Question

If mean and variance of a Binomial variable X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

• A. $\frac{2}{3}$
• B. $\frac{4}{5}$
• C. $\frac{7}{8}$
• D. $\frac{11}{16}$

Answer 1

The correct option is D $\frac{11}{16}$

Solution:

Given that:

$np=2,npq=1$

$⟹2\times q=1⟹q=\frac{1}{2}$

$⟹p+q=1⟹p=\frac{1}{2}$

$⟹n\times \frac{1}{2}\times \frac{1}{2}=1⟹n=4$

$\therefore P(X>1)={}^4{C}_2{p}^2{q}^2+{}^4{C}_3{p}^3{q}^1+{}^4{C}_4{p}^4{q}^0$

$=6{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2$$+4{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^1$$+1{\left(\frac{1}{2}\right)}^{4}1$

$={\left(\frac{1}{2}\right)}^4(6+4+1)=\frac{11}{16}$

Hence, D is the correct option.