Question

Each of the $n$ urns contains $4$ white and $6$ black balls. The $(n+1)th$ urn contains $5$ white and $5$ black balls. Out of the $(n+1)$ urns is chosen an urn at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the $(n+1)th$ urn was chosen to draw the ball is $\frac{1}{6}$, then the value of $n$ is

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is A $10$

Let $E_1:$ event that one of the first $n$ urn is chosen
$E_2:$ event that $(n+1)th$ urn is chosen
$A:$ The vent that the two balls drawn from the urn without replacement are black.
Then we have $P\left(E_1\right)=\frac{n}{n+1},P\left(E_2\right)=\frac{1}{n+1}$
$P\left(\frac{A}{E_1}\right)=\frac{{}^6{C}_2}{{}^{10}{C}_2}=\frac{1}{3}$ and $P\left(\frac{A}{E_2}\right)=\frac{{}^5{C}_2}{{}^{10}{C}_2}=\frac{2}{9}$