Question

A fair coin is tossed n times and let X denotes the number of heads obtained. If P(X = 4), P(X = 5), and P(X = 6) are in A.P., then n is equal to

• A. Only 7
• B. 14
• C. 7 or 14
• D. 8

Answer 1

Answer: (B), (C)

The correct options are
B 14
C 7 or 14

Probability of r successes in n trials is equal to ${}^n{C}_r{p}^r{q}^{n-r}$, where p is the probability of success in one trial and q is the probability of failure in one trial.
This means $n{C}_4{\left(\frac{1}{2}\right)}^n,n{C}_5{\left(\frac{1}{2}\right)}^n,n{C}_6{\left(\frac{1}{2}\right)}^{n}areA.P.$
$\Rightarrow 2\times n{C}_5=n{C}_4+n{C}_6$
$\Rightarrow 2\times \frac{n!}{(n-5)!5!}=\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!}!$
$\Rightarrow 6\times 5+(n-4)(n-5)=2\times 6(n-4)$
$\Rightarrow n^2-9n+50=2(6n-24)$
$\Rightarrow n^2-21n+98=0$
$\Rightarrow n=7orn=14.$