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 \rightarrow $ selected coin got exactly $3$ heads
$B_{1} \rightarrow $ coin $B_{1}$ is selected
$B_{2} \rightarrow $ coin $B_{2}$ is selected
$B_{3} \rightarrow $ 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,
$P \left(\frac{B_{1}}{A}\right) = \frac{P \left(A \cap B_{1}\right)}{P \left(A\right)} = \frac{P \left(B_{1}\right) \cdot P \left(\frac{A}{B_{1}}\right)}{P \left(B_{1}\right) P \left(\frac{A}{B_{1}}\right) + P \left(B_{2}\right) P \left(\frac{A}{B_{2}}\right) + P \left(B_{3}\right) P \left(\frac{A}{B_{3}}\right)}$
$=\frac{\frac{1}{3} \times \left(\frac{1}{2}\right)^{3}}{\frac{1}{3} \times \left(\frac{1}{2}\right)^{3} + \frac{1}{3} \times \left(\frac{1}{4}\right)^{3} + \frac{1}{3} \left(\frac{1}{8}\right)^{3}}=\frac{64}{64 + 8 + 1}=\frac{64}{73}$