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Q.

For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is p. If he fails in one of the exams, then the probability of his passing in the next exam is p/2, otherwise it remains the same. Then the probability that he will qualify, is

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a

2p³ – p

b

2p³ – 2p

c

none of these

d

2p³ – p³

answer is C.

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Detailed Solution

Let P1, P2, P3 be the events of passing 1^st, 2^nd, 3^rd, tests respectively. Probability of passing atleast two tests
$\begin{array}{l} = P (P_{1}, P_{2}, {\bar{P}}_{3}) + P (P_{1} + {\bar{P}}_{2} + P_{3}) + P (P_{1} + P_{2} + P_{3}) \\ p \cdot p (1 - p) + p (1 - p) \cdot (\frac{p}{2}) + (1 - p) (\frac{p}{2}) \cdot p + p \cdot p \cdot p \\ = 2 p^{2} (1 - p) + p^{3} = (2 p^{2} - p^{3}) \end{array}$

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