Class 11 Maths Chapter 14: 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: (d) 0.25
The probabilities of all the outcomes add to 1: P(ω₃) = 1 – 0.3 – 0.45 = 0.25.
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Answer: (a) 8
S = {1, 2, ..., 8} has 8 sample points, and each singleton is a simple event, so there are exactly 8 simple events.
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Answer: (b) 0.17
The two events are mutually exclusive, so P(cooker or mixer) = 0.05 + 0.12 = 0.17.
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Answer: (c) the ball is neither green nor white
No outcome is neither green nor white, so that event is the empty set φ. 'Green or white' is the whole sample space, the sure event.
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Answer: (c) 35
k + 2k + 3k + 4k = 1 gives k = 1/10. P({ω₂, ω₄}) = 2k + 4k = 6/10 = 3/5.
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n(S) = 5! = 120. Ravi and Sana take the two ends in 2 ways and the other three fill the middle in 3! = 6 ways: 12 arrangements. P = 12/120 = 1/10.
- n(S) = 120 and favourable 2 × 3! = 12 (1)
- P = 1/10 (1)
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A = {1, 2, 3, 4, 6}, B = {1, 3, 5, 7, 9}. A ∪ B = {1, 2, 3, 4, 5, 6, 7, 9}. A′ = {5, 7, 8, 9, 10}, so A′ ∩ B = {5, 7, 9}.
- A ∪ B = {1, 2, 3, 4, 5, 6, 7, 9} (1)
- A′ ∩ B = {5, 7, 9} (1)
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There are 90 two-digit numbers (10 to 99), all equally likely. Equal digits: 11, 22, ..., 99, that is 9 numbers. P = 9/90 = 1/10.
- n(S) = 90 and 9 favourable numbers (1)
- P = 1/10 (1)
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With H, E, R the three events, P(H ∪ E ∪ R) = P(H) + P(E) + P(R) – P(H ∩ E) – P(E ∩ R) – P(H ∩ R) + P(H ∩ E ∩ R) = 1.1 – 0.3 + 0.03 = 0.83. (ii) P(none) = 1 – 0.83 = 0.17.
- writes the formula for P(H ∪ E ∪ R) (1)
- P(at least one) = 0.83 (1)
- P(none) = 0.17 (1)
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A = (A ∩ B′) ∪ (A ∩ B), and these two are mutually exclusive, so P(A) = P(A ∩ B′) + P(A ∩ B), giving P(A ∩ B′) = P(A) – P(A ∩ B). Likewise P(A′ ∩ B) = P(B) – P(A ∩ B). 'Exactly one' is (A ∩ B′) ∪ (A′ ∩ B), mutually exclusive, so P = P(A) + P(B) – 2P(A ∩ B) = 0.5 + 0.35 – 0.2 = 0.65.
- splits A into the disjoint A ∩ B′ and A ∩ B and applies axiom (iii) (1)
- P(exactly one) = P(A) + P(B) – 2P(A ∩ B) (1)
- value 0.65 (1)
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n(S) = ⁵²C₂ = 1326. Choose the rank in 13 ways and two of its four cards in ⁴C₂ = 6 ways: 78 favourable pairs. P = 78/1326 = 1/17.
- n(S) = ⁵²C₂ = 1326 (1)
- favourable 13 × ⁴C₂ = 78 (1)
- P = 1/17 (1)