Two bags A and B each contain a mixture of red balls and black balls. Bag A cont
ID: 3086900 • Letter: T
Question
Two bags A and B each contain a mixture of red balls and black balls. Bag A contains a total of 10 balls of which 5 are red and 5 are black. Bag B contains a total of 16 balls of which 8 are red and 8 are black. In step 1 of a game, a blind folded person chooses one of the two bags with equal probability. Still blind folded, in step 2 of the game he chooses 3 balls without replacement from the bag he chose in the step 1 of the game. (a) What is the probability that he will get 3 red balls in step 2, if he chose bag A in step 1? (b) If he got 3 red balls in step 2, what is the probability that he chose bag B in step 1?.Explanation / Answer
probability of choosing the bag = 1/2
probability of getting 3 red balls from bag A = 5C3/10C3 = 10/120 = 1/12
probability of getting 3 red balls from bag B = 8C3/16C3 = 1/10
Probability that bag A was chosen in step 1 & 3 red balls are drawn = [1/2*1/12]/[1/2*1/10 + 1/2*1/12]
P = 1/12 * 120/22
P = 5/11................................(a)
Probability that bag B was chosen in step 1 & 3 red balls are drawn = [1/2*1/10]/[1/2*1/10 + 1/2*1/12]
P = 1/10 * 120/22
P = 6/11...............................(b)