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NCERT Class 12 Mathematics Chapter 13 Probability
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Probability
Chapter – 13
| 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 (EIF) and P (FIE).
Ans:
2. Compute P(A/B), if P(B) = 0.5 and P(A ∩ B) = 0.32
Ans:
3. If P(A) = 0.8 P(B) = 0.5 and P (BIA) = 0.4, find
(i) P(A ∩ B)
Ans:
(ii) P(A/B)
Ans:
(iii) P(A U B)
Ans: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒P(A ∪ B)= 0.8 + 0.5 – 0.32
= 0.98
4. Evaluate P(A U B), if 2P(A) = P(B) = 5/13 and P(A/B) = 2/5
Ans:
It is known that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
5. If P(A) = 6/11 , P(B) = 5/11 and P(A U B) = 7/11, find
(i) P(A ∩ B)
Ans:
(ii) P(A/B)
Ans:
(iii) P(BIA)
Ans:
Determine P(E/F) in Exercises 6 to 9
6. A coin is tossed three times, where
(i) E : head on third toss, F: heads on first two tosses.
Ans: E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴ E ∩ F = {HHH}
(ii) E : at least two heads, F: at most two heads.
Ans: E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ E ∩ F = {HHT, HTH, THH}
Clearly, P(E ∩ F) = ⅜ and P(F) = ⅞
(iii) E : at most two tails, F: at least one tail.
Ans: E = {HHH, HHT, HTH, HTT, THH, THT, TTH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ E ∩ F = {HHT, HTH, THH, HTT, THT, TTH}
7. Two coins are tossed once, where
(i) E : tail appears on one coin, F : one – coin shows head
Ans: E = {HT, TH}
F = {HT, TH}
∴ E ∩ F = {HT, TH}
P(F) = 2/4 = 1/2
P(E ∩ F)= 2/4 = 1/2
(ii) E : not tail appears, F : no head appers.
Ans: E = {HH}
F = {TT}
∴ E ∩ F = ⍉
P(F) = 1 and P(E ∩ F) = 0
8. A die is thrown three times,
E: 4 appears on the third toss.
F: 6 and 5 appears respectively on first two tosses.
Ans: 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
Ans: Let 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.
Ans: 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)
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Ans: 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(EIF) and P(FIE)
Ans: E ∩ F = {3}
∴ P(E ∩ F) = 1/6
(ii) P(EIG) and P (GIE)
Ans: E ∩ G = {3, 5}
∴ P(E ∩ G) = 2/6 = 1/3
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)
Ans: 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,
Ans: Let E’ be the event that the youngest child is a girl.
∴ E’ = {(B,G), (G,G)}
⇒ E ∩ E’ = {(G,G)}
∴ P(E’) = 2/4 = 1/2
P(E ∩ E’) = 1/4
The conditional probability that both are girls, given that the youngest child is a girl, is given by P(E/E’).
Thus, the required probability is 1/2
(ii) at least one is a girl?
Ans: Let A be the event that at least one child is a girl.
∴ A = {(B,G), (G,B), (G,G)}
⇒ E ∩ A = {G,G}
⇒ P(A)= 3/4
P(E ∩ A) = 1/4
The conditional probability that both are girls, given that at least one child is a girl, is given by P(E|A).
Thus,
13. 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?
Ans: The given data can be represented 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
Thus, 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.
Thus, the required probability is 5/9
14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Ans: Number of observations in the sample space S = 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)}
Let P(A|B) represents 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.
Thus, the required probability is 1/15
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’.
Ans: 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 shows 3.
∴ A = {(1, T), (2, T), (4, T), (5, T)}
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)}
⇒ A ∩ B = ⍉
∴ P(A ∩ B) = 0
Then, P(B) = P {(3, 1)} + P{(3, 2)} + P{(3, 3)} + P{(3, 4)} + P{(3, 5)} + P{(3, 6)} + P{(6, 3)}
Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).
Therefore,
In each of the Exercises 16 and 17 choose the correct answer:
16. If P(A) = 1/2 , P(B) = 0, then P( A |B) is
(A) 0
(B) 1/2
(C) not defined.
(D) 1
Ans:
17. If A and B are events such that P(A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B.
(B) A = B.
(C) A ∩ B = ∅
(D) P(A) = P(B).
Ans:
⇒ P(A) = P(B)
Thus, the correct answer is D.
| 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.
Ans:
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.
Ans: Since 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 second draw.
Since the card is not replaced,
∴ P(B) = 25/51
Therefore, 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.
Ans: Let A, B, and C be the respective events that the first, second, and the third drawn orange is good.
Thus, probability that first drawn orange is good, P(A) = 12/15
The oranges are not replaced.
Thus, probability of getting second orange good, P(B) = 11/14
Similarly, probability of getting third orange good, P(C) = 10/13
Since 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
Thus, 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.
Ans:
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?
Ans: 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}
⇒ P(B) = 3/6 = 1/2
∴ A ∩ B = {2}
P(AB) = P(A ∩ B) = 1/6
P(A) × P(B) = 1/2 × 1/2 = 1/4 ≠ 1/6
P(A) × P(B) ≠ P(AB)
Therefore, A and B are not independent events.
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?
Ans: Given P(E) = 3/5, P(F) = 3/10 and P(EF) = P(E ∩ F) = 1/5
P(E)×P(F)= 3/5 × 3/10 = 9/50 ≠ 1/5
P(E) × P(F) ≠ P(E ∩ F)
Therefore, E and F are not independent.
7. Given that the events A and B are such that P(A) = 1/2, P(A ∩ B) = 3/5 and P(B) = p Find p if they are:
(i) mutually exclusive.
Ans: When A and B are mutually exclusive, A ∩ B = ⍉
∴ P(A ∩ B) = 0
Since, P(A U B) = P(A)+P(B) – P (A∩B)
⇒ 3/5 = 1/2 + p – 0
⇒ p = 3/5 – 1/2 = 1/10
(ii) independent.
Ans: When A and B are independent, P(A ∩ B) = P(A) × P(B) = 1/2 p
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)
Ans: If A and B are independent events, then
P(A ∩ B) = P(A) × P(B)
= 0.3 × 0.4
= 0.12
(ii) P( A ∪ B)
Ans: Since, P(A∪B) = P(A) + P(B) – P(A∩B)
⇒ P(A∪B) = 0.3 + 0.4 – 0.12 = 0.58
(iii) P( A |B)
Ans: since,
⇒ P(A|B) = 0.12/0.4 = 0.3
(iv) P( B |A)
Ans: since,
⇒ 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 P(A ∩ B)= 1/3, find P (not A and not B).
Ans: Given, P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8
P (not on A and not on B) = P( A’ ∩ B’)
P (not on A and not on B) = P(( A ∪ B)’) [A’ ∩ B’ = (A ∪ B)’]
= 1 – P(A ∪ B)
= 1 – [ P(A) + P(B) – P(A ∩ B)]
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?
Ans: Given, P(A) = 1/2, P(B) = 7/12 and P (not A or not B) = 1/4
⇒ P(A’ ∪ B’) = 1/4
⇒ P((A ∩ B)’) = 1/4 [A’ ∪ B’ = (A ∩ B)’]
⇒ 1 – P(A ∩ B) = 1/4
⇒ P(A ∩ B) = 3/4 …….(1)
But, P(A) × P(B) = 1/2 × 7/12 = 7/24 ……(2)
Here, 3/4 ≠ 7/24
∴ P(A ∩ B) ≠ P(A) × P(B)
Therefore, A and B are not independent events.
11. Given two independent events A and B such that P(A) = 0.3 P(B) = 0.6 Find
(i) P(A and B)
Ans: P(A and B) = P(A) × P(B)
⇒ P(A ∩ B) = 0.3 × 0.6 = 0.18
(ii) P(A and not B)
Ans: P(A and not B) = P(A ∩ B’)
= P(A) – P(A ∩ B)
= 0.3 – 0.18
= 0.12
(iii) P(A or B)
Ans: P(A or B) = P(AUB)
= P(A) + P(B) – P(A ∩ B)
= 0.3 + 0.6 – 0.18
= 0.72
(iv) P(neither A nor B)
Ans: P(neither A nor B) = P(A’ ∩ B’)
= P((A U B)’) [A’ ∩ B’ = (A ∪ B)’]
= 1 – P (A U B)
= 1 – 0.72
= 0.28
12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Ans: 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.
Ans: 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) first ball is black and second is red.
Ans: Probability of getting first ball black = 10/18 = 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) one of them is black and other is red.
Ans: 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 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that:
(i) the problem is solved.
Ans: Probability that the problem is solved = P( A U B)
= P(A) + P(B) – P(AB)
= 1/2 + 1/3 – 1/6
= 4/6 = 2/3
(ii) exactly one of them solves the problem.
Ans: Probability that exactly one of them solves the problem = P(A). P(B’) + P(B). P(A’)
= 1/2 × 2/3 + 1/2 × 1/3
= 1/3 + 1/6 = 1/2
15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’.
Ans: 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 in 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) E: ‘the card drawn is black’
F: ‘the card drawn is a king’.
Ans: In a deck of 52 cards, 26 cards are black and 4 cards are kings.
∴ P(E) = P (the card drawn is a black) = 26/52 = 1/2
∴ P(F) = P( the card drawn in an ace) = 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 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) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’.
Ans: 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 in 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 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 newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
Ans: Probability that a student reads Hindi or English newspaper
P(H U E’) = 1 – P(H U E)
= 1- {P(H) + P(E) – P(H ∩ E)}
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
Ans: Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by P( E |H)
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Ans: Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (HE).
Choose the correct answer in Exercises 17 and 18.
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
Ans: 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.
(Β) Ρ(Α’Β’) = [1-P(A)] [1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1
Ans: Two events A and B are said to be independent, if P(AB) = P(A) × P(B)
Consider the result given in alternative B.
P(A’ B’) = [1 – P(A)][1 – P(B)]
⇒ P(A’ ∩ B’) = 1 – P(A) – P(B) + P(A). P(B)
⇒1- P(A U B) = 1 – P(A) – P(B) + P(A). P(B)
⇒ P(A U B) = P(A) + P(B) – P(A). P(B)
⇒ P(A) + P(B) – P(AB) = P(A) + P(B) – P(A). P(B)
⇒ P(AB) = P(A). P(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.
∴ A ∩ B = ⍉
⇒ P(AB) = 0
However, P(A). P(B) = mn ≠ 0
∴ P(A). P(B) ≠ P(AB)
C. Let A: Event of getting an odd number on throw of a die = {1, 3, 5}
⇒ P(A) = 3/6 = 1/2
B: Event of getting an even number on throw of a die = {2, 4, 6}
P(B) = 3/6 = 1/2
Here, A ∩ B = ⍉
∴ P(AB) = 0
P(A).P(B) = 1/4 ≠0
⇒ P(A). P(B) ≠ P(AB)
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?
Ans: 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.
Ans: 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 it 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 a hostlier?
Ans: 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(E₁) = 60% = 60/100 = 0.6
P(E₂) = 40% = 40/100 = 0.4
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 3 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?
Ans: 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 (i.e. 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?
Ans: Let E₁ and E₂ be the respective events that a person has a disease and a person has no disease.
Since E₁ and E₂ 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) = 99% = 0.99
P( A |E₂ ) = P (result is positive given the person has no disease) = 0.5% = 0.005
Probability that a person has a disease has a disease, given that his test result is positive, is given by P( E₁|A)
By using Bayes’ theorem, we get
6. There are three coins. One is a 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?
Ans: 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 biased coin = 75%
∴ P(A|E₂) = P(coin showing heads, given that it is a 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 an unbiased 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 get
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?
Ans: 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 get
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 it was produced by machine B?
Ans: Let E₁ and E₂ be the respective events of items produced by machines A and B. Let X be the event that the produced items 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 items was from machine B, given that it is defective, is given by P(E₂|X).
By using Bayes’ theorem, we get
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.
Ans: 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.
Probability that the first group wins the competition, P(E₁) = 0.6
Probability that the second group wins the competition, P(E₂) = 0.4
Probability of introducing a new product if the first group wins, P(A|E₁) = 0.7
Probability of introducing a new product if the second group wins, P(A|E₂) = 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 get
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?
Ans: 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.
Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6,
P(A|E₁) = 3/8
Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4,
P(A|E₂) = 1/2
The probability that the girl threw 1, 2, 3, or 4 with die, if she obtained exactly one head, is given by P(E₂|A).
By using Bayes’ theorem, we get
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 pro- duce 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 it was produced by A?
Ans: Let E₁, E₂ and E₃ be the respective events that the time consumed by machine A, B, and C for the job.
P(E₁) = 50% = 50/100 = 1/2
P(E₂) = 30% = 30/100 = 3/10
P(E₃) = 20% = 20/100 = 1/5
Let X be the event of producing defective items.
P(X|E₁) = 1% = 1/100
P(X|E₂) = 5% = 5/100
P(X|E₃) = 7% = 7/100
The probability that the defective item was produced by A is given by P( E₁|X)
By using Bayes’ theorem, we get
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.
Ans: Let E₁ and E₂ be the respective events of choosing a diamond cards and a card which is not diamond.
Let A denote the lost card.
Out of 52 cards, 13 cards are diamonds and 39 cards are not diamonds.
∴ P(E₁) = 13/52 = 1/4
P(E₂) = 39/52 = 3/4
When one diamond card is lost, there are 12 diamonds cards out of 51 cards.
Two cards can be drawn out of 12 diamonds cards in ¹²C₂ ways .
Similarly, 2 diamonds 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 diamonds cards out of 51 cards.
Two cards can be drawn out of 13 diamonds 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).
The probability that the lost card is diamond is given by P(E₁|A).
By using Bayes’ theorem, we get
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
Ans: 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) = 4/5
∴ P(E₂) = 1 – P(E₁) = 1 – 4/5 = 1/5
If 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).
Therefore, the correct answer is A.
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.
Ans: If A ⊂ B, then A ∩ B = A
⇒ A ∩ B = P(A)
Also, P(A) < P(B)
From (2), we get
⇒ P(A|B) ≥ P(A) ….(3)
Thus, option (C) is correct.
| Miscellaneous Exercise on Chapter 13 |
1. A and B are two events such that P (A) ≠0. Find P(BIA), if
(i) A is a subset of B
Ans: A is a subset of B.
⇒ A ∩ B = A
∴ P(A ∩ B) = P(B ∩ A) = P(A)
(ii) A ∩ B = ⍉
Ans: A ∩ B = ⍉
⇒ P (A ∩ B) = 0
2. A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
Ans: Let E and F respectively denote the events that both children are male and at least one children is a male.
∴ E ∩ F= {(G, G)} ⇒ P(E ∩ F)= 1/4
P(E) = 1/4
P(F) = 3/4
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Ans: Let C and D respectively denote the events that both children are females and the elder child is a female.
3. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Ans: Given, 5% of men and 0.25% of women have grey hair.
Thus, percentage of people with grey hair = (5 + 0.25)% = 5.25%
Probability that the selected haired person is a male = 5/5.25 = 20/21
4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Ans: A person can be either right-handed or left-handed.
Given, 90% of the people are right-handed.
∴ p = P(right-handed) = 9/10
∴ q = P(left-handed) = 1 – p = 1- 9/10 – 1/10
Using binominal distribution, the probability that more than 6 people are right-handed is given by
5. If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
Ans: In a leap year, there are 366 days i.e., 52 weeks and 2 days.
In 52 weeks, there are 52 Tuesdays.
Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining 2 days will be Tuesdays.
The remaining 2 days can be any of the following:
Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday and Sunday and Monday.
Total number of cases = 7
Favourable cases = 2
Probability that a leap year will have 53 Tuesdays = 2/7
6. Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
| Box | Marble colour | ||
| Red | White | Black | |
| A | 1 | 6 | 3 |
| B | 6 | 2 | 2 |
| C | 8 | 1 | 1 |
| D | 0 | 6 | 4 |
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?
Ans: Let R be the event of drawing the red marble.
7. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Ans: Let A, E₁ and E₂ respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
∴ P(A) = 0.40
P(E₁) = P(E₂) = 1/2
⇒ P(A|E₁) = 0.40 × 0.70 = 0.28
⇒ P(A|E₂) = 0.40 × 0.75 = 0.30
Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by P(E₁|A)
8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).
Ans: The total number of determinants of second order with each element being 0 or 1 is (2)⁴ = 16
The value of determinant is positive in the following cases
Therefore, Required probability = 3/16
9. An electronic assembly consists of two subsystems, say, A and B. From previ- ous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities
(i) P(A fails|B has failed).
Ans:
(ii) P(A fails alone).
Ans:
10. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Ans: Let E₁ and E₂ respectively denote the event that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.
P(E₁) = 3/7 and P(E₂) = 4/7
Let A be the event that the ball drawn is red. When a red ball is transferred from bag I to II,
P(A|E₁)= 5/10 = 1/2
When a black ball is transferred from bag I to II,
P(A|E₂) = 4/10 = 2/5
Choose the correct answer in each of the following:
11. If A and B are two events such that P (A) ≠ 0 and P (B|A) = 1, then
(A) A ⊂ B
(B) B ⊂ A
(C) B = ⍉
(D) A = ⍉
Ans: Given,
P(A) ≠ 0 and P(A|B) = 1
P(A) = P(B ∩ A)
⇒ A ⊂ B
Thus, the correct answer is A.
12. If P(A|B) > P(A) then which of the following is correct:
(A) P(B|A) < P(B)
(B) P(A ∩ B) < P(A) .P(B)
(C) P(B|A) > P(B)
(D) P(B|A) = P(B)
Ans: Given,
P(A|B) > P(A)
⇒ P(B|A) > P(B)
Thus, the correct answer is C.
13. If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
Ans: Given,
P(A) + P(B) – P(A and B) = P(A)
⇒ P(A) + P(B) – P (A ∩ B) = P(A)
⇒ P(B) – P(A ∩ B) = 0
⇒ P(A ∩ B) = P(B)
Thus, the correct answer is B.
