Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. If the first draw is red, what is the probability that among the next two draws there is at least one red?

Assume a box contains 2 red balls and 2 black balls. One black ball has been drawn and not replaced. If the first draw is red, what is the probability that among the next two draws there is at least one red?

• A. 1
• B. 1/2
• C. 0
• D. 2/3

Answer: (A)

The situation after removing one black ball leaves 2 red and 1 black. The two draws in question are without replacement from this 3-ball set, and we are told the first of these two draws is red. If the first draw is red, then there is at least one red in the pair of draws automatically—the red you just drew counts toward the two-draw sequence. So the event “at least one red among the next two draws” occurs with certainty given that the first is red. Hence the probability is 1.

The situation after removing one black ball leaves 2 red and 1 black. The two draws in question are without replacement from this 3-ball set, and we are told the first of these two draws is red. If the first draw is red, then there is at least one red in the pair of draws automatically—the red you just drew counts toward the two-draw sequence. So the event “at least one red among the next two draws” occurs with certainty given that the first is red. Hence the probability is 1.