The probability that a certain kind of component will survive a given shock test is 3 4 . Find the probability that among 5 components tested exactly 2 will survive .

Let X denote the number of components that survive shock. Then, X follows a binomial distribution with n = 5 . Let p be the probability that a certain kind of component will survive a given shock test. ∴ p = 3 4 and q = 1 - p = 1 4 Hence, the distribution is given by P ( X = r ) = 5 C r 3 4 r 1 4 5 - r , r = 0 , 1 , 2 , 3 , 4 , 5 Thus, the required probability is P exactly 2 will survive = P ( X = 2 ) = 5 C 2 3 4 2 1 4 5 - 2 = 10 × 9 1024 = 0 . 0879 .

The probability that a certain kind of component will survive a given shock test is 3 4 . Find the probability that among 5 components tested at most 3 will survive.

Let X denote the number of components that survive shock. Then, X follows a binomial distribution with n = 5 . Let p be the probability that a certain kind of component will survive a given shock test. ∴ p = 3 4 and q = 1 4 Hence, the distribution is given by P ( X = r ) = 5 C r 3 4 r 1 4 5 - r , r = 0 , 1 , 2 , 3 , 4 , 5 Now, P atmost 3 will survive = P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) = 5 C 0 3 4 0 1 4 5 - 0 + 5 C 1 3 4 1 1 4 5 - 1 + 5 C 2 3 4 2 1 4 5 - 2 + 5 C 3 3 4 3 1 4 5 - 3 = 1 4 5 + 5 3 4 1 4 4 + 10 3 4 2 1 4 3 + 10 3 4 3 1 4 2 = 1 + 15 + 90 + 270 1024 = 376 1024 = 0 . 3672 .