Question

Five coins whose faces are marked $2,3$ are tossed. The chance of obtaining a total of $12$ is

• A. $\frac{1}{32}$
• B. $\frac{1}{16}$
• C. $\frac{3}{32}$
• D. $\frac{5}{16}$

Answer 1

The correct option is D $\frac{5}{16}$

Let $2,3,2,3,2=12$

$\therefore$ Favourable case $=\frac{5!}{3!2!}$

$=\frac{5\times 4\times 3!}{3!2!}$

$=\frac{5\times 4^2}{2\times 1}=10$

Total cases $=2^5=32$

Probability (of obtaining a total of $12$)

$=\frac{10}{32}=\frac{5}{16}$

Probability for obtaining $2$ is $P=\frac{1}{2}$

$q=\frac{1}{2}$

$\therefore$ probability for obtaining $2$ or $3$ coins

out of $5$ coins is

$P\left(A\right)={}^5{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^2$

$=\frac{10}{32}=\frac{5}{15}$

Option $D$ is correct