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

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

The explanation for the correct option:

Step1. Define events:

Number of times coin tossed$=n$

Number of getting heads $=X$

Probability of getting heads $\text{'}p\text{'}=\frac{1}{2}$

Probability of getting tails $\text{'}q\text{'}=\frac{1}{2}$ $\left(\because p+q=1\right)$

Step2.Calculate the value of 'n':

Given

Probability of getting seven heads $=$ Probability of getting nine heads

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

Step3. Probability of getting two heads:

Probability of getting two heads is $P(X=2)={}^{n}C_r{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{16-2}$

$$\begin{gathered} ={}^{16}C_2{\left(\frac{1}{2}\right)}^{16} \\ =\frac{16!}{\left(16-2\right)!2!}\frac{1}{2^{16}} \\ =\frac{\mathbf{15}}{{\mathbf{2}}^{\mathbf{13}}} \end{gathered}$$

Hence, option(B) is the correct answer.