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 last two draws among the next three 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 last two draws among the next three are both red?

• A. 1/3
• B. 2/3
• C. 1
• D. 0

Answer: (A)

After the initial black ball is drawn, three balls remain: two red and one black. To have the last two of the next three draws be red, the first draw of that trio must be the black ball. If the first draw is black, the remaining two balls are both red, so the second and third draws will both be red. The chance the first of the three draws is black is 1 out of 3, since there is one black ball among three. Therefore the probability that the last two draws are red is 1/3.

After the initial black ball is drawn, three balls remain: two red and one black. To have the last two of the next three draws be red, the first draw of that trio must be the black ball. If the first draw is black, the remaining two balls are both red, so the second and third draws will both be red. The chance the first of the three draws is black is 1 out of 3, since there is one black ball among three. Therefore the probability that the last two draws are red is 1/3.