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
(a) 15/2⁸
(b) 2/15
(c) 15/2¹³
(d) None of these

Answer 1

(c) 15/2¹³

Let X denote the number of heads in a fixed number of tosses of a coin .Then, X is a binomial variate
with parameters $n\mathrm{and}p=\frac{1}{2}$

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

$$\begin{gathered} P(X=r)={}^{\mathit{n}}C_{\mathit{r}}(0.5{)}^r(0.5{)}^{n\mathit{-}r}={}^{n}C_r(0.5{)}^n \\ \therefore P(X=7)={}^{n}C_7(0.5{)}^n \\ \mathrm{and}P\mathit{(}X=9)\mathit{=}{}^{\mathit{n}}C_{\mathit{9}}\mathit{(}0.5{)}^{\mathit{n}} \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}P(X=7)=P\mathit{(}X=9) \\ \therefore {}^{n}C_7(0.5{)}^n={}^{n}C_{\mathit{9}}\mathit{(}0.5{)}^n \\ \Rightarrow \frac{n!}{7!\left(n-7\right)!}=\frac{n!}{9!\left(n-9\right)!} \\ \Rightarrow 9\times 8=\left(n-7\right)\left(n-8\right) \\ \Rightarrow n^2-8n-7n+56=72 \\ \Rightarrow n^2-15n-16=0 \\ \Rightarrow \left(n+1\right)\left(n-16\right)=0 \\ \Rightarrow n=-1orn=16 \\ \Rightarrow n=-1(\mathrm{Not}\mathrm{possible}\mathrm{as}\mathrm{n}\mathrm{denotes}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{coin}) \\ \therefore \mathit{}n=16 \\ \mathrm{Hence},P(X=2)={}^{16}C_2(0.5{)}^{16} \\ =\frac{16.15}{2}\times \frac{1}{2^{16}} \\ =\frac{15}{2^{13}} \end{gathered}$$