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Let $X$ be binomial variate with parameters $n=6$ and $p$. If $4 P(X=4)=P(X=2)$, then $p$ equals
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The correct answer is:
$1 / 3$
We have, $n=6$.
It is given that,
$$
\begin{aligned}
& 4 P(X=4)=P(X=2) \\
& \Rightarrow \quad 4 \cdot{ }^6 C_4 p^4 q^2={ }^6 C_2 p^2 q^4 \\
& {\left[\because P(X=r)={ }^n C_r p^r q^{n-r}\right]} \\
& \Rightarrow \quad 4 \times \frac{6 \times 5}{2 \times 1} p^2=\frac{6 \times 5}{2 \times 1} q^2 \\
& \Rightarrow \quad 4 p^2=q^2 \\
& \Rightarrow \quad 2 p=q \\
& \Rightarrow \quad 2 p=1-p \quad[\because q=1-p] \\
& \Rightarrow \quad p=\frac{1}{3} \\
&
\end{aligned}
$$
It is given that,
$$
\begin{aligned}
& 4 P(X=4)=P(X=2) \\
& \Rightarrow \quad 4 \cdot{ }^6 C_4 p^4 q^2={ }^6 C_2 p^2 q^4 \\
& {\left[\because P(X=r)={ }^n C_r p^r q^{n-r}\right]} \\
& \Rightarrow \quad 4 \times \frac{6 \times 5}{2 \times 1} p^2=\frac{6 \times 5}{2 \times 1} q^2 \\
& \Rightarrow \quad 4 p^2=q^2 \\
& \Rightarrow \quad 2 p=q \\
& \Rightarrow \quad 2 p=1-p \quad[\because q=1-p] \\
& \Rightarrow \quad p=\frac{1}{3} \\
&
\end{aligned}
$$
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