If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Step-by-step solution

Let p = probability of success and q = probability of failure, Now given, Sum n p + n p q = 24             . . . i Product n p × n p q = 128                 . . . ii Also p = 1 - q .... iii Solving i , ii and iii we get, p = 1 2 , q = 1 2 , n = 32 Now, P X = 1 + P X = 2 = C 1 32 p q 31 + C 2 32 p 2 q 30 = 33 2 28