Search any question & find its solution
Question:
Answered & Verified by Expert
If $X$ has a binomial distribution, $B(n, p)$ with parameters $n$ and $p$ such that $P(X=2)=P(X=3)$, then $\mathrm{E}(\mathrm{X})$, the mean of variable $\mathrm{X}$, is
Options:
Solution:
1797 Upvotes
Verified Answer
The correct answer is:
$3-p$
$3-p$
Since $\mathrm{X}$ has a binomial distribution, $\mathrm{B}(n$,
p)
$$
\begin{aligned}
&\therefore \mathrm{P}(X=2)={ }^n \mathrm{C}_2(p)^2(1-p)^{n-2} \\
&\text { and } \mathrm{P}(X=3)={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\text { Given } \mathrm{P}(X=2)=\mathrm{P}(X=3) \\
&\Rightarrow{ }^n \mathrm{C}_2 p^2(1-p)^{n-2}={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\Rightarrow \frac{n !}{2 !(n-2) !} \cdot \frac{p^2(1-p)^n}{(1-p)^2} \\
&=\frac{n !}{3 !(n-3) !} \cdot \frac{p^3(1-p)^n}{(1-p)^3}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{1}{n-2}=\frac{1}{3} \cdot \frac{p}{1-p} \\
&\Rightarrow 3(1-p)=p(n-2) \\
&\Rightarrow 3-3 p=n p-2 p \\
&\Rightarrow n p=3-p \\
&\Rightarrow \mathrm{E}(X)=\text { mean }=3-p \\
&(\because \text { mean of } \mathrm{B}(n, p)=n p)
\end{aligned}
$$
p)
$$
\begin{aligned}
&\therefore \mathrm{P}(X=2)={ }^n \mathrm{C}_2(p)^2(1-p)^{n-2} \\
&\text { and } \mathrm{P}(X=3)={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\text { Given } \mathrm{P}(X=2)=\mathrm{P}(X=3) \\
&\Rightarrow{ }^n \mathrm{C}_2 p^2(1-p)^{n-2}={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\Rightarrow \frac{n !}{2 !(n-2) !} \cdot \frac{p^2(1-p)^n}{(1-p)^2} \\
&=\frac{n !}{3 !(n-3) !} \cdot \frac{p^3(1-p)^n}{(1-p)^3}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{1}{n-2}=\frac{1}{3} \cdot \frac{p}{1-p} \\
&\Rightarrow 3(1-p)=p(n-2) \\
&\Rightarrow 3-3 p=n p-2 p \\
&\Rightarrow n p=3-p \\
&\Rightarrow \mathrm{E}(X)=\text { mean }=3-p \\
&(\because \text { mean of } \mathrm{B}(n, p)=n p)
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.