Question

Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X ≤ 2) = k/(2)¹⁶, then k is equal to :
(A) 17 (B) 137 (C) 1 (D) 121

Answer 1

$\textbf{Concept used:}$

$\text{The probability mass function of binomial distribution is}$

$P(X=x)=n_{C_x}\;p^x\;q^{n-x}$

$x=0,1,2.......n$

$\text{Given:}$

$\text{Mean=8 and variance=4}$

$\implies\,np=8\;\text{and}\;npq=4$

$\implies\,(8)q=4$

$\implies\,q=\frac{4}{8}$

$\implies\bf\,q=\frac{1}{2}$

$\text{But, p+q=1}\implies\bf\;p=\frac{1}{2}$

$np=8\implies\bf\;n=16$

$\therefore\text{The probability mass function is}$

$P(X=x)={16}_{C_x}(\frac{1}{2})^{16}$

$\text{Given condition is}$

$P(X{\leq}2)=\frac{k}{2^{16}}$

$\implies\,P(X=0)+P(X=1)+P(X=2)=\frac{k}{2^{16}}$

$\implies\,{16}_{C_0}(\frac{1}{2})^{16}+{16}_{C_1}(\frac{1}{2})^{16}+{16}_{C_2}(\frac{1}{2})^{16}=\frac{k}{2^{16}}$

$\implies\,[16}_{C_0}+{16}_{C_1}+{16}_{C_2}]\,(\frac{1}{2})^{16}=\frac{k}{2^{16}}$

$\implies\,1+16+\frac{16{\times}15}{1{\times}2}=k$

$\implies\,1+16+(8{\times}15)=k$

$\implies\,1+16+120=k$

$\implies\boxed{\bf\,k=137}$

$\therefore\textbf{Option (B) is correct}$