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

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

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A)
Total number of cases = $2^{100}$
Number of favourable cases = $100C_1+100C_3+....100C_{99}$
$= 2^{100-1}$
$= 2^{99}$
$\therefore$ Required probability = $\large\frac{2^{99}}{2^{100}}$
$\large\frac{1}{2}$
Ans : (A)