Question

A fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to the probability of getting $9$ heads, then the probability of getting $2$ heads is:

• A. $\frac{15}{2^{12}}$
• B. $\frac{15}{2^{13}}$
• C. $\frac{15}{2^{14}}$
• D. $\frac{15}{2^8}$

Answer 1

The correct option is B

$\frac{15}{2^{13}}$

Explanation for the correct option:

Step1. Calculate the value of $\mathrm{n}$:

According to the given condition, we have

Probability of getting $9$ heads $=$ Probability of getting $7$ heads

$$\begin{gathered} \Rightarrow P(x=9)=P(x=7) \\ \Rightarrow {}^{n}C_9{{\left(\frac{1}{2}\right)}^{n-9}\left(\frac{1}{2}\right)}^9={}^{n}C_7{{\left(\frac{1}{2}\right)}^{n-7}\left(\frac{1}{2}\right)}^7 \\ \Rightarrow {}^{n}C_9{\left(\frac{1}{2}\right)}^{n-9+9}={}^{n}C_7{\left(\frac{1}{2}\right)}^{n-7+7} \\ \Rightarrow {}^{n}C_9{\left(\frac{1}{2}\right)}^n={}^{n}C_7{\left(\frac{1}{2}\right)}^n \\ \Rightarrow {}^{n}C_9={}^{n}C_7 \\ \Rightarrow n=9+7\mathbf{}\mathbf{\left(}\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{p}}\mathbf{\Rightarrow }\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{r}\mathbf{}\mathbf{+}\mathbf{p}\mathbf{\right)} \\ \Rightarrow n=16 \end{gathered}$$

Step2. Calculate the probability of getting $2$ heads:

We have $P(x=2)={}^{16}C_2{\left(\frac{1}{2}\right)}^{16-2}{\left(\frac{1}{2}\right)}^2$

$$\begin{gathered} ={}^{16}C_2{\left(\frac{1}{2}\right)}^{16-2+2} \\ ={}^{16}C_2{\left(\frac{1}{2}\right)}^{16} \\ =\frac{16!}{\left(16-2\right)!2!}{\left(\frac{1}{2}\right)}^{16}\left(\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right) \\ =\frac{16!}{14!2!}{\left(\frac{1}{2}\right)}^{16} \\ \mathbf{=}\frac{\mathbf{15}}{{\mathbf{2}}^{\mathbf{13}}} \end{gathered}$$

Hence, Option(B) is the correct answer.