Question

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

• A. 6 and 7
• B. 6 and 14
• C. 7 or 14
• D. None

Answer 1

The correct option is C 7 or 14

${}^n{C}_4{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^{n-4}{,}^n{C}_5{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{n-5},$
${}^n{C}_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^{n-6}$ are in A.P.
$\Rightarrow \frac{n!}{(n-4)!4{!}^{\prime }}\frac{n!}{5(n-5){!}^{\prime }}\frac{n!}{6(n-6){!}^{\prime }}$ are in A.P.
$\Rightarrow 6\times 5+(n-4)(n-5)=2\times 6(n-4)\Rightarrow n^2-9n+50=(6n-24)\times 2\Rightarrow n^2-21n+98=0$
$\Rightarrow n=7$ or 14.