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, an urn is chosen 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 balls is $\frac{1}{16}$then the value of$\frac{\mathrm{n}}{2}$ is

Let A1= Event of choosing one of the first n urns which
contain 4 white, 6 black balls
A2= Event of getting (n+1)th urn which contain 5 white,
5 black balls
Let E = event of getting both balls black when 2 balls
are drawn from the urn chosen.
Given $\mathrm{P}\left(\frac{{\mathrm{A}}_2}{\mathrm{E}}\right)=\frac{1}{16}\Rightarrow$by using Baye’s Theorem,
$\frac{10}{15\mathrm{n}+10}=\frac{1}{16}\Rightarrow \frac{2}{3\mathrm{n}+2}=\frac{1}{16}\Rightarrow \mathrm{n}=10\text{ or }\frac{\mathrm{n}}{2}=5$
We have$\mathrm{P}\left(\mathrm{E}\right)=\frac{75}{100}=\frac{3}{4}\text{ and }\mathrm{P}\left(\mathrm{F}\right)=\frac{80}{100}=\frac{4}{5}$