Alice has two bags of marbles: Bag A contains 8 red and 3 blue marbles, and Bag B contains 5 red and 7 blue marbles. She randomly selects a bag and then draws one marble at random. It is blue. What is the probability that it came from Bag B? Can you explain why the answer is 77/113? and actually answer the question if you don't know, dont post an answer

1

It might help to draw out a probability tree. Check out the diagram below. The first two branches at the top represent bags A and B. The probability of picking either bag is 1/2 assuming each bag is equally likely to be picked.

If we go with bag A, then the probability of getting blue is 3/11 since there are 3 blue out of 8+3 = 11 total. The probability of getting bag A and a blue marble is (1/2)*(3/11) = 3/22 which I've marked in the diagram as well.

If we go with bag B, then the probability of getting blue is 7/12 since there are 7 blue out of 5+7 = 12 total. The probability of getting bag B and blue is (1/2)*(7/12) = 7/24

Add up the results found getting

P(blue) = P(blue & bag A) + P(blue & bag B)

P(blue) = 3/22 + 7/24

P(blue) = 36/264 + 77/264

P(blue) = 113/264

This is shown in the diagram as well.

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Recall that conditional probability in general is defined as follows

P(A given B) = P(A and B)/P(B)

In this problem, we'll have

P(bag B given blue) = P(bag B and blue)/P(blue)

Therefore, we can say...

P(bag B given blue) = P(bag B and blue)/P(blue)

P(bag B given blue) = (7/24)/(113/264)

P(bag B given blue) = (7/24)*(264/113)

P(bag B given blue) = (7*264)/(24*113)

P(bag B given blue) = (7*24*11)/(24*113)

P(bag B given blue) = (7*11)/(113)

P(bag B given blue) = 77/113

In short, the probability of getting bag B, given the marble is blue, is 77/113.