Question

A coin is tossed $(2n+1)$ times, the probability that head appear odd number of times is

• A. $\frac{n}{2n+1}$
• B. $\frac{n+1}{2n+1}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}$

$P\left(H\right)=P\left(T\right)=\frac{1}{2}$ (with a single coin)
The required probability $=p$(that head occurs $1,3,...or(2n+1)$ times)
We know $P(X=r)={}^{2n+1}{\text{C}}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{2n+1-r}$
$\therefore P\left(1\right)+P\left(3\right)+P\left(5\right)+\cdots +P(2n+1)$

$={}^{2n+1}{\text{C}}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{2n}+{}^{2n+1}{\text{C}}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{2n-2}+\cdots +{}^{2n+1}{\text{C}}_{2n+1}{\left(\frac{1}{2}\right)}^{2n}$
$=\frac{1}{2^{2n+1}}[{}^{2n+1}{\text{C}}_1+{}^{2n+1}{\text{C}}_3+\cdots +{}^{2n+1}{\text{C}}_{2n+1}]$
$=\frac{1}{2^{2n+1}}\times 2^{2n}=\frac{1}{2}$

Hence, option C is correct.