A fair coin is tossed 100 times. The probability of getting a head for even number of times is
A fair coin is tossed 100 times. The probability of getting a head for even number of times is
$\frac{1}{2}$
$\frac{3}{8}$
$\frac{1}{8}$
$\frac{3}{4}$
Solution
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 $=$
$\begin{aligned}
& \mathrm{P}(\mathrm{X}=2)+(\mathrm{X}=4)+\ldots . .+(\mathrm{X}=100)] \\
& =\left[{ }^{100} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{98}+\ldots .+{ }^{100} \mathrm{C}_{100}\left(\frac{1}{2}\right)^{100}\left(\frac{1}{2}\right)^{\circ}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_4+\ldots .+{ }^{100} \mathrm{C}_{100}\right] \\
& =\left(\frac{1}{2}\right)^{100}\left[2^{100-1}\right]=\frac{1}{(2)^{100}} \times(2)^{99}=\frac{1}{2}
\end{aligned}$