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If two dice are thrown simultaneously then find the probability that the sum of the numbers coming up on them is 11, given that the number 5 always occurs on the first dice.1. 1/32. 1/363. 1/124. 1/6 |
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Answer» Correct Answer - Option 4 : 1/6 Concept: P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A. Calculation: Let S be the sample space ∴ n(S) = 36 Let, A = the event that the sum of the numbers on the two dice is 11. ∴ A = {(5,6), (6,5)} ∴ n(A) = 2 Let, B = the event of the occurrence of 5 on the first dice. B = {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)} ∴ n(B) = 6 Now, P(B) = n(B) / n(S) = 6/36 = 1/6. ⇒ A ∩ B = {(5,6)} ∴ n(A∩B) = 1 Now, P(A ∩ B) = n(A ∩ B) / n(S) = 1/36. ⇒ P(A | B) = P(A ∩ B) / P(B) = 1/6 Hence, option 4 is correct. |
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