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)
$={}^{100}{C}_1(\frac{1}{2}{)}^1(\frac{1}{2}{)}^{99}+{}^{100}{C}_3(\frac{1}{2}{)}^3(\frac{1}{2}{)}^{97}$
$+{}^{100}{C}_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^{95}+...+{}^{100}{C}_{99}{\left(\frac{1}{2}\right)}^{99}{\left(\frac{1}{2}\right)}^1$
$=\frac{1}{2^{100}}\left[{}^{100}{C}_1+{}^{100}{C}_3+{}^{100}{C}_5+...+{}^{100}{C}_{99}\right]$
$=\frac{1}{2^{100}}\left[{}^{100}{C}_1+{}^{100}{C}_3+{}^{100}{C}_5+...{+}^{100}{C}_{99}\right]$
$=\frac{1}{2^{100}}\times 2^{99}=\frac{1}{2}$