Consider the following events:
${\mathit{E}}_1$ = Getting 5 OR 6 in a single throw of the die
${\mathit{E}}_2$ = Getting 1, 2, 3 OR 4 in a single throw of the die
A = Getting exactly 2 heads
We have to find, P(${\mathit{E}}_2$/A).
Since P (${\mathit{E}}_2$/A) =
Now, $\mathit{P}\left({\mathit{E}}_1\right)$ = $\frac{2}{6}$ = $\frac{1}{3}$ and $\mathit{P}\left({\mathit{E}}_2\right)$ = $\frac{4}{6}$ = $\frac{2}{3}$
Also,
P (A|${\mathit{E}}_1$) = Probability of getting exactly 2 heads when a coin is tossed 3 times = $\frac{3}{8}$
And, P (A|${\mathit{E}}_2$) = Probability of getting 2 heads when a coin is tossed 2 times = $\frac{1}{2}$
∴ P (${\mathit{E}}_2$|A) = $\frac{\frac{1}{2}\times \frac{2}{3}}{\frac{3}{8}\times \frac{1}{3}+\frac{1}{2}\times \frac{2}{3}}$ = $\frac{\frac{1}{3}}{\frac{1}{3}\left(\frac{3}{8}+1\right)}$ = $\frac{\frac{1}{3}}{\frac{1}{3}\left(\frac{8+3}{8}\right)}$ = $\frac{8}{11}$