MCQ
Let $X$ be a binomially distributed random variable with mean $4$ and variance $\frac{4}{3}$. Then $54 P ( X \leq 2)$ is equal to.
- A$\frac{73}{27}$
- B$\frac{146}{27}$
- C$\frac{146}{81}$
- D$\frac{126}{81}$
$npq =4 / 3$
$n =6, p =2 / 3, q =1 / 3$
$54( P ( X =2)+ P ( X =1)+ P ( X =0))$
$54\left({ }^{6} C _{2}\left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)^{4}+{ }^{6} C _{1}\left(\frac{2}{3}\right)^{1}\left(\frac{1}{3}\right)^{5}+{ }^{6} C _{0}\left(\frac{2}{3}\right)^{0}\left(\frac{1}{3}\right)^{6}\right)$
$=\frac{146}{27}$
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