A bucket contains 5 red balls and 2 green balls. A ball is chosen at random and removed from the bucket. A ball of the opposite color is then placed in the bucket. Then a second ball is chosen from the bucket. If the second ball is green, what is the probability that the first ball was green?

Question

anonymous · Answer

Since the second is known to be green, there is only 1 way out of 7 in which the first can be green, so that the probability is 1/7

anonymous · Answer

Luis_Rivera, I'm getting 2/17, not 1/7.

The prob. of choosing green-green is 2/7 * 1/7 = 2/49
The prob. of choosing green-red is 2/7 * 6/7 = 12/49
The prob. of choosing red-green is 5/7 * 3/7 = 15/49
The prob. of choosing red-red is 5/7 * 4/7 = 20/49

After you know you chose green for the second draw, you know you either chose green-green or red-green. The comparitive probability between those two options is 2/17.

anonymous · Answer

no, no, theres only 1 green left, and it asks for the first !!!

anonymous · Answer

Did you miss the part about adding a ball of opposite color in between the draws? If the first ball chosen is a red ball, for the second draw there will be 4 red and 3 green balls in the bucket.

anonymous · Answer

yes, thats why it cant be green cuz its the opposite color !!

i aint got time for this