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Question:

bowl 1 contains 6 red chips and 4 blue chips. 5 chips are selected at random and placed in bowl 2. then 1 chips is drawn from bowl 2. Relative to the hypothesis that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl 1 to 2.

Solution:

in lamens terms, the question is asking "what is the probability of 2red and 3 blue chips, given 1 blue" or P(2r3b | 1b)

p(2r3b | 1b) = ( P(2r3b) * P(1b | 2r3b) ) / ( P(2r3b) * P(1b | 2r3b) + P(4r1b) * P(1b | 4r1b) + P(3r2b) * P(1b | 3r2b) + P(1r4b) * P(1b | 1r4b))

Ok great. so i calculate all that, i get 5/14... but these questions take FOREVER!! my question: is there a quicker way to do this? I thought about the following:

The denominator in the above equation should all add up to P(of picking any 5 random chips with at least 1 blue), lets say P(A). So does it not make sense to say P(A) = 1 - P(no blue chips) = 1 - P(5red).

P(5red) = 1/42

so P(A) = 41/42 which is not equal to the denominator from above (which btw is 6/15).

Any insight?

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# Probabilities with Chips and Bowls

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