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

Answer: (C)

$P(H)-P(T)-1/2$ (with a single coin)
The required probability $=P$(head occurs $1$ or $3$ ... or $(2n+1)$ times)
$\therefore P\left(X=r\right)={}^{2n+1}{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{2n+1-r}$
$\therefore P(1)+P(3)+P(5)+....+P(2n+1)$
$={}^{2n+1}{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{2n}+{}^{2n+1}{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{2n-2}+$
$...+{}^{2n+1}{C}_{2n+1}{\left(\frac{1}{2}\right)}^{2n+1}$
$=\frac{1}{2^{2n+1}}\left[{}^{2n+1}{C}_1+{}^{2n+1}{C}_3+...+{}^{2n+1}{C}_{2n+1}\right]$
$\frac{1}{2^{2n+1}}\times 2^{2n}=\frac{1}{2}$
[As ${\left(1+x\right)}^{2n+1}={}^{2n+1}{C}_0+{}^{2n+1}{C}_1{x}^1+$
$...+{}^{2n+1}{C}_{2n+1}{x}^{2n+1}$
putting $x=1,-1$ and then adding, we get the value of
${}^{2n+1}{C}_1+{}^{2n+1}{C}_3+....+{}^{2n+1}{C}_{2n+1}=2^{2n}]$