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Let $\mathrm{ X}$ be a binomially distributed random variable with mean 4 and variance $\mathrm{ \frac{4}{3}}$ . Then, $\mathrm{ 54 \, \, P(X \leq 2)}$ is equal to
Option: 1
$\frac{73}{27}$

Option: 2
$\frac{146}{27}$

Option: 3
$\frac{146}{81}$

Option: 4
$\frac{126}{81}$

Answers (1)

best_answer

$\mathrm{n p =4} \\$

$\mathrm{n p q =\frac{4}{3} }\\$

$\mathrm{\Rightarrow q =\frac{1}{3} \Rightarrow 1-p=\frac{1}{3} \Rightarrow p=\frac{2}{3}} \\$

$\mathrm{\Rightarrow n =6}$

$\mathrm{\therefore\; 54 P(x \leq 2)} \\$

$\mathrm{=54[P(0)+P(1)+P(2)]} \\$

$\mathrm{=54\left[{ }^{6} C_{0} \cdot\left(\frac{1}{3}\right)^{6}+{ }^{6} C_{1}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)\right. \left.+{ }^{6} C_{2}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}\right] }$

$=54\left[\frac{1}{3^{6}}(1+6 \cdot 32+15 \cdot 16)\right] \\$

$=\frac{146}{27}$

Hence correct option is 2

Posted by

vishal kumar

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