Browse Questions

# A fair coin is tossed 100 times. The probability of getting tails on odd number of times is

$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{1}{8} \\ (C)\;\large\frac{3}{8} & \quad (D)\;None\: of \: these \end {array}$

Let X denote the number of tails. Then X is a binomial variate with parameters
$n = 100\: \: \: \: P = \large\frac{1}{2}$
$\therefore P(x=r)=100C_r \bigg( \large\frac{1}{2} \bigg) ^{100}\: \: \: \: r=0,1,2,.....100$
Required probability = $P(x=1) +P(x=3)+.....P(x=99)$
$= \bigg( \large\frac{1}{2} \bigg)^{100} [100C_1+100C_3+.....+100C_{99} ]$
$\large\frac{1}{2^{100}} [2^{99} ]$
$\large\frac{1}{2}$
Hence (A)