#### Question

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is

15/2

^{8}2/15

15/2

^{13}None of these

#### Solution

15/2^{13}

Let *X *denote the number of heads in a fixed number of tosses of a coin .Then, *X *is a binomial variate with parameters ,

\[n \text{ and } p = \frac{1}{2}\]

Given that

*P*(*X*=7) =*P*(*X*= 9).Also, p = q = 0.5

\[P(X = r) = ^{n}{}{C}_r (0 . 5 )^r (0 . 5 )^{n - r} = ^{n}{}{C}_r (0 . 5 )^n \]

\[ \therefore P(X = 7) = ^{n}{}{C}_7 (0 . 5 )^n \]

\[ \text{ and} \ P(X = 9) =^{n}{}{C}_9 (0 . 5 )^n \]

\[\text{ It is given that } P(X = 7) = P(X = 9)\]

\[ \therefore ^{n}{}{C}_7 (0 . 5 )^n =^{n}{}{C}_9 (0 . 5 )^n \]

\[ \Rightarrow \frac{n!}{7! \left( n - 7 \right) !} = \frac{n!}{9! \left( n - 9 \right) !}\]

\[ \Rightarrow 9 \times 8 = \left( n - 7 \right)\left( n - 8 \right)\]

\[ \Rightarrow n^2 - 8n - 7n + 56 = 72\]

\[ \Rightarrow n^2 - 15n - 16 = 0\]

\[ \Rightarrow \left( n + 1 \right)\left( n - 16 \right) = 0\]

\[ \Rightarrow n = - 1 \ or \ n = 16 \]

\[ \Rightarrow n = - 1 (\text{ Not possible as n denotes the number of tosses of a coin } )\]

\[ \therefore n = 16 \]

\[\text{ Hence,} \ P(X = 2) = ^{16}{}{C}_2 (0 . 5 )^{16} \]

\[ = \frac{16 . 15}{2} \times \frac{1}{2^{16}}\]

\[ = \frac{15}{2^{13}}\]

\[ \therefore P(X = 7) = ^{n}{}{C}_7 (0 . 5 )^n \]

\[ \text{ and} \ P(X = 9) =^{n}{}{C}_9 (0 . 5 )^n \]

\[\text{ It is given that } P(X = 7) = P(X = 9)\]

\[ \therefore ^{n}{}{C}_7 (0 . 5 )^n =^{n}{}{C}_9 (0 . 5 )^n \]

\[ \Rightarrow \frac{n!}{7! \left( n - 7 \right) !} = \frac{n!}{9! \left( n - 9 \right) !}\]

\[ \Rightarrow 9 \times 8 = \left( n - 7 \right)\left( n - 8 \right)\]

\[ \Rightarrow n^2 - 8n - 7n + 56 = 72\]

\[ \Rightarrow n^2 - 15n - 16 = 0\]

\[ \Rightarrow \left( n + 1 \right)\left( n - 16 \right) = 0\]

\[ \Rightarrow n = - 1 \ or \ n = 16 \]

\[ \Rightarrow n = - 1 (\text{ Not possible as n denotes the number of tosses of a coin } )\]

\[ \therefore n = 16 \]

\[\text{ Hence,} \ P(X = 2) = ^{16}{}{C}_2 (0 . 5 )^{16} \]

\[ = \frac{16 . 15}{2} \times \frac{1}{2^{16}}\]

\[ = \frac{15}{2^{13}}\]

Is there an error in this question or solution?

Solution A Fair Coin is Tossed a Fixed Number of Times. If the Probability of Getting Seven Heads is Equal to that of Getting Nine Heads, the Probability of Getting Two Heads is (A) 15/28 (B) 2/15 (C) 15/213 Concept: Bernoulli Trials and Binomial Distribution.