Class 12 Maths Chapter 13: Probability worksheet with answer key
A 20-mark worksheet on Probability for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) 0.18
With R the event 'it rains' and L 'the bus is late', P(R ∩ L) = P(R) P(L|R) = 0.3 × 0.6 = 0.18. Option (a) adds; (c) divides; (d) uses P(R′) = 0.7.
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Answer: (c) 6
P(first wins) = 2/n and P(second wins | first wins) = 1/(n − 1), so 2/(n(n − 1)) = 1/15, n(n − 1) = 30 and n = 6. Check: 2/6 × 1/5 = 1/15. Option (a) gives 1/10; (b) treats the draws as one draw of a winner.
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Answer: (c) P(E | the number is a perfect square)
Each clue reduces the sample space. Primes up to 40: 12, of which only 2 is even, 1/12. Multiples of 3: 13, of which the 6 multiples of 6 are even, 6/13. Perfect squares {1, 4, 9, 16, 25, 36}: 3 even, 1/2. Less than 6: {1, 2, 3, 4, 5}, 2 even, 2/5. The greatest is 1/2, after the perfect-square clue.
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Answer: (a) 0.6
For independent events P(A ∩ B) = P(A) P(B), so P(B) = 0.12/0.3 = 0.4 and P(B′) = 1 − 0.4 = 0.6. Option (b) is P(B); (c) is 1 − P(A ∩ B).
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Answer: (a) 0.26
P(E|F) = 1 − P(E′|F) = 0.65, so P(E ∩ F) = P(F) P(E|F) = 0.4 × 0.65 = 0.26. Option (b) uses 0.35 in place of 0.65; (c) stops at P(E|F); (d) subtracts.
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P(A′ ∩ B′) = P((A ∪ B)′) = 1 − 0.7 = 0.3 and P(B′) = 1 − 0.4 = 0.6. So P(A′|B′) = P(A′ ∩ B′)/P(B′) = 0.3/0.6 = 0.5.
- P(A′ ∩ B′) = 1 − P(A ∪ B) = 0.3 (1)
- P(B′) = 0.6 and P(A′|B′) = 0.5 (1)
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A′ and B are independent when A and B are, so P(A′ ∩ B) = P(A′) P(B) = 0.4 P(B) = 0.24 and P(B) = 0.6. Then P(A ∪ B) = P(A) + P(B) − P(A) P(B) = 0.6 + 0.6 − 0.36 = 0.84.
- uses P(A′ ∩ B) = P(A′) P(B) to get P(B) = 0.6 (1)
- P(A ∪ B) = 0.84 (1)
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P(A) = P(B) = P(C) = 1/2. A ∩ B = A ∩ C = B ∩ C = {1}, each of probability 1/4 = 1/2 × 1/2, so each pair is independent. But A ∩ B ∩ C = {1} has probability 1/4, while P(A) P(B) P(C) = 1/8; one of the four conditions fails, so the three events are not mutually independent.
- each pairwise intersection is {1}, probability 1/4 = 1/2 × 1/2 (1)
- P(A ∩ B ∩ C) = 1/4 ≠ 1/8, so not mutually independent (1)
| Mode | On time | Late |
|---|---|---|
| Metro | 72 | 8 |
| Bus | 84 | 36 |
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(i) Of 80 metro commuters, 72 are on time: 72/80 = 9/10. (ii) Of 156 commuters on time, 72 used the metro: 72/156 = 6/13. (iii) P(M) = 80/200 = 0.4, P(T) = 156/200 = 0.78, P(M ∩ T) = 72/200 = 0.36, while P(M) P(T) = 0.312 ≠ 0.36; so the events are not independent (equally, P(T|M) = 0.9 ≠ P(T) = 0.78).
- (i) 9/10 (1)
- (ii) 6/13 (1)
- (iii) compares P(M ∩ T) = 0.36 with P(M) P(T) = 0.312: dependent (1)
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(i) P(C) = 0.9 × 0.8 + 0.1 × 0.3 = 0.72 + 0.03 = 0.75. (ii) P(missed) = 0.25 and P(rang and missed) = 0.9 × 0.2 = 0.18, so P(rang | missed) = 0.18/0.25 = 0.72.
- (i) total probability 0.75 (1)
- (ii) P(rang ∩ missed) = 0.18 and P(missed) = 0.25 (1)
- 0.72 (1)
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Let E₁, E₂ be 'billing' and 'network', and A 'not resolved in the first call'. P(E₁ ∩ A) = 0.6 × 0.1 = 0.06 and P(E₂ ∩ A) = 0.25(1 − r). By Bayes' theorem P(E₁|A) and P(E₂|A) have the same denominator P(A), so they are equal exactly when P(E₁ ∩ A) = P(E₂ ∩ A): 0.25(1 − r) = 0.06, 1 − r = 0.24, r = 0.76.
- P(billing ∩ not resolved) = 0.06 (1)
- equal posteriors reduced to 0.25(1 − r) = 0.06 (1)
- r = 0.76 (1)