Question

Urn A contains seven white balls and three black balls. Urn B contains five white balls and four black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was black given that the second ball drawn was black? (Round your answer to three decimal places.)

Answer 1

I got it

Answer 2

Sorry for being a bit late/slow. Here is how I did it (and I'm sure you did it much the same way)


Define the following events

E: you transfered a white ball from urn A to urn B
F: you transfered a black ball from urn A to urn B


The probabilities of each event are

P(E) = 7/10 = 0.7
P(F) = 3/10 = 0.3



Now define two more events

G: You pick a white ball from urn B
H: You pick a black ball from urn B


The conditional probabilies based on event E given to happen are

P(G|E) = 6/10 = 0.6
P(H|E) = 4/10 = 0.4


The conditional probabilies based on event F given to happen are

P(G|F) = 5/10 = 0.5
P(H|F) = 5/10 = 0.5


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We want to compute P(F|H), but we have P(H|F) instead. So we can use Baye's Theorem


P(F|H) = (P(H|F)*P(F))/(P(H))


Now use the law of total probability to break up P(H). So that means P(H) = P(H|E)*P(E) + P(H|F)*P(F)


Therefore, we now have this


P(F|H) = (P(H|F)*P(F))/(P(H|E)*P(E) + P(H|F)*P(F))


Plug in the given values and compute


P(F|H) = (P(H|F)*P(F))/(P(H|E)*P(E) + P(H|F)*P(F))


P(F|H) = (0.5*0.3)/(0.4*0.7 + 0.5*0.3)


P(F|H) = 0.34883720930232


So the probability of a black ball of being transfered from urn A to urn B (given a black ball was selected from urn B) is roughly 0.34883720930232