A fair coin is tossed ‘n’ number of times. The probability that head will turn up an even number of times, is equal to
Answer & explanation
Correct answer: option 2
Total outcomes $=2^{n}$
Total number of favourable outcomes
$={ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots .{ }^n C_{\left[\frac{n}{2}\right]}$
$=2^{n-1}$
Thus, required probability $=\frac{2^{n-1}}{2^n}=\frac{1}{2}$