If the mean and standard deviation of a binomial distribution are 8 a…

The correct answer is Option (1) → $\frac{15}{4096}$

Mean (μ) = n.p

Variance $(σ)^2=n.p(1-p)$

Standard deviation = $\sqrt{σ^2}=σ$

From $μ=n.p=8$

From $σ^2=n.p(1-p)$

$n.p(1-p)=4$ (since $σ=2⇒σ^2=4$)

$⇒8.(1-p)=4$

$8-8p=4$

$=4=8p$

$p=\frac{1}{2}=0.5$

From $μ=n.p$

$n.0.5=8$

$n=16$

Probability mass function of a binomial distribution

$P(X=r)={^nC}_r.p^r.(1-p)^{n-r}$

$r=2,n=16,p=0.5$

${^{16}C}_r.(0.5)^2.(0.5)^{16-2}$

$⇒\frac{16!}{14!2!}.(0.5)^{16}$

$⇒\frac{16.15}{2}.\frac{1}{2^{16}}$

$⇒\frac{120}{2^{16}}≃\frac{15}{4096}$