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 three-draw sequence ends with 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 three-draw sequence ends with red?

• A. 2/3
• B. 1/3
• C. 1
• D. 0

Answer: (A)

With one black ball already drawn, the box now has 2 red and 1 black to be drawn three times without replacement. The last draw is simply the last ball in a random order of these three balls, and each ball is equally likely to appear last. Since there are two red balls among the three, the probability that the last ball drawn is red is 2 out of 3, i.e., 2/3.

With one black ball already drawn, the box now has 2 red and 1 black to be drawn three times without replacement. The last draw is simply the last ball in a random order of these three balls, and each ball is equally likely to appear last. Since there are two red balls among the three, the probability that the last ball drawn is red is 2 out of 3, i.e., 2/3.