Question

A fair coin is tossed $n$ times. If the probability that head occurs $6$ times is equal to the probability that head occurs $8$ times, then $n$ is equal to

• A. $15$
• B. $14$
• C. $12$
• D. $7$

Answer 1

The correct option is B

$14$

The explanation for the correct option:

Step1: Define events:

Number of times coin tossed$=n$

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

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

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

As per the condition given,

The probability of getting seven heads is equal to the probability of getting nine heads.

${}^{n}C_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^{n-6}={}^{n}C_8{\left(\frac{1}{2}\right)}^8{\left(\frac{1}{2}\right)}^{n-8}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{}{\mathbf{p}}^{\mathbf{r}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{r}}\mathbf{]}$

$\Rightarrow {}^{n}C_6{\left(\frac{1}{2}\right)}^n={}^{n}C_8{\left(\frac{1}{2}\right)}^n$

$$\begin{gathered} \Rightarrow {}^{n}C_6={}^{n}C_8 \\ \Rightarrow n=6+8\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{a}}\mathbf{}\mathbf{=}{}^{\mathbf{n}}\mathit{C}_{\mathbf{b}}\mathbf{}\mathbf{\Rightarrow }\mathbf{n}\mathbf{=}\mathbf{}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{]} \\ \Rightarrow n=14 \end{gathered}$$

Hence, option(B) is the correct answer.