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

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

• A. 0
• B. 1/3
• C. 1/2
• D. 2/3

Answer: (A)

After removing one black ball, three balls remain: two red and one black. We’re about to draw three times without replacement. If the first two draws are both red, that uses up both red balls, leaving only the black ball for the third draw. Therefore, it’s impossible for the third draw to be red, making the probability zero.

After removing one black ball, three balls remain: two red and one black. We’re about to draw three times without replacement. If the first two draws are both red, that uses up both red balls, leaving only the black ball for the third draw. Therefore, it’s impossible for the third draw to be red, making the probability zero.