Box $A$ contains 2 black and 3 red balls, while Box $B$ contains 3 black and 4 red balls. Out of these two…
Box $A$ contains 2 black and 3 red balls, while Box $B$ contains 3 black and 4 red balls. Out of these two boxes one is selected at random; and the probability of choosing Box $A$ is double that of Box $B$. If a red ball is drawn from the selected box, then the probability that it has come from Box $B$, is
$\frac{21}{41}$
$\frac{10}{31}$
$\frac{12}{31}$
$\frac{13}{41}$
Solution
Let $P(B)=p$ according to given condition
$\begin{aligned}
P(A) & =2 P(B)=2 p \\
P\left(\frac{R}{A}\right) & =\frac{{ }^3 C_1}{{ }^5 C_1}=\frac{3}{5}
\end{aligned}$
and
$P\left(\frac{R}{B}\right)=\frac{{ }^4 C_1}{{ }^7 C_1}=\frac{4}{7}$
Using Baye's theorem
$\begin{aligned}
P\left(\frac{B}{R}\right) & =\frac{P(B) \cdot P\left(\frac{R}{B}\right)}{P(A) \cdot P\left(\frac{R}{A}\right)+P(B) \cdot P\left(\frac{R}{B}\right)} \\
& =\frac{p \cdot \frac{4}{7}}{2 p \cdot \frac{3}{5}+p \cdot \frac{4}{7}} \\
& =\frac{\frac{4}{7}}{\frac{6}{5}+\frac{4}{7}}=\frac{\frac{4}{7}}{\frac{42+20}{35}} \\
& =\frac{20}{62}=\frac{10}{31}
\end{aligned}$