Question

A fair coin is tossed a fixed number of items. If the probability of getting $7$ heads is equal to that of getting $9$ heads, then the probability of getting$3$ heads is

• A. $\frac{35}{2^{12}}$
• B. $\frac{35}{2^{14}}$
• C. $\frac{7}{2^{12}}$
• D. None of these

Answer 1

The correct option is A

$\frac{35}{2^{12}}$

The explanation for the correct option:

Step 1. Define events:

Number of times coin tossed $=n$

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

Probability of not getting heads $\text{'}\mathrm{q}\text{'}=1-\mathrm{p}$

$=\frac{1}{2}$

Step2.Calculate the value of $\text{'}n\text{'}$:

As per the condition given

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

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

Step3. Calculate the probability of getting $3$ heads:

Probability of getting $3$ heads is ${}^{16}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{16-3}$

$$\begin{gathered} ={}^{16}C_3{\left(\frac{1}{2}\right)}^{16} \\ =\frac{16!}{\left(16-3\right)!3!}{\left(\frac{1}{2}\right)}^{16}\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right] \\ =\frac{35}{2^{12}} \end{gathered}$$

Hence, Option(A) is the correct answer.