Question

A coin is tossed $2n$ times.

The chance that the number of times one gets head is not equal to the number of times one gets tail is

• A. $\frac{2n!}{{(n!)}^2}{\left(\frac{1}{2}\right)}^{2n}$
• B. $1-\frac{2n!}{{(n!)}^2}$
• C. $1-\left(\frac{2n!}{{(n!)}^2}\times \frac{1}{4^n}\right)$
• D. None of these

Answer 1

The correct option is C

$1-\left(\frac{2n!}{{(n!)}^2}\times \frac{1}{4^n}\right)$

The explanation for the correct option:-

Step 1: First find the probability of success and failure for the given event:

We have been given that, a coin is tossed $2n$ times.

We need to find the probability of the number of times one gets head is not equal to the number of times one gets tail.

Here,$p=\frac{1}{2},q=\frac{1}{2},N=2n,x=n$

Step 2: To find the required probability:

Use the binomial distribution formula

The formula for binomial distribution:

$\mathit{P}\mathbf{(}\mathit{x}\mathbf{:}\mathit{N}\mathbf{,}\mathit{p}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{N}}{\mathit{C}}_{\mathbf{x}}\mathbf{}{\mathit{p}}^{\mathbf{x}}\mathbf{}{\mathit{q}}^{\mathbf{(}\mathit{N}\mathbf{-}\mathbf{x}\mathbf{)}}$

Where,

$N=$ the number of experiments

$x=0,1,2,3,4,\dots$

$p=$Probability of Success in a single experiment

$q=$ Probability of Failure in a single experiment $=1–p$

By using, binomial distribution, the required probability is,

$=1-$( the probability of getting an equal number of heads or tails )

$$\begin{gathered} =1-\left({}^{2n}C_n\times {\left(\frac{1}{2}\right)}^n\times {\left(\frac{1}{2}\right)}^{2n-n}\right) \\ =1-\left(\frac{2n!}{n!(2n-n)!}\times \frac{1}{2^n}\times {\left(\frac{1}{2}\right)}^n\right) \\ =1-\left(\frac{2n!}{n!n!}\times \frac{1}{2^n}\times \frac{1}{2^n}\right) \\ =1-\left(\frac{2n!}{{(n!)}^2}\times \frac{1}{4^n}\right) \end{gathered}$$

Hence, the correct option is (C).