91Ó°ÊÓ

The probability that \(A\) speaks truth is \(4 / 5\) and the probability that \(B\) speaks truth is \(3 / 4\). The probability that they contradict each other when asked to speak on a fact is (a) \(3 / 10\) (b) \(7 / 20\) (c) \(1 / 4\) (d) \(2 / 5\)

A box contains 3 white and 2 red balls. If the first drawn ball is not replaced, then the probability that the second drawn ball will be red is (a) \(8 / 25\) (b) \(2 / 5\) (c) \(3 / 5\) (d) \(21 / 25\)

The mean and the variance of a binomial distribution are 4 and 2, respectively, then the probability of 2 successes is (a) \(28 / 256\) (b) \(42 / 256\) (c) \(56 / 256\) (d) \(72 / 256\)

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is (a) \(1 / 3\) (b) \(1 / 6\) (c) \(1 / 2\) (d) \(1 / 4\)

If three students \(A, B, C\) can solve a problem with probabilities \(1 / 3,1 / 4\) and \(1 / 5\) respectively, then the probability that the problem will be solved is a) \(3 / 5\) (b) \(4 / 5\) (c) \(2 / 5\) (d) \(47 / 60\)

There are \(n\) urns each containing \((n+1)\) balls such that the \(i^{\text {th }}\) urn contains \(i\) white balls and \((n+1-i)\) red balls. Let \(u_{i}\) be the event of selecting the \(i^{\text {th }}\) urn, \(i=1,2,3 \ldots, n\), and \(W\) denote the event of getting a white ball. (i) If \(P\left(u_{j}\right) \alpha i\), where \(i=1,2,3, \ldots, n\), then \(\lim _{n \rightarrow \infty} P(W)\) is equal to (a) 1 (b) \(1 / 4\) (c) \(2 / 3\) (d) \(3 / 4\)