Question

If a fair coin is tossed $5$ times, then the probability that we get exactly $2$ heads is
$\left(a\right)\frac{5}{16}\left(b\right)\frac{5}{32}$
$\left(c\right)\frac{3}{8}\left(d\right)\frac{1}{4}$

Answer 1

Let $p$ and $q$ denote the probability of getting head and tail respectively in each toss.
$\therefore p=\frac{1}{2},q=1-\frac{1}{2}=\frac{1}{2}$

By using Bernoulli trials,
$P(X=x)={}^n{C}_x{p}^x{q}^{n-x}$
$={}^n{C}_x{\left(\frac{1}{2}\right)}^n(\because p=q=\frac{1}{2})$

$\therefore$ Probability of getting exactly $2$ heads, when the coin is tossed $5$ times$(n=5)$ is,
$P(X=2){=}^5{C}_2{\left(\frac{1}{2}\right)}^5=10{\left(\frac{1}{2}\right)}^5=\frac{5}{16}$
Ans : $\left(a\right)$