Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. If the last of the next three draws is red, what is the probability that the first is red?

Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. If the last of the next three draws is red, what is the probability that the first is red?

• A. 1/3
• B. 1/2
• C. 2/3
• D. 1

Answer: (B)

After removing one black ball, the box has two red and one black. We will draw all three without replacement, so the possible color orders are Red-Red-Black, Red-Black-Red, and Black-Red-Red, each with equal likelihood (1/3). If the last draw is red, the remaining possible sequences are Red-Black-Red and Black-Red-Red. In these two scenarios, the first draw is red in only one of them (Red-Black-Red). Since both are equally likely, the probability that the first draw is red given that the last draw is red is 1/2.

After removing one black ball, the box has two red and one black. We will draw all three without replacement, so the possible color orders are Red-Red-Black, Red-Black-Red, and Black-Red-Red, each with equal likelihood (1/3).

If the last draw is red, the remaining possible sequences are Red-Black-Red and Black-Red-Red. In these two scenarios, the first draw is red in only one of them (Red-Black-Red). Since both are equally likely, the probability that the first draw is red given that the last draw is red is 1/2.