Eight coins are tossed together. The probability of getting exactly $3$ heads is

There are $2$ possible outcomes in a coin, which are head and tail,
$p=P\left(\text{Getting a head}\right)=\frac{1}{2}$ ...$\left(1\right)$
$q=P\left(\text{Not getting a tail}\right)=1-p$
$q=1-\frac{1}{2}=\frac{1}{2}$ ...$\left(2\right)$
The binomial distribution formula for any random variable $x$ is given by
$P(x=r)={}^n{C}_r{p}^r{q}^{n-r}$

where,
$n=\text{the number of experiments}$
$x=0,1,2,3,4,...$
$p=\text{Probability of success in a single experiment}$
and $q=\text{Probability of failure in a single experiment}(=1-p)$
Here eight coins are tossed together. The probability of getting exactly $3$ heads is to be find
Hence, $n=8$, $r=3$
$p=\text{probability of getting a head}$
$p=\frac{1}{2}$ and $q=\frac{1}{2}$ $\left[\text{From}\left(1\right)\&\left(2\right)\right]$
$\therefore$ The required probability
$={}^8{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{8-3}=\frac{8!}{5!\times 3!}{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^5$
$=\frac{8\times 7\times 6\times 5!}{5!\times 3\times 2}{\left(\frac{1}{2}\right)}^8$
$=\frac{7}{32}$

HEnce, the correct option is (b).