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 two red balls appear consecutively before the black ball appears?

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 two red balls appear consecutively before the black ball appears?

• A. 1/3
• B. 2/3
• C. 1
• D. 0

Answer: (A)

After removing one black ball, three balls remain: two red and one black. We’ll draw them in some order, and every order of these three balls is equally likely. The possible color sequences are: red-red-black, red-black-red, and black-red-red. Only the red-red-black sequence has the two red balls appearing consecutively before the black ball, so there are 3 equally likely outcomes and 1 favorable one. This gives a probability of 1/3. (If you prefer counting distinct reds, there are 6 permutations; 2 of them meet the condition, 2/6 = 1/3.)

After removing one black ball, three balls remain: two red and one black. We’ll draw them in some order, and every order of these three balls is equally likely. The possible color sequences are: red-red-black, red-black-red, and black-red-red. Only the red-red-black sequence has the two red balls appearing consecutively before the black ball, so there are 3 equally likely outcomes and 1 favorable one. This gives a probability of 1/3. (If you prefer counting distinct reds, there are 6 permutations; 2 of them meet the condition, 2/6 = 1/3.)