Question 1119373: Two containers A and B are different. The Container A contains three white balls and two red balls and container B contains only one white ball. A coin is thrown and if faced, a ball is randomly withdrawn from container A and placed in container B, but if it yields a crown, then two balls are randomly withdrawn from container A and placed in container B. Now a ball is withdrawn randomly from container B. Question: Which the probability of the ball to be white?
Answer by greenestamps(13200)   (Show Source):
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For a problem like this, it is a good exercise to calculate the probability of every possible outcome and verify that the sum of those probabilities is 1. That gives you some confidence that you have done all the calculations correctly.
There are three steps in the process: (1) flip a coin; (2) move balls from A to B; and (3) draw a ball from B. Label the different cases with ordered triples -- for example, (F,R,W) to represent the coin being a face, the ball moved from A to B being red, and the ball drawn from B being white.
(1) (F,W,W): P(F) = (1/2); P(W) = (3/5); P(W) = (2/2) --> P(F,W,W) = 6/20
(2) (F,W,R): P(F) = (1/2); P(W) = (3/5); P(R) = (0/2) --> P(F,W,R) = 0
(3) (F,R,W): P(F) = (1/2); P(R) = (2/5); P(W) = (1/2) --> P(F,R,W) = 2/20
(4) (F,R,R): P(F) = (1/2); P(R) = (2/5); P(R) = (1/2) --> P(F,R,R) = 2/20
That takes care of all the cases where the coin comes up face. The probability that the coin comes up face is 1/2, and the sum of the probabilities for all these cases is 10/20 = 1/2; so our calculations are probably correct to this point.
(5) (C,WW,W): P(C) = (1/2); P(WW) = (3/10); P(W) = 3/3 --> P(C,WW,W) = 9/60
(6) (C,WW,R): P(C) = (1/2); P(WW) = (3/10); P(R) = 0/3 --> P(C,WW,R) = 0
(7) (C,WR,W): P(C) = (1/2); P(WR) = (6/10); P(W) = 2/3 --> P(C,WR,W) = 12/60
(8) (C,WR,R): P(C) = (1/2); P(WR) = (6,10); P(R) = 1/3 --> P(C,WR,R) = 6/60
(9) (C,RR,W): P(C) = (1/2): P(RR) = (1/10); P(W) = 1/3 --> P(C,RR,W) = 1/60
(10)(C,RR,R): P(C) = (1/2); P(RR) = (1/10); P(R) = 2/3 --> P(C,RR,R) = 2/60
Again these last 6 cases take care of all the cases where the coin comes up crown, and the sum of the probabilities is 1/2. So it is very likely that all our calculations are correct.
So now to answer the question -- the probability that the ball drawn at the end is white, we add up the probabilities of all the cases where the last step is "W":
(6/20)+(2/20)+(9/60)+(12/60)+(1/60) = (18+6+9+12+1)/60 = 46/60 = 23/30
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