Question

The contents of urn I and II are as follows,
Urn I: $4$ white and $5$ black balls
Urn II:$ 3$ white and $6$ black balls
One urn is ehosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn, colour is noted and replaced. The process is repeated $4$ times and as a result one ball of white colour and $3$ of black colour are noted. Find the probability the chosen urn was $I$.

• A. $\frac{125}{287}$
• B. $\frac{64}{127}$
• C. $\frac{25}{287}$
• D. $\frac{79}{192}$

Answer 1

Answer: (A)

$A : 1$ ball is $W$ & $3$ black balls
$B_1$ : Urn 1 is chosen
$B _2$ : Urn 2 is chosen
$ P\left(B_1 / A\right)=\frac{P\left(A / B_1\right) P\left(B_1\right)}{P\left(A / B_1\right) P\left(B_1\right)+P\left(A / B_2\right) P\left(B_2\right)} $
$P\left(B_1 / A\right)=\frac{\frac{1}{2} \times \frac{4}{9} \times\left(\frac{5}{9}\right)^3 \times{ }^4 C_3}{\frac{1}{2} \times \frac{4}{9} \times\left(\frac{5}{9}\right)^3 \times{ }^4 C_3+\frac{1}{2} \times \frac{3}{9} \times\left(\frac{6}{9}\right)^3 \times{ }^4 C_3}=\frac{125}{287}$