A fair coin is tossed 100 times. The probability of getting a head for even number of times is

We have \(n=100\) and probability of getting head \(=1 / 2\)
Let \(\mathrm{p}=1 / 2 \Rightarrow \mathrm{q}=1 / 2\)
Probability of getting head even number of times \(=\)
\( \mathrm{P}(\mathrm{X}=2)+(\mathrm{X}=4)+\ldots . .+(\mathrm{X}=100)] \)
\( =\left[{ }^{100} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{98}+\ldots .+{ }^{100} \mathrm{C}_{100}\left(\frac{1}{2}\right)^{100}\left(\frac{1}{2}\right)^{\circ}\right] \)
\( =\left(\frac{1}{2}\right)^{100}\left[{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_4+\ldots .+{ }^{100} \mathrm{C}_{100}\right] \)
\( =\left(\frac{1}{2}\right)^{100}\left[2^{100-1}\right]=\frac{1}{(2)^{100}} \times(2)^{99}=\frac{1}{2} \)