Question

A fair coin is tossed $n$ times. Let $X=$ the number of times head occurs $P(X=4),P(X=5)$ and $P(X=6)$ are in A.P., then the value of $n$ can be

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

Answer 1

Answer: (A), (D)

The correct options are
A $7$
D $14$

P(X=4), P(X=5), P(X=6) is given by $\frac{C_4^n}{2^n},\frac{C_5^n}{2^n},\frac{C_6^n}{2^n}$ respectively.

Since they are in AP, we have: $C_4^n,C_5^n,C_6^n$ in AP.
Thus, $C_4^n+C_6^n=2\ast C_5^n$
This is satisfied when $n=7$ or $14.$