Question 8

Initially, the box contains 3 green balls ($G$) and 2 orange balls ($O$), making 5 balls in total.

Let $E_1$ be the color of the first drawn ball and $O_2$ be the event of drawing an orange ball in the second draw:

• Case 1: First ball drawn is green ($G_1$). The probability is $P(G_1) = \frac{3}{5}$. Since green balls are not replaced, the box now contains 2 green balls and 2 orange balls (4 balls total). The conditional probability of drawing orange on the second draw is $P(O_2 \mid G_1) = \frac{2}{4} = \frac{1}{2}$.
• Case 2: First ball drawn is orange ($O_1$). The probability is $P(O_1) = \frac{2}{5}$. Since orange balls are replaced with another orange ball, the box still contains 3 green balls and 2 orange balls (5 balls total). The conditional probability of drawing orange on the second draw is $P(O_2 \mid O_1) = \frac{2}{5}$.

By the law of total probability: