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

#### Options

7, 14

10, 14

12, 7

14, 12

#### Solution

7, 14

\[\text{ Here } , p = \frac{1}{2}\text{ and } q = \frac{1}{2}\]

\[\text{ Binomial distribution is given by } \]

\[P(X = r) =^{n}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{n - r}\]

*P* (*X* = 4), *P *(*X *= 5), *P*(*X *= 6) are in A.P.

\[\therefore ^{n}{}{C}_4 +^{n}{}{C}_6 = 2 ^{n}{}{C}_5 \]

\[ \Rightarrow \frac{n(n - 1)(n - 2)(n - 3)}{2\left( 4! \right)} + \frac{n(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)}{2\left( 6! \right)} = \frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{5!}\]

\[\text{ By simplifying, we get} \]

\[\frac{1}{2} + \frac{(n - 4)(n - 5)}{2(30)} = \frac{n - 4}{5}\]

\[\text{ Taking LCM as 60, we get}\]

\[30 + n^2 - 9n + 20 = 12n - 48 \]

\[ \Rightarrow n^2 - 21n + 98 = 0\]

\[ \Rightarrow (n - 7)(n - 14) = 0\]

\[ \Rightarrow n = 7, 14\]