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InterviewSolution
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The probability of A, B, C solving a problem are \(\frac{1}{3},\) \(\frac{2}{7}\) and \(\frac{3}{8}\) respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it. |
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Answer» Let E1, E2, E3 be the eventsthat the problem issolved by A, B, C respectively and let p1, p2, p3 be corresponding probabilities. Then, p1 = P(E1) = \(\frac{1}{3}\), p2 = P(E2) = \(\frac{2}{7}\), p3 = P(E3) = \(\frac{3}{8}\), q1 = P(\(\bar{E}_1\)) = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\), q2 = P(\(\bar{E}_2\)) = 1 - \(\frac{2}{7}\) = \(\frac{5}{7}\), q3 = P(\(\bar{E}_3\)) = 1 - \(\frac{3}{8}\) = \(\frac{5}{8}\). The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways: (1) A solves and B, and C do not solve; (2) B solves and A, and C do not solve; (3) C solves and A, and B do not solve; Required probability = p1 q2 q3 + q1 p2 q3 + q1 q2 p3 = \(\frac{1}{3}\) x \(\frac{5}{7}\) x \(\frac{5}{8}\) + \(\frac{2}{3}\) x \(\frac{2}{7}\) x \(\frac{5}{8}\) + \(\frac{2}{3}\) x \(\frac{5}{7}\) x \(\frac{3}{8}\) = \(\frac{25}{168}\) + \(\frac{5}{42}\) + \(\frac{5}{28}\) = \(\frac{25}{56}.\) |
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