NCERT Solutions Class 12 Mathematics Chapter 13 Probability

Exercise 13.1

1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E∩F) = 0.2, find P (E|F) and P(F|E).
Solution
It is given that P(E) = 0.6, P(F) = 0.3, and P(E⋂F) 0.2 

2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
Solution
It is given that P(B) = 0.5 and P(A ∩ B) = 0.32

3. If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(A ∪ B)

Solution

It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4
(i) P (B|A) = 0.4 

4. Evaluate P (A ∪ B), if 2P (A) = P (B) =5/13 and P(A|B) =2/5

Solution

It is given that, 2P(A) = P(B) = 5/13 
⇒ P(A) = 5/26 and P(B) = 5/13 
P(A/B) = 2/5 

5. If P(A) = 6/11, P(B) = 5/11 and and P(A ∪ B) = 7/11 find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(B|A) 
Solution
It is given that P(A) = 6/11, P(B) = 5/11, and P(A ∪ B) = 7/11 
(i) P(A ∪ B) = 7/11 
 ∴ P(A) + P(B) - P(A ∩ B) = 7/11 

6. A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads, F: at most two heads
(iii) E: at most two tails, F: at least one tail 
Solution
If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
E ∩ F = {HHH}

7. Two coins are tossed once, where
(i) E: tail appears on one coin, F: one coin shows head
(ii) E: not tail appears, F: no head appears 
Solution
If two coins are tossed once, then the sample space S is 
S = { HH, HT, TH, TT} 
(i) E = { HT, TH} 
F = { HT, TH}

8. A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses.
Solution
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216

9. Mother, father and son line up at random for a family picture
E: son on one end, F: father in middle
Solution
If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
∴ E ∩ F = {MFS, SFM}

10. A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.
Let
A: Obtaining a sum greater than 9
= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5.
= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).
PA|B = PA∩BPB = 236636 = 26 = 13
(b) E: Sum of the observations is 8.
= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F: Red die resulted in a number less than 4.

11. A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find
(i) P (E|F) and P (F|E)
(ii) P (E|G) and P (G|E)
(ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)

Solution

When a fair die is rolled, the sample space S will be
S = {1, 2, 3, 4, 5, 6}
It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}
∴ P(E) = 3/6 = 1/2 
P(F) = 2/6 = 1/3 
P(G) = 4/6 = 2/3 
(i) E ∩ F = {3} 

(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}
E ∩ F = {3}
(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is a girl?

Solution 

Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let A be the event that both children are girls.
∴ A = {(g, g)} 
(i) Let B be the event that the youngest child is a girl. 
∴ B = [(b, g), (g, g)] 

Therefore, the required probability is 1/2. 
(ii) Let C be the event that at least one child is a girl. 
∴ C = {(b, g), (g, b), (g, g)} 
⇒ A ∩ C = { g, g} 
⇒ P(C) = 3/4 
P(A∩C) = 1/4 
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A/C). 

3. An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution
The given data can be tabulated as

	True/False	Multiple choice	Total
Easy	300	500	800
Difficult	200	400	600
Total	500	900	1400

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900
Therefore, probability of selecting an easy multiple choice question is
P (E ∩ M) = 500/1400 = 5/14
Probability of selecting a multiple choice question, P (M), is 900/1400 = 9/14
P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

14. Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Solution

When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different. ∴ A = {(1, 3), (2, 2), (3, 1)}
B=1, 21, 31, 41, 51, 62, 12, 32, 42, 52, 63, 13, 23, 43, 53, 64, 14, 24, 34, 54, 65, 15, 25, 35, 45,66, 16, 26, 36, 46, 5 A∩B = 1, 3, 3, 1
∴ P(B) = 3036 =56 and
PA∩B = 236 = 118Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Solution

The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is, 

Let A be the event that the coin shows a tail and B be the event that at least one die show 3. 
∴ A = { (1,T), (2, T), (4, T) , (5, T)}
B = {(3, 1), (3, 2), (3, 3), (3, 4), 
⇒ A ∩ B = ∅

16. If P(A) = 1/2, P(B) = 0, then P(A|B) is
(A) 0
(B) 1/2
(C) not defined
(D) 1
Solution
It is given that P(A) = 1/2 and P(B) = 0 

17. If P(A) = 1/2, P(B) = 0, then P(A|B) is
(A) 0
(B) 1/2
(C) not defined
(D) 1

Solution

It is given that, P(A|B) = P(B|A) 

Exercise 13.2

1. If P(A) = 3/5 and P(B) = 1/5, find P(A ∩ B) if A and B are independent events. 
Solution
It is given that P(A) = 3/5 and P(B) = 1/5 
A and B are independent events. Therefore, 
P(A ∩ B) = P(A) . P(B) = 3/5. 1/5 = 3/25

2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution
There are 26 black cards in a deck of 52 cards.
Let P (A) be the probability of getting a black card in the first draw. 
∴ P(A) = 26/52 = 1/2 
Let P(B) be the probability of getting a black card on the second draw. 
Since the card is not replaced, 
∴ P(B) = 25/51 
Thus, probability of getting both the cards black = (1/2) × (25/51) = 25/102

3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale
Solution
Let A, B, and C be the respective events that the first, second, and third drawn orange is good.
Therefore, probability that first drawn orange is good, P (A) = 12/15
The oranges are not replaced.
Therefore, probability of getting second orange good, P (B) = 11/14
Similarly, probability of getting third orange good, P(C) = 10/13
The box is approved for sale, if all the three oranges are good.
Thus, probability of getting all the oranges good = 12/15 × 11/14 × 10/13 = 44/91
Therefore, the probability that the box is approved for sale is = 44/91.

4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Solution
If a fair coin and an unbiased die are tossed, then the sample space S is given by,  

5. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event "number obtained is even" and B be the event "number obtained is red". Find if A and B are independent events.
Solution
When a die is thrown, the sample space (S) is
S = {1, 2, 3, 4, 5, 6}
Let A: the number is even = {2, 4, 6} 
⇒ P(A) = 3/6 = 1/2 
B : the number is red = {1, 2, 3}
⇒ PB =36=12∴ A ∩ B = {2} 

6. Let E and F be events with P(E) = 3/5, P(F) = 3/10 and P(E∩ F ) = 1/5. Are E and F independent ? 
Solution
It is given that P(E) = 3/5, P(F) = 3/10 and P(EF) = P(E∩ F) = 1/5 

7. Given that the events A and B are such that P(A) = 12, PA∪B=35 and P (B) = p. Find p if they are (i) mutually exclusive (ii) independent.
Solution
It is given that P(A) = 1/2, P(A∪B) = 3/5, and P(B) = p
(i) When A and B are mutually exclusive, A∩B = Ø 
P(A∩B) = 0
It is known that, P(A∪B) = P(A) + P(B) - P(A∩B) 

8. Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
(i) P (A ∩ B)
(ii) P (A ∪ B)
(iii) P (A | B)
(iv) P (B | A)
Solution
It is given that P (A) = 0.3 and P (B) = 0.4
(i) If A and B are independent events, then
P(A ∩ B) = P(A).P(B) = 0.3 × 0.4 = 0.12
(ii) It is known that, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)  
⇒ P(A ∪ B) = 0.3 + 0.4 - 0.12 = 0.58
(iii) It is known that, P(A|B) = P(A ∩ B)/P(B) 
⇒ P(A|B) = 0.12/0.4 = 0.3
(iv) It is known that, P(B|A) = P(A ∩ B)/P(A)  
⇒ P(B|A) = 0.12/0.3 = 0.4

9. If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and and P(A ∩ B) = 1/8, find P(not A and not B) 
Solution

10. Events A and B are such that P(A) = 1/2, P(B) = 7/12 and P(not A or not B) = 1/4 . State whether A and B are independent ? 
Solution
It is given that P(A) = 1/2, P(B) = 7/12, and P(not A or not B) = 1/4 

11. Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find
(i) P (A and B)
(ii) P (A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Solution
It is given that P(A) = 0.3 and P(B) = 0.6 
Also, A and B are independent events,  
(i) ∴ P(A and B) = P(A). P(B) 
⇒ P(A  ∩ B) = 0.3 × 0.6 = 0.18

(ii) P(A and not B) = P(A ∩ B')
= P(A) - P(A ∩ B)
= 0.3 - 0.18 
= 0.12

(iii) P( A or B) = P(A ∪ B) 
= P(A) + P(B)  - P(A∩B) 
= 0.3 + 0.6 - 0.18 
= 0.72

(iv) P (neither A nor B) = P(A' ∩ B')
= P[(A ∪ B)'] 
= 1 - P(A ∪ B)
= 1 - 0.72 
= 0.28

12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution
Probability of getting an odd number in a single throw of a die = 3/6 = 1/2
Similarly, probability of getting an even number = 3/6 = 1/2
Probability of getting an even number three times = 1/2 × 1/2 × 1/2 = 1/8
Therefore, probability of getting an odd number at least once
= 1 − Probability of getting an odd number in none of the throws
= 1 − Probability of getting an even number thrice
= 1 – 1/8
= 7/8

13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red. 
Solution
Total number of balls = 18 
Number of red balls = 8 
Number of black balls = 10 
(i) Probability of getting a red ball in the first draw  = 8/18 = 4/9 
The ball is replaced after the first draw. 
∴ Probability of getting a red ball in the second draw = 8/18 = 4/9 
Therefore, probability of getting both the balls red = 4/9 × 4/9 = 16/81

(ii) Probability of getting first ball black  = 10/8 = 5/9 
The ball is replaced after the first draw. 
Probability of getting second ball as red. = 8/18 = 4/9
Therefore, probability of getting first ball as black and second ball as red = 5/9 × 4/9 = 20/81

(iii) Probability of getting first ball as red = 8/18 = 4/9
The ball is replaced after the first draw.
Probability of getting second ball as black = 10/18 = 5/9
Therefore, probability of getting first ball as black and second ball as red = 4/9 × 5/9 = 20/81
Therefore, probability that one of them is black and other is red
= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black
= 20/81 + 20/81 = 40/81

14. Probability of solving specific problem independently by A and B are and 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem
Solution 
Probability of solving the problem by A, P(A) = 1/2 
Probability of solving the problem by B, P(B) = 1/3
Since the problem is solved independently by A and B, 

15. One card is drawn at random from a well-shuffled deck of 52 cards. In which of the following case is the events E and F independent?
(i) E : ‘the card drawn is a spade’
F : ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’
Solution
(i) In a deck of  52 cards, 13 cards are spades and 4 cards are aces. 
∴ P(E) = P(the card drawn is a spade) = 13/52 = 1/4 
∴ P(F) = P(the card drawn is an ace) = 4/52 = 1/13 
In the deck of cards, only 1 card is an ace of spades. 
P(EF) = P(the card drawn is spade and an ace) = 1/52 
P(E) × P(F) = 1/4 ×  1/13 = 1/52 = P(EF)
⇒ P(E) × P(F) = P(EF)
Therefore, the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.
∴ P(E) = P(the card drawn is black) = 26/52 = 1/2
∴ P(F) = P(the card drawn is a king) = 4/52 = 1/13
In the pack of 52 cards, 2 cards are black as well as kings.
∴ P (EF) = P(the card drawn is a black king) = 2/52 = 1/26
P(E) × P(F) = 1/2 × 1/13 = 1/26 = P(EF)
Therefore, the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.
∴ P(E) = P(the card drawn is a king or a queen) = 8/52 = 2/13
∴ P(F) = P(the card drawn is a queen or a jack) = 8/52 = 2/13
There are 4 cards which are king or queen and queen or jack.
∴ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)

= 4/52 = 1/13 
P(E) × P(F) = 2/13. 2/13 = 4/169 ≠ 1/13 
⇒ P(E). P(F) ≠ P(EF) 
Therefore, the given events E and F are not independent.

16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English news papers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.
Solution
Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper. 
It is given that, 

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0
(B) 1/3
(C) 1/12
(D) 1/36
Solution 
When two dice are rolled, the number of outcomes is 36.
The only even prime number is 2.
Let E be the event of getting an even prime number on each die. 
∴ E = {(2, 2)} 
⇒ P(E) = 1/36
Therefore, the correct answer is D.

18. Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A'B') = [1 - P(A)][1-P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1
Solution 
Two events A and B are said to be independent, if p(AB) = P(A) × P(B) 
Consider the result given in alternative B. 

This implies that A and B are independent, if P(A'B') = [1 - P(A)] [1 - P(B)] 
Distracter Rationale 
A. Let P(A) = m, P(B) = n, 0 < m, n < 1
A and B are mutually exclusive. 

D. From the above example, it can be seen that, 
P(A) + P(B) = 1/2 + 1/2 = 1 
However, it cannot be inferred that A and B are independent. 
Thus, the correct answer is B.

Exercise 13.3

1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution
The urn contains 5 red and 5 black balls.
Let a red ball be drawn in the first attempt.
P (drawing a red ball) = 5/10 = 1/2
If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.
P (drawing a red ball) = 7/12
Let a black ball be drawn in the first attempt.
P (drawing a black ball in the first attempt) = 5/10 = 1/2
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
P (drawing a red ball) = 5/12
Therefore, probability of drawing second ball as red is

2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution
Let E₁ and E₂ be the events of selecting first bag and second bag respectively.
∴ P(E₁) = P(E₂) = 1/2 
Let A be the event of getting a red ball. 
⇒ P(A|E₁) = P(drawing a red ball from first bag) = 4/8 = 1/2 
⇒ P(A|E₂ ) = P (drawing a red ball from second bag) = 2/8 = 1/4 
The probability of drawing a ball from the first bag, given that is is red, is given by P (E₂|A) .
By using Bayes' theorem, we obtain 

3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?
Solution
Let E₁ and E₂ be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.

P(A|E₁) = P(student getting an A grade is a hostler) = 30% = 0.3 
P(A|E₂) = P(student getting an A grade is a day scholar) = 20% = 0.2 
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P (E₁ |A) .
By using Bayes' theorem, we obtain  

4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4 What is the probability that the student knows the answer given that he answered it correctly?
Solution 
Let E₁ and E₂ be the respective events that the student knows the answer and he guesses the answer.
Let A be the event that the answer is correct.
∴ P(E₁) =  3/4 
 P(E₂ ) = 1/4 
The probability that the student answered correctly, given that he knows the answer , is 1. 
∴ P (A|E₁ )  = 1
Probability that the student answered correctly, given that he guessed, is 1/4.
∴ P (A|E₂ ) = 1/4 
The probability that the student knows the answer, given that he answered it correctly, is given by P (E₁ |A). 
By using Bayes' theorem, we obtain

5. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution
Let E₁ and E₂ be the respective events that a person has a disease and a person has no disease.
Since E₁ and E₂ are events complimentary to each other,
∴ P (E₁) + P (E₂) = 1
⇒ P (E₂) = 1 − P (E₁) = 1 − 0.001 = 0.999
Let A be the event that the blood test result is positive.
P(E₁) = 0.1% = 0.1/100 = 0.001 
P(A|E₁) = P(result is positive given the person has disease) = 90% = 0.99 P(A|E₂) = P(result is positive given that the person has no disease) = 0.5% = 0.005
Probability that a person has a disease, given that his test result is positive, is given by
P (E₁|A).
By using Bayes’ theorem, we obtain 

6. There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Solution
Let E₁, E₂, and E₃ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
∴ P (E₁) = P (E₂) = P (E₃) = 1/3
Let A be the event that the coin shows heads.
A two-headed coin will always show heads.
∴ P (A|E₁ )  = P(coin showing heads, given that it is a two - headed coin) = 1 
Probability of heads coming up, given that it is a two - headed coin) = 1 
Probability of heads coming up, given that it is a biased coin = 75% 
∴ P (A|E₂)  = P(coin showing heads, given that it is biased coin) = 75/100 = 3/4 
Since the third coin is unbiased, the probability that it shows heads is always 1/2. 
∴ P (A|E₃)  = P(coin showing heads, given that it is biased coin) = 1/2 
The probability that the coin is two - headed, given that it shows heads, is given by
 P (E₁ |A). 
By using Bayes' theorem, we obtain  

7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution
Let E₁, E₂, and E₃ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.
Total number of drivers = 2000 + 4000 + 6000 = 12000
P (E₁) = P (driver is a scooter driver) = 2000/12000 = 1/6
P (E₂) = P (driver is a car driver) = 4000/12000 = 1/3
P (E₃) = P (driver is a truck driver) = 6000/12000 = 1/2 
P (A|E₁)= P(scooter driver met with an accident) = 0.01 = 1/100
P (A|E₂)= P (car driver met with an accident) = 0.03 = 3/100 
P (A|E₃)= P (truck driver met with an accident) = 0.15 = 15/100 
The probability that the driver is a scooter driver, given that he met with an accident, is given by P (E₁ |A). 
By using Bayes' theorem, we obtain 

8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?

Solution

Let E₁ and E₂ be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.
∴ Probability of items produced by machine A, P (E₁) = 60% = 3/5
Probability of items produced by machine B, P (E₂) = 40% = 2/5
Probability that machine A produced defective items, P (X|E₁) = 2% = 2/100
Probability that machine B produced defective items, P (X|E₂) = 1% = 1/100
The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E₂|X).
By using Bayes’ theorem, we obtain

9. Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution

Let E₁ and E₂ be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.
P (E₁) = Probability that the first group wins the competition = 0.6
P (E₂) = Probability that the second group wins the competition = 0.4
P (A|E₁) = Probability of introducing a new product if the first group wins = 0.7
P (A|E₂) = Probability of introducing a new product if the second group wins = 0.3
The probability that the new product is introduced by the second group is given by
P (E₂|A).
By using Bayes’ theorem, we obtain

10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Solution

Let E₁ be the event that the outcome on the die is 5 or 6 and E₂ be the event that the outcome on the die is 1, 2, 3, or 4.
∴ P(E₁) = 2/6 = 1/3 and P(E₂) = 4/6 = 2/3
Let A be the event of getting exactly one head.
P (A|E₁) = Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8
P (A|E₂) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 = 1/2
The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E₂|A).
By using Bayes’ theorem, we obtain 

11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?

Solution

Let E₁, E₂, and E₃ be the respective events of the time consumed by machines A, B, and C for the job.

12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Solution

 

Let E₁ and E₂ be the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card.
Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.
∴ P(E₁) = 13/52 = 1/4
P(E₂) = 39/52 = 3/4
When one diamond card is lost, there are 12 diamond cards out of 51 cards.
Two cards can be drawn out of 12 diamond cards in ¹²C₂ ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in ⁵¹C₂ ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E₁).

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.
Two cards can be drawn out of 13 diamond cards in ¹³C₂ ways whereas 2 cards can be drawn out of
51 cards in ⁵¹C₂ ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by P (A|E₂).

13. Probability that A speaks truth is 4/5 . A coin is tossed. A reports that a head appears. The probability that actually there was head is
A. 4/5
B. 1/2
C. 1/5
D. 2/5 
Solution
Let E₁ and E₂ be the events such that
E₁: A speaks truth
E₂: A speaks false
Let X be the event that a head appears.
P(E₁ ) = 45Therefore,
P(E₂) = 1-P(E₁)=1-45=15If a coin is tossed, then it may result in either head (H) or tail (T).
The probability of getting a head is 1/2 whether A speaks truth or not.
∴ P(X|E₁) = P(X|E₂ ) = 1/2
The probability that there is actually a head is given by P (E₁|X).

14. If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is correct?
(A). P(A|B) = P(B)/P(A) 
(B). P(A|B) < P(A) 
(C) P(A|B) ≥ P(A) 
(D) None of these 
Solution
If A ⊂ B, then A ∩ B = A
⇒ P (A ∩ B) = P (A)
Also, P (A) < P (B)

From (2), we obtain  
⇒ P(A|B) ≥ P(A) ...(3)
∴ P(A|B) is not less than P(A).
Thus, from (3), it can be concluded that the relation given in alternative C is correct.

Exercise 13.4

1. State which of the following are notthe probability distributions of a random variable. Give reasons for your answer.

(i)

X	0	1	2
P (X)	0.4	0.4	0.2

(ii)

X	0	1	2	3	4
P (X)	0.1	0.5	0.2	− 0.1	0.3

(iii)

Y	−1	0	1
P (Y)	0.6	0.1	0.2

(iv)

Z	3	2	1	0	−1
P (Z)	0.3	0.2	0.4	0.1	0.05

Solution
It is known that the sum of all the probabilities in a probability distribution is one.
(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1
Therefore, the given table is a probability distribution of random variables.
(ii) It can be seen that for X = 3, P (X) = −0.1
It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.
(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Therefore, the given table is not a probability distribution of random variables.
(iv) Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 ≠ 1
Therefore, the given table is not a probability distribution of random variables.

2. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. What are the possible values of X? Is X a random variable?
Solution
The two balls selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball.
X represents the number of black balls.
∴ X (BB) = 2
X (BR) = 1
X (RB) = 1
X (RR) = 0
Therefore, the possible values of X are 0, 1, and 2.
Yes, X is a random variable.

3. Let X represent the difference between the number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution
A coin is tossed six times and X represents the difference between the number of heads and the number of tails.
∴ X (6 H, 0T) = |6 – 0| = 6
X (5 H, 1 T) = |5 – 1| = 4
X (4 H, 2 T) = |4 – 2| = 2
X (3 H, 3 T) = |3 – 3| = 0
X (2 H, 4 T)= |2 – 4| = 2
X (1 H, 5 T) = |1 – 5| = 4
X (0H, 6 T) = |0 – 6| = 6
Thus, the possible values of X are 6, 4, 2, and 0.

4. Find the probability distribution of
(i) number of heads in two tosses of a coin
(ii) number of tails in the simultaneous tosses of three coins
(iii) number of heads in four tosses of a coin
Solution
(i) When one coin is tossed twice, the sample space is
{HH, HT, TH, TT}
Let X represent the number of heads.
∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0
Therefore, X can take the value of 0, 1, or 2.
It is known that,
P(HH) = P(HT) = P(TH) = P(TT) = 1/4
P (X = 0) = P (TT) = 1/4
P (X = 1) = P (HT) + P (TH)= 1/4 + 1/4 = 1/2
P (X = 2) = P (HH) = 1/4
Thus, the required probability distribution is as follows.

X	0	1	2
P (X)	1/4	1/2	1/4

(ii) When three coins are tossed simultaneously, the sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let X represent the number of tails.
It can be seen that X can take the value of 0, 1, 2, or 3.
P (X = 0) = P (HHH) = 1/8
P (X = 1) = P (HHT) + P (HTH) + P (THH) = 1/8 + 1/8 + 1/8 = 3/8
P (X = 2) = P (HTT) + P (THT) + P (TTH) = 1/8 + 1/8 + 1/8 = 3/8
P (X = 3) = P (TTT) = 1/8
Thus, the probability distribution is as follows.

X	0	1	2	3
P (X)	1/8	3/8	3/8	1/8

(iii) When a coin is tossed four times, the sample space is

Let X be the random variable, which represents the number of heads.
It can be seen that X can take the value of 0, 1, 2, 3, or 4.
P (X = 0) = P (TTTT) = 1/16
P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)
= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4
P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT) + P (THTH)
= 1/16 + 1/16 + 1/16 + 1/16 + 1/16 +1/16 = 6/16 = 3/8
P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)
= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4
P (X = 4) = P (HHHH) = 1/16
Thus, the probability distribution is as follows.

X	0	1	2	3	4
P (X)	1/16	1/4	3/8	1/4	1/16

5. Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Solution 
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
(i) Here, success refers to the number greater than 4.
P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 × 4/6 = 4/9
P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss) = 4/6 × 2/6 + 4/6 × 2/6 = 4/9
P (X = 2) = P (number greater than 4 on both the tosses)
= 2/6 × 2/6 = 1/9
Thus, the probability distribution is as follows.

X	0	1	2
P (X)	4/9	4/9	1/9

(ii) Here, success means six appears on at least one die.
P (Y = 0 ) = P (six appears on none of the dice) =
56 × 56 = 2536P (Y = 1) = P (six appears on at least one of the dice) =
16 × 56 + 56 × 16 +16 × 16 = 1136
Thus, the required probability distribution is as follows.

Y	0	1
P (Y)	2536	1136

6. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution 
It is given that out of 30 bulbs, 6 are defective.
⇒ Number of non-defective bulbs = 30 − 6 = 24
4 bulbs are drawn from the lot with replacement.
Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

Therefore, the required probability distribution is as follows.

X	0	1	2	3	4
P (X)	256/625	256/625	96/625	16/625	1/625

7. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Solution 
Let the probability of getting a tail in the biased coin be x.
∴ P (T) = x
⇒ P (H) = 3x
For a biased coin, P (T) + P (H) = 1
⇒ x + 3x = 1 
⇒ 4x = 1 
⇒ x = 1/4 
∴ P(T) = 1/4 and P(H) = 3/4 
When the coin is tossed twice, the sample space is { HH, TT, HT, TH}. 
Let X be the random variable representing the number of tails.  
∴ P(X = 0) = P (no tail) = P(H) × P(H) = 3/4 × 3/4 = 9/16 
P(X = 1) = P (one tail) = P(HT) + P(TH) 

P(X = 2) = P (two tails) = P(TT) = 1/4 × 1/4 = 1/16
Therefore, the required probability distribution is as follows.

X	0	1	2
P (X)	9/16	3/8	1/16

8. A random variable X has the following probability distribution.

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Determine
(i) k
(ii) P (X < 3)
(iii) P (X > 6)
(iv) P (0 < X < 3)
Solution
(i) It is known that the sum of probabilities of a probability distribution of random variables is one. 
∴ 0 + k + 2k + 2k + 3k + k² + 2k² + (7k² + k) = 1
⇒ 10k² + 9k - 1 = 0 
⇒ (10k -1)(k + 1) = 0 
⇒ k = -1, 1/10 
k = -1 is not possible as the probability of an event is never negative.  
⇒ k = 1/10

(ii) P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2) 
= 0 + k + 2k 
= 3k 
= 3 × 1/10
= 3/10

(iii) P(x > 6) = P(x = 7)
7k² + k 

(iv) P(0 < x < 3) = P(x = 1) + P(X = 2) 
= k + 2k 
= 3k 
= 3 × 1/10 
= 3/10

9. The random variable X has probability distribution P(X) of the following form, where k is some number:

(i) Determine the value of 'k'. 
(ii) Find P(X < 2), P(X ≥ 2), P(X ≤ 2). 
Solution
(i) It is known that the sum of probabilities of a probability distribution of random variables is one.
∴ k + 2k + 3k + 0 = 1
⇒ 6k = 1
⇒ k = 1/6

(ii) P(X < 2) = P(X = 0) + P(X = 1)
= k + 2k 
= 3k 
= 3/6
= 1/2
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) 
= k + 2k + 3k 
= 6k 
= 6/6 
= 1 
P(X ≥ 2) = P(X = 2) + P(X > 2) 
= 3k + 0
= 3k 
= 3/6
= 1/2

10. Find the mean number of heads in three tosses of a fair coin.
Solution
Let X denote the success of getting heads.
Therefore, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that X can take the value of 0, 1, 2, or 3.

∴ P(X = 3) = P(HHH) 
=1/2 × 1/2 × 1/2 
= 1/8 
Therefore, the required probability distribution is as follows.  

X	0	1	2	3
P(X)	1/8	3/8	3/8	1/8

Mean of X E(X), µ = Σ X_i P(X_i )

11. Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X. 
Solution
Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.
∴ P (X = 0) = P (not getting six on any of the dice) = 25/36
P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)
= 2 (1/6 × 5/6) = 10/36
P (X = 2) = P (six on both the dice) = 1/36
Therefore, the required probability distribution is as follows.

X	0	1	2
P(X)	25/36	10/36	1/36

Then, expectation of X = E(X) = Σ Xi P(Xi )
= 0 × 25/36 + 1 × 10/36 + 2 × 1/36 
= 1/3

12. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denotes the larger of the two numbers obtained. Find E(X).

Solution

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways
X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.
For X = 2, the possible observations are (1, 2) and (2, 1).
∴ P(X = 2) = 2/30 = 1/15
For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).
∴ P(X = 3) = 4/30 = 2/15
For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).
∴ P(X = 4) = 6/30 = 1/5
For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).
∴ P(X = 5) = 8/30 = 4/15
For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6,5) , (6, 4), (6, 3), (6, 2), and (6, 1).
∴ P(X = 6) = 10/30 = 1/3
Therefore, the required probability distribution is as follows.

X	2	3	4	5	6
P(X)	1/15	2/15	1/5	4/15	1/3

13. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Solution
When two fair dice are rolled, 6 × 6 = 36 observations are obtained.
P(X = 2) = P(1, 1) = 1/36
P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18
P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12
P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9
P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36
P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6
P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9
P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12
P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18
P(X = 12) = P(6, 6) = 1/36
Therefore, the required probability distribution is as follows.

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	1/36	1/18	1/12	1/9	5/36	1/6	5/36	1/9	1/12	1/18	1/36

Then, E(X) = Σ X_i P(X_i )

14. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.
Solution
There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.
The given information can be compiled in the frequency table as follows.

X	14	15	16	17	18	19	20	21
f	2	1	2	3	1	2	3	1

P(X = 14) = 2/15, P(X = 15) = 1/15, P(X = 16) = 2/15, P(X = 16) = 3/15,
P(X = 18) = 1/15, P(X = 19) = 2/15, P(X = 20) = 3/15, P(X = 21) = 1/15
Therefore, the probability distribution of random variable X is as follows.

X	14	15	16	17	18	19	20	21
f	2/15	1/15	2/15	3/15	1/15	2/15	3/15	1/15

Then, mean of X = E(X)

15. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).
Solution
It is given that P(X = 0) = 30% = 30/100 = 0.3 
P(X = 1) = 70% = 70/100 = 0.7
Therefore, the probability distribution is as follows.

X	0	1
P(X)	0.3	0.7

Then, E(X) = Σ X_i P(X_i)
= 0 × 0.3 + 1 × 0.7 
= 0.7
E(X²) = ΣX²_i P (X_i )
= 0² × 0.3 + (1)² × 0.7 
= 0.7 
It is known that, Var (X) = E(X²) - [E(X)]² 
= 0.7 - (0.7)² 
= 0.7 - 0.49 
= 0.21

16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 5
(D) 8/3
Solution
Let X be the random variable representing a number on the die. 
The total number of observations is six. 
∴ P(X = 1) = 3/6 = 1/2 
P(X = 2) = 2/6 = 1/3 
P(X = 5) = 1/6
Therefore, the probability distribution is as follows.

X	1	2	5
P(X)	1/2	1/3	1/6

Mean = E(X) = Σ p_i  x_i 

17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
(A) 37/221
(B) 5/13
(C) 1/13 
(D) 2/13 
Solution
Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.
In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.