A coin is tossed $5$ times. The probability of head is $\frac{1}{2}$. The probability of exactly $2$ heads is:

This follows a Binomial Distribution $P(X=r)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^r{q}^{n-r}$.Given:$n=5$ (number of trials)$r=2$ (number of successes)$p=\frac{1}{2}$ (probability of head)$q=1-p=\frac{1}{2}$ (probability of tail)Calculation:
$$P(2)=\left(\genfrac{}{}{0ex}{}{5}{2}\right){\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{5-2}$$
$$\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5\times 4}{2\times 1}=10$$
$$P(2)=10\times {\left(\frac{1}{2}\right)}^5=10\times \frac{1}{32}$$
$$P(2)=\frac{10}{32}=\frac{5}{16}$$