Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. What is the probability that the next two draws are both red?

Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. What is the probability that the next two draws are both red?

• A. 1/3
• B. 1/2
• C. 2/3
• D. 0

Answer: (A)

When draws happen without replacement, the outcome of one draw affects the next, so you multiply the probabilities of each sequential event. After removing one black ball, the box has 2 red and 1 black, a total of 3 balls. For two reds in a row, the first draw must be red: probability 2/3. If that happens, there are now 2 balls left: 1 red and 1 black, so the second draw being red has probability 1/2. Multiply these steps: 2/3 × 1/2 = 1/3. So the chance of both of the next draws being red is 1/3. The other numbers correspond to single-step chances (2/3) or a conditional second-step chance (1/2), not their joint probability.

When draws happen without replacement, the outcome of one draw affects the next, so you multiply the probabilities of each sequential event.

After removing one black ball, the box has 2 red and 1 black, a total of 3 balls. For two reds in a row, the first draw must be red: probability 2/3. If that happens, there are now 2 balls left: 1 red and 1 black, so the second draw being red has probability 1/2. Multiply these steps: 2/3 × 1/2 = 1/3. So the chance of both of the next draws being red is 1/3. The other numbers correspond to single-step chances (2/3) or a conditional second-step chance (1/2), not their joint probability.