A fair coin is tossed ‘n’ number of times. The probability that head will turn up an even number of times, is equal to

$\frac{n-1}{2 n}$ $\frac{1}{2}$ $\frac{n+1}{2 n}$ $\frac{2^{n-1}-1}{2^n}$

$\frac{1}{2}$

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 $=\frac{2^{n-1}}{2^n}=\frac{1}{2}$