Each of the ‘n’ urns contains four white and six black balls. The (n + 1)^th urn contains five white and five black balls. One of the (n + 1) urns 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 1/16,then the value of n/5 is ____

Let E₁ denote the event that one of the first "n" urns is chosen and

 E₂denote the event that (n +1)^th urn is selected.
A denotes the event that two balls drawn are black, then
$\begin{array}{l}\mathrm{p}\left({\mathrm{E}}_1\right)=\frac{\mathrm{n}}{\mathrm{n}+1},\mathrm{p}\left({\mathrm{E}}_2\right)=\frac{1}{\mathrm{n}+1}\\ \mathrm{p}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)=\frac{{ }^6{\mathrm{C}}_2}{{ }^{10}{\mathrm{C}}_2}=\frac{1}{3}\mathrm{\&}\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_2}\right)=\frac{{ }^5{\mathrm{C}}_2}{{ }^{10}{\mathrm{C}}_2}=\frac{2}{9}\end{array}$
The probability that the (n + 1)^th urn was chosen to draw the balls is 1/16,

Using Bayes theorem
$$\begin{gathered} \Rightarrow \frac{1}{16}=\frac{\left(\frac{1}{\mathrm{n}+1}\right)\frac{2}{9}}{\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{\mathrm{n}+1}\right)\left(\frac{2}{9}\right)} \\ \Rightarrow \frac{1}{16}=\frac{2}{3n+2} \\ \Rightarrow 3n+2=32\Rightarrow 3n=30 \\ \Rightarrow \mathrm{n}=10\therefore \frac{n}{5}=2 \end{gathered}$$