Question

The mean and variance of a random variable $X$ having a binomial probability distribution are $6$ and $3$ respectively, then the probability $P\left(X \geq 2\right)$ is

• A. $\frac{13}{4096}$
• B. $\frac{4083}{4096}$
• C. $\frac{3}{1024}$
• D. $\frac{13}{2048}$

Answer 1

Answer: (B)

$np=6,npq=3$
$q=\frac{1}{2},p=\frac{1}{2},n=12$
$P\left(X \geq 2\right)=1-P\left(X = 0\right)-P\left(X = 1\right)$
$=1-^{12}C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{12}-^{12}C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{11}$
$=1-\frac{13}{2^{12}}=\frac{4083}{4096}$