An unbiased coin is tossed $n$ times. Let the random variable X denote the number of times the head occurs. If $P(X = 1), P(X = 2)$ and $P(X = 3)$ are in AP, find the value of $n$.

7

The correct answer is Option (3) → 7

When an unbiased coin is tossed, the probability of occurring a head = $p =\frac{1}{2}$

so $q=1-\frac{1}{2}=\frac{1}{2}$

A coin is tossed n times, so there are n Bernoullian trials.

Thus, we have a binomial distribution with $p=\frac{1}{2},q=\frac{1}{2}$ and number of trials = $n$.

$P(X = r) = {^nC}_r p^r q^{n - r} = {^nC}_r (\frac{1}{2})^r (\frac{1}{2})^{n-r} ={^nC}_r (\frac{1}{2})^n$.

Given $P(X = 1), P(X = 2)$ and $P(X = 3)$ are in AP

$⇒{^nC}_1 (\frac{1}{2})^n,{^nC}_2 (\frac{1}{2})^n$ and ${^nC}_3 (\frac{1}{2})^n$ are in AP

$⇒{^nC}_1,{^nC}_2$ and ${^nC}_3$ are in AP

$⇒2.{^nC}_2={^nC}_1+{^nC}_3$

$⇒2.\frac{n(n-1)}{1.2}=\frac{n}{1}+\frac{n(n - 1)(n-2)}{1.2.3}$

$⇒n-1=1+\frac{(n-1)(n-2)}{6}$   $(∵n≠0)$

$⇒n-2=\frac{(n-1)(n-2)}{6}$   $(∵n≠2)$

$⇒1=\frac{n-1}{6}$

$⇒n-1=6⇒n=7$

Hence, the value of n is 7.