Probabilities with Chips and Bowls

[wown]

A better way to do Probabilities with Chips and Bowls?

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), let's 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?

 

[Stephen Tashi]

To compute the probability that the chip drawn from the second bowl is blue, compute it as if you drew it from the first bowl. The intermediate step of transferring chips to a second bowl shouldn't affect the bottom line probability of getting a blue chip on one draw.

 

[wown]

so you are implying that the probability is .4?

 

[Stephen Tashi]

The probability that the chip drawn from the bowl 2 is blue = 0.4.

The calculation wouldn't be so simple if you had to draw more than 1 chip from bowl 2.