible
2. A card is randomly drawn from a standard deck of 52 playing
cards. Find theprobability that the card drawn is
a) A king
b) A diamond
c) Not a diamond
d) A king of diamonds
Solution
a) There are 4 kings in the deck
Probability getting a king =
4
52
=
1
13
b) The are 13 diamond is the deck
Probability getting diamond =
13
52
=
1
4
c) Probability of not getting a diamond
= 1 − probability of getting a diamond
=1−
=
1
4
3
𝑜𝑟 75%
4
d) The probability of getting a king of diamond =
1
52
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AL-IMRAN HOSH ACADEMY
3. A letter is randomly chosen from the world NATIONAL. What
is the probability that it will be;
a) The letter A
b) A vowel
c) The letter S
Solution
a) The probability of getting the letter A is
2 1
=
8 4
b) The probability of a vowel is
3
=
8
=
c) The probability of the letter S is
0
= =0
8
4. A box contains 3 red discs, 4 green discs, and 5 blue discs.
One discs israndomly drawn from the box. What is
probability that will be;
a) A red discs
b) A green discs
c) A red discs or green discs
d) Blue discs
Not a blue discs
Solution
a. A red discs
3
12
b. A green discs
=
4
12
1
4
=
1
3
c. A red discs or green discs
d. Blue discs =
3
12
+
4
12
1
1
7
4
3
12
= + =
5
12
Not a blue discs = 1 − 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑏𝑙𝑢𝑒 𝑑𝑖𝑠𝑐𝑠
5
7
= 1−
=
12 12
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AL-IMRAN HOSH ACADEMY
5. An unbiased die is tossed. What is probability of getting,
a) A number not divisible by 2
b) A number greater than 2
c) A number less than 7
Solution
a) The numbers of not divisible by 2 (1, 3, 5)
3
Probability of not divisible by 2
6
b) The numbers greater than 2 (3, 4, 5, 6)
Probability greater then 2
4
6
=
=
1
2
2
3
c) The number less then 7 (1, 2, 3, 4, 5, 6)
Probability of getting a number less than 7 =
6
6
=
1 𝑜𝑟 100%
6. A card is randomly drawn from standard deck of 52 playing
cards. What isthe probability that the card drawn is
a) A picture card
b) A spade
c) Not a spade
d) An ace of spades
Solution
a) There are 12 picture cards in the deck
Probability of getting a picture card =
12
52
=
3
13
b) There are 13 spade in the deck
∴ Probability of getting a spade =
13 1
=
52 4
c) Probability of not getting a spade
= 1 − Probability of getting a spade
1
=1−
4
3
=
4
1
d) The probability of getting an ace of spades =
52
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
7. A letter is randomly chosen from the words SAMPLE
SPACE. Find theprobability that the letter chosen is
a) The letter S or P
b) A vowel
c) A constant
Solution
a) The probability of getting S =
2
11
Probability of getting the letter P=
2
11
∴ Total probability of getting S or P =
b) The probability of getting a vowel =
c) The probability of constant =
2
11
+
2
=
11
4
11
4
11
7
11
8. The pages of thin book are numbers 1 to 48. A page is chosen at
randomly. Write down, given your answers as fraction, the probability that
the page numbers
a) Contains a single digit
b) Contains more than a single digit
c) Contains at least one digit than is 3
Solution
a) The number contain a single digit (1, 2, 3, 4, 5, 6, 7, 8, 9)
∴ The probability of getting a single digit
9
48
=
3
13
b) The number contains more than single digit is 10 up to 48
∴ The probability of getting more than a single digit
c) The probability of getting at least one digit
48
48
39
48
=
1 𝑜𝑟 100%
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
16
=
13
16
AL-IMRAN HOSH ACADEMY
EXERCISE 3.5
1.
A die and a coin are tossed together. The probability diagram is show.
(a) How many possible outcomes are there altogether?
(b) What is probability of getting a tail on the coin?
(c) What is the probability of getting an odd number?
(d) What is the probability of getting tail and an odd number?
SOLUTION
𝑠 = [(𝐻, 1), (𝐻, 2), (𝐻, 3), (𝐻, 4), (𝐻, 5), (𝐻, 6),
(𝑇, 1), (𝑇, 2), (𝑇, 3), (𝑇, 4), (𝑇, 5)(𝑇, 6)]
(a) The total possible outcomes is = 2 × 6 = 12
(b) Let A be the event of getting a tail on the coin
𝐴 = [(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)]
6
1
𝑃(𝑡𝑎𝑖𝑙 ) =
=
12 2
(c) Let B the event of getting on odd number
𝐵 = (𝐻, 1), (𝐻, 3)(H, 5), (T, 1), (T, 3), (T, 5),
𝑃 (𝐵 ) =
6
1
=
12 2
1
1
1
(d) The probability of getting tail and an odd number= × =
2
2
4
2. A spinner board consists of four equal quadrants numbered 1 to 4 shown.
The spinner is spun twice, find the probability that
a. Both numbers are even
b. The sum is less than 4
c. The product of the numbers is odd
Solution
𝑆1/𝑆2
1
2
3
4
(1,2)
1
(1,1)
(1,3)
(1,4)
(2,2)
(2,3)
2
(2,1)
(2,4)
(3,2)
(3,3)
3
(3,1)
(3,4)
(4,2)
4
(4,1)
4.3)
(4,4)
Sample space of data is 16
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AL-IMRAN HOSH ACADEMY
(a) Let A be the event that both numbers are even
4
1
The probability of getting both numbers 𝑃(𝐴)16 = 4
(b) Let B be the event that sum is less than 4
3
The probability of getting a sum that is less than 4 𝑃(𝐵) 16
4
1
(c) Let C the event that product is odd 𝑃(𝐶) 16 = 4
3. Four cards are numbered 2, 3, 5, and 7. Respectively.
Two cards are chosenat random
(a) State the total number of possible outcomes.
(b) Find the probability that the
(i)
Number on both cards are odd.
(ii) Sum of the two numbers is odd
(iii) Product of the two numbers is odd.
Solution
(a) Total number of possible outcomes
𝑆 = (2,3), (2,5), (2,7), (3,5), (3,7), (5,7)
𝑃(𝑆) = 6
(b) The probability that
i.
The numbers on both cards are odd
ii.
The sum of the two numbers is odd
iii.
The product of the two numbers is odd
4. Group A consists of the numbers 2, 3, and 4 while group B
consists of the numbers 5, 6, and 7. A numbers is chosen at
random from group A and is denoted by a. Another number is
then chosen at random from the group B andis denoted by b. find
the probability that
(a) The sum of 𝑎 + 𝑏 is even
(b) The sum of 𝑎 + 𝑏 is prime
(c) The product of 𝑎𝑏 is even.
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
Solution
Sample space= 9
(a) The probability that the sum of 𝑎 + 𝑏 is even is
(2,6), (3,5), (3,10), (4,6)
4
𝑃 (𝐴) =
9
(b) Let B be the event that the sum of 𝑎 + 𝑏 is prime
The probability that the sum of 𝑎 + 𝑏 is prime is (2,7), (4,7)
𝑃 (𝐴 ) =
2
9
(c) Let C be the event that product is even
The probability that the product is even
= (2,5), (2,6), (2,7), (3,6), (4,5), (4,6), (4,7)
7
𝑃 (𝐴) =
9
5. Group C consist of the numbers, 2, 4 and 6 while the group D
consists of thenumbers 3, 5 and 7. A number is chosen at random
from group C and is denoted by c. Another number is chosen at
random from group D and denoted
by d. find the probability that
(a) Is less than
𝑐
𝑑
1
2
(b) Is greater than 1
1
(c) Is more than but less than 1?
2
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
19
AL-IMRAN HOSH ACADEMY
Solution
Probability Diagram of 2 Groups
𝒄
𝒅
2
3
2
3
4
4
3
6
6
3
Let S be the sample space = 9
(a) The probability that
𝑐
𝑑
5
7
2
5
4
5
6
5
2
7
4
7
6
7
1
2
2
2
5
7
is less than is ( 𝑎𝑛𝑑 )
𝑃 (𝐴) =
2
9
4 6 6
(b) Let B the event that grater than 1= ( , , )
3 3 5
𝑃 (𝐵 ) =
(c) Let the probability that
𝑐
𝑑
3 1
=
9 3
1
is more than but less than
2
4 2 4 6
1=( , . , )
5 3 7 7
4
9
6. Two unbiased dice are thrown together. Find the probability that
(a) The two numbers are equal
(b) One number is not a multiple of the other number
(c) The difference between the two numbers is 3.
Solution
𝑃 (𝐶 ) =
The sample space of the two dice is 36
(a) Let A the event that the two numbers are equal
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
20
AL-IMRAN HOSH ACADEMY
𝐴 = (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
𝑃 (𝐴 ) =
6
1
=
36 6
(b) Let B be the event that one number is multiple of the other
number.
𝐵 = (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2),
(2,4), (2,6),
(3,1), (3,3), (3,6), (4,1), (4,2), (4,4), (5,1), (5,5),
(6,1), (6,2), (6,3), (6,6)
𝑃 (𝐵 ) =
22 11
=
36 18
The probability that the number is not multiple of other
7
= 1 − 𝑃(𝐵) = 1 − 11
=
18
18
(c) Let C be the event that different between the two numbers is
3
C = [(1,4), (2,5), (3,6), (4,1), (5,2), (6,1), ]
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
21
AL-IMRAN HOSH ACADEMY
EXERCISE 3.6
1. If P (A)= 0.3, P (A ∩ B) =0.15 and P (A 𝖴 B) =0.95, find P (B).
Solution
𝑃(𝐴 𝖴 𝐵) = 𝑃(𝐴) + 𝑃(𝐵 − 𝑃(𝐴 ∩ 𝐵)
0.95 = 0.3 + 𝑃(𝐵) − 0.15
0.95 = 𝑃(𝐵) + 0.3
𝑃(𝐵) = 0.95 + 0.15 − 0.3
4
𝑃 (𝐵 ) = = 0.8
5
2.
Given that the events A and B are mutually exclusive with
3
2
𝑃 (𝐴) =
and 𝑃(𝐵 ) = find the value of 𝑃(𝐴 ∪ 𝐵)
10
5
Solution
𝑃 (𝐴 ∪ 𝐵 ) = 0 Since the event is mutually exclusive
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴) + 𝑃 (𝐵 ) − 𝑃(𝐴 ∪ 𝐵)
3 2
𝑃 (𝐴 ∪ 𝐵 ) =
+ −0
10 5
7
𝑃 (𝐴 ∪ 𝐵 ) =
10
2
2
1
3.
Given 𝑃(𝐴) = , 𝑃 (𝐵 ) = and 𝑃 (𝐴 ∩) = , find the value of
3
3
2
𝑃(𝐴 ∪ 𝐵).
4.
Solution
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴) + 𝑃 (𝐵 ) − 𝑃(𝐴 ∪ 𝐵)
2 2 1
𝑃 (𝐴 ∪ 𝐵 ) = + −
3 3 2
5
𝑃 (𝐴 ∪ 𝐵 ) =
6
Given 𝑃(𝑋) = 0.37, 𝑃(𝑌) = 0.48, and 𝑃(𝑋 𝖴 𝑌) = 0.69, find the
value of 𝑃(𝑋 ∩ 𝑌).
Solution
𝑃(𝑋 𝖴 𝑌) = 𝑃(𝑋) + 𝑃(𝑌) − 𝑃(𝑋 ∩ 𝑌)
0.69 = 0.37 + 0.48 − 𝑃(𝑋 ∩ 𝑌)
0.69 = 0.85 − 𝑃(𝑋 ∩ 𝑌)
𝑃(𝑋 ∩ 𝑌) = 0.85 − 0.69
4
𝑃 (𝑋 ∩ 𝑌) = 0.16 =
25
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
5.
6.
7.
Given 𝑃(𝐴) =
7
10
, 𝑃(𝐴 ∪ 𝐵 ) =
9
10
and 𝑃 (𝐴 ∩) =
3
, find the value of
20
𝑃(𝐵)
Solution
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴) + 𝑃 (𝐵 ) − 𝑃(𝐴 ∪ 𝐵)
9
7
3
=
+ 𝑃 (𝐵 ) −
10 10
20
9
7
3
𝑃 (𝐵 ) =
−
+
10 10 20
7
𝑃 (𝐵 ) =
20
Given 𝑃(𝑆) = 0.34; 𝑃(𝑇) = 0.49 and 𝑃(𝑆 𝖴 𝑇) = 0.83, show that the
events S and T are mutually exclusive.
Solution
In mutually exclusive the 𝑃(𝑆 ∩ 𝑇) = 0
𝑃(𝑆 𝖴 𝑇) = 𝑃(𝑆) + 𝑃(𝑇) − 𝑃(𝑆 ∩ 𝑇)
0.83 = 0.34 + 0.49
0.83 = 0.83
Hence, it is a mutually exclusive events
From the 500 employees in company, 200 participate in the
company profilesharing scheme (𝑃), 400 employees have
medical insurance protection scheme (𝑀) and 200 employees
have both the scheme
a) Draw a Venn diagram to illustrate the above information.
b) What is the probability that an employee chosen at random
i. Participates in at least one of the schemes?
ii. Does not participate in any one of the scheme?
Solution
𝑃 (𝑝) =
200
500
=
2
𝑃(𝑚) =
5
400
500
=
4
5
𝑃(𝑝 ∩ 𝑚) =
(a) Vein diagram
(b) Does not participate in any one of the scheme?
(i) 𝑃 (𝑝 ∪ 𝑚) = 𝑃 (𝑝) + 𝑃(𝑚) − 𝑃(𝑝 ∩ 𝑚)
2
4
2
5
5
= + −
5
=
4
5
(ii) 𝑃 (does not participate) =
100
500
=
1
5
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
23
200
500
=
2
5
AL-IMRAN HOSH ACADEMY
8.
In class of 150 students, 50 of them study the Somali
language, 100 study the German language and 30 study both
languages. If a student is randomly selected, find the
probability that the student
a. Studies the Somali or German languages.
b. Studies only one of the languages.
Solution
𝑃 (𝐴 ) =
50
1
= ,
150 3
𝑃 (𝐵 ) =
100 1
30
1
= , and P(A ∩ B) =
=
200 2
150 5
a) 𝑃(𝐴 ∪ 𝐵 ) = 𝑃(𝐴) + 𝑃(𝐵 ) − 𝑃 (𝐴 ∩ 𝐵 )
1 2 1
+ −
3 3 5
4
𝑃(𝐴 ∪ 𝐵 ) =
5
𝑃 (𝐴 ∪ 𝐵 ) =
b) 30 of the 50 students studying Somali also study German
so of those 50, (50 − 30 = 20), study somali only
While of the 100 students studying German language
Some (100 − 30 = 70). Study German only.
So the number studying just one language is (20 + 70) = 90
And the probability of such student randomly selected from 150 is
90
3
= = 60%
150 5
9.
If 𝑃(𝐴) = 0.3 𝑃(𝐵 ) = 0.8 , and 𝑃(𝐴 ∩ 𝐵 ) = 0.15,
Calculate P(A ∪ B)
Solution
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴) + 𝑃 (𝐵 ) − 𝑃(𝐴 ∩ 𝐵)
𝑃 (𝐴 ∪ 𝐵 ) = 0.3 + 0.8 − 0.15 = 0.95
19
𝑃 (𝐴 ∪ 𝐵 ) =
20
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
10.
If the probability that Ali will fail in a mathematics test 0.5, the
probability that Nor will fail the test is 0.2 and the probability that both
of them will failtest is 0.1, what is the probability that at least one of
them will fail the test? What is the probability that both of them will
pass the test?
Solution
𝑃 (𝐴𝑙𝑖 ) = 0.5 = 50%
𝑃(𝑁𝑜𝑢𝑟) = 0.2 = 20%
𝑃 (𝐴 ∩ 𝑁) = 0.1 = 10%
Probability that at least one of them will fail the test is given by
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴) + 𝑃 (𝐵 ) − 𝑃(𝐴 ∩ 𝐵)
𝑃 (𝐴 ∪ 𝐵 ) = 0.5 + 0.2 − 0.1 = 0.6
The probability that both of them will pass =
1 − (probability that none ofthem will fail the test)
= 1 − (1 − 𝑃 (𝐴 ∪ 𝐵 ))
= 1 − (1 − 0.6)
1−
2 3
=
5 5
= 0.6
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
25
AL-IMRAN HOSH ACADEMY
11.
12.
From a set of seven cards numbered 1 to 7, two cards are drawn at
random. Find the probability that
a) The sum of the numbers is 14
b) The sum of the numbers is less than 6
c) Both numbers are odd
d) The sum of the numbers is less than 6 or both the numbers are odd.
Solution
Sample space=
(1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (2,3), (2,4), (2,5), (2,6), (2,7),
(3,4), (3,5), (3,6), (3,7), (4,5), (4,6), (4,7), (5,6), (5,7), (6,7).
a) The probability that the sum of the numbers is
b) The probability that the sum of the numbers is less than
c) The probability that both numbers are odd
d) The probability that the both numbers is less than 6 or both the
number are odd
The balls are numbered for 1 to 10 and put into a container. The balls
numbered from 1 to 5 are painted red while the balls numbered 6 to 10
aepainted blue. A ball is selected at random from the container. Find the
probability that
a) A blue ball is selected
b) The number on the ball is
c) The ball is blue or with a number less than 4.
Are the event blue is selected, and the event number is less than 4,
mutually exclusive?
Solution
Out of 10 integers a number can be chosen in 10𝐶1 = 10
𝑤𝑎𝑦𝑠
∴Total number of elementary events= 10
a) The probability that blue ball can be select is
5
10
=
1
2
b) The probability that the number is less than 4 is
c) Probability of blue ball or less than 4 is
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
26
1
2
+
3
10
3
10
4
=
5
AL-IMRAN HOSH ACADEMY
13.
A boy throws a fair coin and a regular tetrahedron with its four face
marked1, 2, 3, and 4. Find the probability that he gets a 3 on the
tetrahedron and a head on the coin.
Solution
The probability of getting head on the coin is
1
2
1
The probability of getting 3 on the tetrahedron is and
4
The probability of getting 3 on the tetrahedron and a head on the coin is
1
1
1
2
4
8
= × =
14.
Three different machines in a factory have a different probability of
breaking down during a shift as shown in table below.
Machine
Probability of breaking down
4
A
15
3
B
10
2
C
11
Find:
(a) The probability that all machines will break down during one shift.
(b) The probability that none of the matches will break down in a
particular shift.
Solution
Let M the probability of braking machines in one shift.
4
3
2
𝑃(𝑀) = 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶 ) = × ×
15
10
11
24
=
𝑑𝑖𝑣𝑖𝑑𝑒 𝑤𝑖𝑡ℎ (6)
1650
4
=
275
And let N the probability of braking machines in particular shift.
𝑃 (𝑀) = 𝑃 (𝐴 ∩ 𝐵 ∩ 𝐶 ) = 𝑃(𝐴) × 𝑃(𝐵) × 𝑃(𝐶)
= (1 − 𝑃 (𝐴)) × (1 − 𝑃 (𝐵 )) × (1 − 𝑃 (𝐶 ))
= [1 −
4
] × [1 −
15
3
10
] × [1 −
2
]
11
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
=
=
=
15.
11
×
7
15
10
693
1650
21
×
9
11
divide with (33)
50
A bag contains 10 black balls and 7 white ones. If two balls are drawn
from the bag one at a time, find the probability of drawing balls of
different colors.
a) Without replacement
b) with replacement
Solution
(a) 𝑃(black and white)=P(black)×P(white) + 𝑃(white × black)
10 7
7 10
140
=( × )+( × )=
17 16
17 16
272
(b) 𝑃(black and white)=P(black)×P(white) + 𝑃(white × black)
10 7
7 10
140
=( × )+( × )=
17 17
17 17
289
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
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AL-IMRAN HOSH ACADEMY
EXERCISE 3.7
1. A bag contains 3 red balls, 3 green balls and 1 blue ball. A ball is chosen
at random and then placed back in the bag. A second ball is chosen at
random.
Find the probability that
(a)The first ball is red
(b)
The second ball is red
(c)Both balls are red
Solution
3
4
7
7
3
4
12
7
9
49
3
7
3
7
7
49
(a) The probability that the first ball is red = × =
12
49
(b) The probability that the second ball is red = × =
(c) The probability that both balls are red = × =
1
2. The probability that Ali scores in any particular game of water polo is ,
4
(a)Find the probability that in the first two games,
(i)
He does not score in either game.
(ii)
He scores in only one of the game
(b)
The probability that Ali scores more than one goal in any
1
particular game is , Find the probability that he scores exactly one
6
goal in the third game.
Solution
1
3
4
4
a) The probability that he does not score is 1 − =
3
3
9
4
4
16
i) The probability that he does not score in either games × =
ii) The probability that he scores in only one of the game
𝑃(𝑠𝑐𝑜𝑟𝑒𝑠) × 𝑃(𝑛𝑜𝑡 𝑠𝑐𝑜𝑟𝑒𝑠) + 𝑃(𝑛𝑜𝑡 𝑠𝑐𝑜𝑟𝑒𝑠) × 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠)
1
3
3
1
3
4
4
4
4
8
= ( × )+( × ) =
b) The probability that he scores exactly one goal in the third game
= The probability that Ali scores− the probability that Ali scored
morethan one goal
1 1
1
= − =
4 6 12
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
29
AL-IMRAN HOSH ACADEMY
3. A box contains 10 sweets. Of these, 6 are red and 4 are green.
Nasra are takes a sweet at random and eats it. Fa’iza then takes
A sweet at random and eats it.
(a) Copy and complete the tree diagram shown above.
(b) Find the probability that
i) They both take green sweets,
ii) They take sweets of different colours.
Solution
(a)
Nasra’s choice in green=
4
10
4
6
3
9
9
9
Fa’iza’s choice in green sweets= , Red = , Green =
(b) The probability that
(i) They both take green sweets=
4
10
3
12
9
90
× =
6
=
2
15
4
4
6
ii) They take sweets of different colours= ( × ) + ( × )
10
9
10
9
24 24 48 24
=
+
=
=
90 90 90 45
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
30
AL-IMRAN HOSH ACADEMY
4. Survey is carried out on a group of 100 students, 60 of whom
are boys, Findthe probability that any two students, chosen at
random from the original group, will
(a)Be both girls
(b)
Both not be girls
Solution
a) The probability that an two students chosen both girls=
b) 𝑃 (both not girls) = 1 − P(both girls) = 1 −
26
165
=
40
100
139
×
39
99
165
=
26
165
3
5. The probability that a football team will win any particular game is
4
(a) Find the probability that team
(i)
Will not win the first game of the season
(ii) Will not win either of the first two games of the
season.
(b) The probability that the team will draw any particular game is
1
6
Find the probability that the team will lose the last game of the season.
Solution
Let A be the event that football team will not win the first game
Let B be the event that football team will not win either of the first
two games of the season.
a) The probability that team
3
1
i) Will not win= 1 − 𝑃 (𝑤𝑖𝑛) = 1 − =
4
4
ii) Will not win either of the first two game of the season
3 1
11
1
= 𝑃 (𝐵 ) = 1 − 𝑃 (𝑤𝑖𝑛) + 𝑃(𝑑𝑟𝑎𝑤 ) = 1 − + = 1 −
=
4 6
12 12
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
31
AL-IMRAN HOSH ACADEMY
6. When a girl wakes up each morning, the probability that she feels happy
is
9
10
(a)
i.
What is rhe probability What is the probability that she will feel
happy when she wakes up, on both Monday and Tuesday next
week, but not on both days?
ii.
What is the probability that she will feel happy when she wakes up,
on either Monday or Tuesday next week, but not on both day?
(b) Write down, but do not evaluate, an expression for the probability that
she will wake up happy on every day of the next week.
Solution
a)
(i) The probability that she will feel happy in both Monday and
Tuesday =
9
10
×
9
81
=
10
100
(ii) The probability that she feel happy on either Monday or Tuesday
next week=
=
=
9
10
9
10
9
× (1 −
×
9
100
1
10
+
+
9
100
) + (1 −
10
1
10
=
×
9
)×
10
9
10
9
10
9
50
b) The probability that she is happy in every day of the week
=
9
10
×
9
10
×
9
10
×
9
10
×
9
10
×
9
10
×
9
10
9 7
=( )
10
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
32
AL-IMRAN HOSH ACADEMY
EXERCISE 3.8
1. Given A and B are independent events such that𝑃(𝐴) = 0.25 and 𝑃(𝐵) =
0.4, Find 𝑃(𝐴/𝐵).
Solution
𝑃 (𝐴 ∩ 𝐵 ) = 𝑃 (𝐴) × 𝑃(𝐵 ) = 0.4 × 0.25 = 0.1
𝐴
𝑃(𝐴∩𝐵)
𝐵
𝑃(𝐵)
𝑃( ) =
=
0.1
0.4
= 0.25
1
2. The probability that Ahmed passed a statistics test is and the probability
5
2
that he passes a computer test is . If the probability that Ahmed passes
1
3
both the tests is . Find the probability that he passes the computer test.
10
Solution
1
𝑃 (𝐴) =
5
2
𝑃 (𝐵 ) =
3
1
𝑃(𝐴 ∩ 𝐵 ) =
10
1
𝑃(𝐴 ∩ 𝐵 ) =
10
1
𝐴
𝑃(𝐴 ∩ 𝐵) 10
1 3
3
𝑃( ) =
=
=
× =
2
𝐵
𝑃(𝐵)
10 2 20
3
1
2
4
3
3. Given X and Y are two events such that 𝑃 (𝑋) = , 𝑃(𝑌) =
1
and 𝑃(𝑋 ∩ 𝑌) = .
2
Solution
Find P(X/Y).
1
𝑃 (𝑋 ∩ 𝑌) 2 1
𝑃(𝑋/𝑌) =
= = ×4 = 2
1 2
𝑃 (𝑌)
4
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
33
AL-IMRAN HOSH ACADEMY
4. A firm has 500 workers. As many as 80% of them play
football, 75% playhockey, 50% rugby and 36% play all the
three games. 35 of workers play only football and 32 of them
play only hockey while 203 of them play bothhockey and
rugby. Show all the information clearly in a Venn diagram. If a
worker is selected at random, what is the probability that
he/she also plays hockey?
Solution
Vein diagram
80
(a)
100
75
(b)
(c)
× 500 = 400 play football
50
100
100
× 500 = 375 play hockey
× 500 = 250 play rugby
36
(d)
× 500 = 180 play all the tree games
100
(e) 35 plays football only
(f) 32 plays only hockey
(g) 203 plays both hockey and rugby
𝑃(𝐴) = the probability that a selected person play football
400
4
=
500
5
𝑃(𝐴) = the probability that a selected person play only hockey
375
3
=
500
4
4
3
3
5
4
5
𝑃 ( 𝐴 ∩ 𝐵 ) = 𝑃 (𝐴 ) × 𝑃 (𝐵 ) = × =
𝑃(𝐴/𝐵) = Probability that he/she plays football given that he/she also
plays hocky
𝐴
𝑃(𝐴 ∩ 𝐵) 3 3 4
𝑃( ) =
= ÷ = = 0.8
𝐵
𝑃(𝐵)
5 4 5
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
34
AL-IMRAN HOSH ACADEMY
1
2
4
3
5. If 𝑃 (𝐷 ) = , 𝑃 (𝐸 ) =
1
and 𝑃(𝐷/𝐸) = , find
2
a) 𝑃(𝐷 ∩ 𝐸)
b) 𝑃(𝐷 ∩ 𝐸̅ )
c) 𝑃(𝐷/𝐸̅ )
Solution
1
2
1
4
1
3
6
(a) 𝑃 (𝐷 ∩ 𝐸 ) = 𝑃 (𝐷 ) × 𝑃 (𝐸 ) = × =
2
1
1
1
3
4
3
12
(b) 𝑃 (𝐷 ∩ 𝐸̅ ) = 𝑃 (𝐷 ) × 𝑃 (𝐸̅ ) = × (1 − ) = × =
4
𝐷
𝑃(𝐷∩𝐸̅ )
𝐸
𝑃(𝐸)
(c) 𝑃 ( ̅ ) =
=
1
12
1
1
3
12
÷ =
×3 =
3
12
=
1
4
6. From 500 students in a school, 200 of them play football(𝑆), 400 play
basketball (𝐾) and 200 play both the games. Find the probability that a
student randomly selected, plays football if it known that he/she plays
basketball.
Solution
200 2
400 4
200 2
𝑃 (𝑆) =
=
𝑃(𝐾 ) =
= ,
𝑃 (𝑆 ∩ 𝐾 ) =
=
500 5
500 5
500 5
The probability that a student randomly selected, plays football if it is
known that he/she plays basket ball
𝑃(𝑆/𝐾) =
𝑃(𝑆∩𝐾)
𝑃(𝐾)
2
4
2
5
10
5
5
5
4
20
= ÷ = × =
=
1
2
Barre: Maxamed Yuusuf Xasan (MY) (0615778479/0613780005)
35