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 \[P(X>n-3)=\frac{k}{2^{n}}\], then k is equal to :
Solutions · 1
Official solution
Mean =np=16
Variance =npq=8
\[ \Rightarrow q = p = {1 \over 2}\] and n=32
\[P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)\]
\[ = \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}\]
\[ = {{529} \over {{2^n}}}\]