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

Question

The mean and variance of a random variable X having a binomial probability distribution are 6 and 3 respectively, then the probability P(X ≥ 2) is (A) (13/4096) (B) (4083/4096) (C) (3/1024) (D) (13/2048)

Tardigrade Admin · Accepted Answer

np=6,npq=3 q=(1/2),p=(1/2),n=12 P(X ≥ 2)=1-P(X = 0)-P(X = 1) =1-12C0((1/2))0((1/2))12-12C1((1/2))1((1/2))11 =1-(13/212)=(4083/4096)

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

$\frac{13}{4096}$

$\frac{4083}{4096}$

$\frac{3}{1024}$

$\frac{13}{2048}$

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

Q. The mean and variance of a random variable X having a binomial probability distribution are 6 and 3 respectively, then the probability P(X≥2) is

409613​

40964083​

10243​

204813​