Let X have a binomial distribution B n , p such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If P X > n - 3 = k 2 n , then k is equal to

Step-by-step solution

Let M = Mean and V = Variance M > V So, M + V = 24 ,   M V = 128 ⇒ M = 16 and V = 8 Now, mean n p = 16 and variance n p q = 8 ⇒ q = 1 2 ∴     p = 1 2 , n = 32 Now P X > n - 3 = 1 2 n C n - 2 n + C n - 1 n + C n n ∴   k 2 n = 1 2 n C 30 32 + C 31 32 + C 32 32 ⇒ k = 32 × 31 2 + 32 + 1 = 496 + 33 = 529