If X sim B(5, p) is a binomial variate such t | vedclass

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(C) Given $X \sim B(5, p)$.
$P(X=3) = P(X=4)$
$\Rightarrow { }^5 C_3 p^3(1-p)^2 = { }^5 C_4 p^4(1-p)$
$\Rightarrow 10(1-p) = 5p$
$\Rightarrow 10 -

Vedclass · Accepted Answer

$\frac{200}{243}$

Vedclass · Answer

(C) Given $X \sim B(5, p)$.
$P(X=3) = P(X=4)$
$\Rightarrow { }^5 C_3 p^3(1-p)^2 = { }^5 C_4 p^4(1-p)$
$\Rightarrow 10(1-p) = 5p$
$\Rightarrow 10 - 10p = 5p \Rightarrow 15p = 10 \Rightarrow p = \frac{2}{3}$.
Now,we need to find $P(|X-3| < 2)$.
$P(|X-3| < 2) = P(-2 < X-3 < 2) = P(1 < X < 5) = P(X=2) + P(X=3) + P(X=4)$.
Since $P(X=3) = P(X=4)$,we have $P(X=2) + 2P(X=3)$.
$P(X=2) = { }^5 C_2 (\frac{2}{3})^2 (\frac{1}{3})^3 = 10 	imes \frac{4}{9} 	imes \frac{1}{27} = \frac{40}{243}$.
$P(X=3) = { }^5 C_3 (\frac{2}{3})^3 (\frac{1}{3})^2 = 10 	imes \frac{8}{27} 	imes \frac{1}{9} = \frac{80}{243}$.
$P(|X-3| < 2) = \frac{40}{243} + 2(\frac{80}{243}) = \frac{40 + 160}{243} = \frac{200}{243}$.

If $X \sim B(5, p)$ is a binomial variate such that $P(X=3)=P(X=4)$,then $P(|X-3| < 2)=$

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