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}} \times2^{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
$\,{}^{2n+1}C_{1} +\,{}^{2n+1}C_{3} + ....+\,{}^{2n+1}C_{2n+1} = 2^{2n}]$