Consider the following events: $E_1$ = Getting 5 OR 6 in a single throw of the die $E_2$ = Getting 1, 2, 3 OR 4 in a single throw of the die A = Getting exactly 2 heads We have to find, P($E_2$/A). Since P ($E_2$/A) = Now, $P(E_1)$ = $\frac{2}{6}$ = $\frac{1}{3}$ and $P(E_2)$ = $\frac{4}{6}$ = $\frac{2}{3}$ Also, P (A|$E_1$) = Probability of getting exactly 2 heads when a coin is tossed 3 times = $\frac{3}{8}$ And, P (A|$E_2$) = Probability of getting 2 heads when a coin is tossed 2 times = $\frac{1}{2}$ ∴ P ($E_2$|A) = $\frac{\frac{1}{2} \times \frac{2}{3}}{\frac{3}{8} \times \frac{1}{3} + \frac{1}{2} \times \frac{2}{3}}$ = $\frac{\frac{1}{3}}{\frac{1}{3} \left( \frac{3}{8} + 1 \right)}$ = $\frac{\frac{1}{3}}{\frac{1}{3} \left( \frac{8+3}{8} \right)}$ = $\frac{8}{11}$