Question

There are two urns$U_1$ and $U_2$. $U_1$ contains $2$ white and $8$ black balls and $U_2$ contains $4$ white and $6$ black balls. One urn is chosen at random and a ball is drawn and its colour noted and replaced. The process is repeated $3$ times and as a result one ball of white colour and $2$ of black are obtained. The probability that the urn chosen was $U_1$ is:

• A. $\frac{8}{17}$
• B. $\frac{7}{17}$
• C. $\frac{6}{17}$
• D. $\frac{5}{17}$

Answer 1

The correct option is A $\frac{8}{17}$

Let $E_1$ be the event of choosing urn $U_1$ and $E_2$ be the event of choosing urn $U_2$.
Let '$A$' be the event of drawing $3$ balls successively with replacement, so that the balls obtained are $1$ white, $2$ black in any order.
$P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$
$P\left(\frac{A}{E_1}\right)=\frac{2}{10}\times \frac{8}{10}\times \frac{8}{10}\times 3\text{and}P\left(\frac{A}{E_2}\right)=\frac{4}{10}\times \frac{6}{10}\times \frac{6}{10}\times 3$
$P\left(E_1/A\right)=\left(\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}\right)=\frac{\frac{1}{2}(\frac{2}{10}.\frac{8}{10}.\frac{8}{10})3}{\frac{1}{2}[(\frac{2}{10}.\frac{8}{10}.\frac{8}{10}).3+(\frac{4}{10}.\frac{6}{10}.\frac{6}{10}).3]}=\frac{\mathrm{2.8.8}}{\mathrm{2.8.8}+\mathrm{4.6.6}}=\frac{16\times 8}{16(8+9)}=\frac{8}{17}$