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The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $\mathrm{P(X=1)=\frac{4}{243}},$ then $\mathrm{P(X=4 \text { or } 5)}$ is equal to:
Option: 1
$\frac{5}{9}$

Option: 2
$\frac{64}{81}$

Option: 3
$\frac{16}{27}$

Option: 4
$\frac{145}{243}$

Answers (1)

best_answer

$\mathrm{ \Rightarrow n p=\alpha }\\$ .........(1)

$\mathrm{n p q=\frac{\alpha}{3} }\\$ .........(2)

$\mathrm{\text { from (1) \& (2) }}\\$

$\mathrm{q=\frac{1}{3} \quad \&\; p=\frac{2}{3}}\\$

$\mathrm{{ }^{n} C_{1} q^{n-1} \cdot p^{1}=\frac{4}{243}}\\$

$\mathrm{\frac{n}{3^{n}}=\frac{2}{243}}\\$

$\mathrm{n=6}\\$

$\mathrm{P(4 \text { or } 5)={ }^{6} C_{4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}+{ }^{6} C_{5}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)^{0}}\\$

$=\frac{16}{27}$

Hence correct option is 3

Posted by

SANGALDEEP SINGH

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