If X is a binomial variate with parameters n and p , where 0< p <1 such that P(X=r) P(X=n-r) is independent of n and r, then p equals

If X is a binomial variate with parameters n and p , where 0< p <…

MCQ

If X is a binomial variate with parameters n and p , where $0< p <1$ such that $\frac{P(X=r)}{P(X=n-r)}$ is independent of n and $r$, then $p$ equals

✓
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$\frac{1}{3}$
D
$\frac{1}{5}$

✓

Answer

Correct option: A.

$\frac{1}{2}$

(a) $\frac{1}{2}$
Explanation: $\because P ( X = r )={ }^{ n } C _{ r }( P )^{ r }( q )^{ n - r }$
$
\begin{array}{l}
=\frac{n!}{(n-r)!r!}(P)^{r}(1-p)^{n-r}[\because q=1-p] \\
P(X=0)=(1-p)^{n}
\end{array}
$
And $P(X=n-r)={ }^n C_{n-r}(P)^{n-r}(q)^{n-(n-r)}$
$
=\frac{n!}{(n-r)!r!}(p)^{n-r}(1-p)^{-r}[\because q=1-p]\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]
$
Now, $\frac{P(x=r)}{P(x=n-r)}=\frac{\frac{n!}{(n-r) w!} p^r(1-p)^{n-r}}{\frac{n!}{(n-r)+!} p^{n-r}(1-p)^{+r}}$ [using Eqs. (i) and (ii)]
$
=\left(\frac{1-p}{p}\right)^{n-\gamma} \times \frac{1}{\left(\frac{1-p}{p}\right)^\tau}
$
The above expression is independent of n and r , if $\frac{1-p}{p}=1 \Rightarrow \frac{1}{p}=2 \Rightarrow p=\frac{1}{2}$

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