Question

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. $\frac{126}{81}$

Answer 1

Answer: (B)

$np =4$
$n pq =4 / 3$
$n =6, p =2 / 3, q =1 / 3$
$54( P ( X =2)+ P ( X =1)+ P ( X =0))$
$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) $
$=\frac{146}{27}$