InterviewSolution
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InterviewSolution
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There are six tickets numbered from O to 5. Two tickets are selected at random from the lot. What is the probability that the sum Is (a) at most 8 (b) at least 8 (c) prime number |
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Answer» n(S) = 6C2 = \(\frac{6\times5}{2\times1}\) = 15 (a) Let A: sum is at most 8 i.e max B A = {(O, 1) (0, 2) (0,3) (0,4) (0,5) (1,2) (1,3) (1.4) (1,5) (2,3) (2, 4) (2, 5) (3, 4) (3, 5)) n(A)=14 P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{14}{15}\) (b) Let B = sum is at least 8 = {(5, 3) (4, 5)) ⇒ n(B) = 2 ⇒ n(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{2}{15}\) (c) Let C : Sum is prime number C = { (0. 1) (0, 2) (0, 3) (0, 5) (1, 2) (1. 4) (2, 3) (3, 4)) ⇒ n(C) = 8 P(C) = \(\frac{n(B)}{n(S)}\) = \(\frac{8}{15}\) |
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