Question

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

• A. $\frac{1}{256}$
• B. $\frac{7}{32}$
• C. $\frac{5}{32}$
• D. $\frac{3}{32}$

Answer 1

Answer: (B)

Throwing of eight coins simultaneously is the same as throwing of one coin $8$ times.
$\therefore$ Probability of getting a head, $p=\frac{1}{2}$
$\Rightarrow q=1-p=1-\frac{1}{2}=\frac{1}{2}$
Thus, we have a binomial distribution with $p=\frac{1}{2}$, $q=\frac{1}{2}$ and $n=8$.
Hence, required probability $={}^8{C}_3{p}^3{q}^5$
$=56\times {\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^5$
$=56\times \frac{1}{8}\times \frac{1}{32}=\frac{7}{32}$