Question

A box contains $3$ coins $B_1,B_2,B_3$ and the probability of getting heads on the coins are $\frac{1}{2},\frac{1}{4},\frac{1}{8}$ respectively. If one of the coins is selected at random and tossed for $3$ times and exactly $3$ heads appeared, then the probability that it was coin $B_1$ is

• A. $\frac{9}{73}$
• B. $\frac{10}{73}$
• C. $\frac{36}{73}$
• D. $\frac{64}{73}$

Answer 1

Answer: (D)

$A\to$ selected coin got exactly $3$ heads
$B_1\to$ coin $B_1$ is selected
$B_2\to$ coin $B_2$ is selected
$B_3\to$ coin $B_3$ is selected
$P\left(B_1\right)=P\left(B_2\right)=P\left(B_3\right)=\frac{1}{3}$
Using Bayes Theorem,
$P\left(\frac{B_1}{A}\right)=\frac{P\left(A\cap B_1\right)}{P\left(A\right)}=\frac{P\left(B_1\right)\cdot P\left(\frac{A}{B_1}\right)}{P\left(B_1\right)P\left(\frac{A}{B_1}\right)+P\left(B_2\right)P\left(\frac{A}{B_2}\right)+P\left(B_3\right)P\left(\frac{A}{B_3}\right)}$
$=\frac{\frac{1}{3}\times {\left(\frac{1}{2}\right)}^3}{\frac{1}{3}\times {\left(\frac{1}{2}\right)}^3+\frac{1}{3}\times {\left(\frac{1}{4}\right)}^3+\frac{1}{3}{\left(\frac{1}{8}\right)}^3}=\frac{64}{64+8+1}=\frac{64}{73}$