Question

A fair coin is tossed $ 100 $ times. The probability of getting tails an odd number of time is

• A. $ \frac{1}{2} $
• B. $ \frac{1}{4} $
• C. $ 0 $
• D. $ 1 $

Answer 1

Answer: (A)

Since, probability of getting tail in single coin is
$p = \frac{1}{2}, q = \frac{1}{2}$
$\therefore $ Required probability $ = p$ (getting tail in one time )
$+ ...+ P$ (getting tail in three times)
$+ P$ (getting tail in five times)
$+ P$ (getting tail in Ninety Nine times)
$= \,{}^{100} C_1 (\frac{1}{2})^1 (\frac{1}{2})^{99} + \,{}^{100}C_3 (\frac{1}{2})^3 (\frac{1}{2})^{97}$
$+\,{}^{100}C_{5} \left( \frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{95} + ... + \,{}^{100}C_{99} \left(\frac{1}{2}\right)^{99} \left(\frac{1}{2}\right)^{1} $
$ = \frac{1}{2^{100}} \left[ \,{}^{100}C_{1} + \,{}^{100}C_{3} +\,{}^{100}C_{5} + ...+\,{}^{100}C_{99} \right]$
$ = \frac{1}{2^{100}} \left[\,{}^{100}C_{1} + \,{}^{100}C_{3} + \,{}^{100}C_{5} + ...+ ^{100}C_{99} \right]$
$= \frac{1}{2^{100}} \times 2^{99} = \frac{1}{2}$