Question

If the mean and variance of a binomial variate $x$ are $8$ and $4$ respectively then$P\left(X<3\right)=$

• A. $\frac{137}{2^{16}}$
• B. $\frac{697}{2^{16}}$
• C. $\frac{265}{2^{16}}$
• D. $\frac{265}{2^{15}}$

Answer 1

Answer: (A)

$\frac{137}{2^{16}}$

Solution -

Mean = $rp=8$

variance = $rpq=4$

$q=\frac{1}{2}$ $p=\frac{1}{2}$ $n=16$.

$P(X<3)=P(X=0)+P(X=1)+P(X=2)$

$={}^{16}{C}_0\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{16}{+}^{16}{C}_p{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{15}{+}^{16}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{14}$

$={(}^{16}{C}_0{+}^{16}{C}_1{+}^{16}{C}_2){\left(\frac{1}{2}\right)}^{16}=\frac{137}{2^{16}}$

A is correct.