A box contains 2 green balls and 3 blue balls. A ball is selected at random from the box and not replaced. A second ball is then selected. What is the probability of selecting two balls of different colours?

There are 2 green balls (G) and 3 blue balls (B), making a total of $2+3=5$ balls.
The probability of selecting a green ball first is $P(G_1) = \frac{2}{5}$.
The probability of selecting a blue ball first is $P(B_1) = \frac{3}{5}$.

Since the first ball is not replaced, there will be 4 balls remaining for the second selection.

• First Selection:
  • Branch 1: Green (G)
    • Probability: $P(G_1) = \frac{2}{5}$
    • Remaining balls: 1 Green, 3 Blue (Total 4)
    • Second Selection (after G1):
      • Sub-branch 1.1: Green (G)
        • Probability: $P(G_2 | G_1) = \frac{1}{4}$
        • Combined outcome: GG
        • Combined probability: $P(G_1 \text{ and } G_2) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20}$
      • Sub-branch 1.2: Blue (B)
        • Probability: $P(B_2 | G_1) = \frac{3}{4}$
        • Combined outcome: GB
        • Combined probability: $P(G_1 \text{ and } B_2) = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20}$
  • Branch 2: Blue (B)
    • Probability: $P(B_1) = \frac{3}{5}$
    • Remaining balls: 2 Green, 2 Blue (Total 4)
    • Second Selection (after B1):
      • Sub-branch 2.1: Green (G)
        • Probability: $P(G_2 | B_1) = \frac{2}{4} = \frac{1}{2}$
        • Combined outcome: BG
        • Combined probability: $P(B_1 \text{ and } G_2) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20}$
      • Sub-branch 2.2: Blue (B)
        • Probability: $P(B_2 | B_1) = \frac{2}{4} = \frac{1}{2}$
        • Combined outcome: BB
        • Combined probability: $P(B_1 \text{ and } B_2) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20}$

The outcomes where two balls of different colours are selected are:

• Green then Blue (GB)
• Blue then Green (BG)

We sum the probabilities of the identified outcomes:

The probability of selecting two balls of different colours is $\boxed{\frac{3}{5}}$.