Question

Each of the urns contains $4$ white and $6$ blackballs. The (n + 1)th urn contains $5$ white and $5$black balls. One of the (n+1) urns is chosen atrandom and two balls are drawn from it withoutreplacement and both the balls turn out to beblack. Then the probability that the (n+1)th urn was chosen to draw the balls is $1/16$, the value of n is

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

Answer 1

The correct option is A $10$

$P\left(\frac{{(n+1)}^{th}urn}{bothblack}\right)=\frac{P({(n+1)}^{th}urn\cap bothblack)}{P\left(bothblack\right)}=\frac{\frac{1}{n+1}\frac{{}_2^{5}C}{{}_2^{10}C}}{\frac{1.n}{n+1}\frac{{}_2^{6}C}{{}_2^{10}C}+\frac{1}{n+1}\frac{{}_2^{5}C}{{}_2^{10}C}}=\frac{{}_2^{5}C}{n_2^{6}C{+}_2^{5}C}=\frac{10}{15n+10}=\frac{1}{16}$

$\Rightarrow 160=15n+10\Rightarrow n=10$