Question

Let X denotes the number of times head occur in n tosses of a fair coin. If $P(X=4),P(X=5)$ and $P(X=6)$ are in AP, then the value of n is

• A. 7, 14
• B. 10, 14
• C. 12, 7
• D. 14, 12

Answer 1

The correct option is A 7, 14

Clearly, $X$ is a binomial variate with parameter and $p=\frac{1}{2}$ such that

$P(X=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}$ ${=}^n{C}_r{\left(\frac{1}{2}\right)}^n$

Now, $P(X=r),P(X=5)$ and $P(X=6)$ are in AP

$\Rightarrow 2P(X=5)=P(X=4)+P(X=6)$
$\Rightarrow {2.}^n{C}_5{\left(\frac{1}{2}\right)}^n{=}^n{C}_4{\left(\frac{1}{2}\right)}^n{+}^n{C}_6{\left(\frac{1}{2}\right)}^n$
$\Rightarrow {2.}^n{C}_5{=}^n{C}_4{+}^n{C}_6$
$\Rightarrow 2.\frac{n!}{(n-5)!5!}=\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}$

$\Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}$
$\Rightarrow n^2-21n+98=0$

$\Rightarrow (n-7)(n-14)=0$

$\Rightarrow n=7$ or $14$