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.

 

[blue_raver22]

I would suggest using a different approach to calculate the conditional probability in this scenario. Instead of calculating the probability of each possible combination of chips and their respective colors, you can use a simpler formula known as Bayes' Theorem. This theorem allows you to calculate the conditional probability by using the prior probability and likelihood of the event.

In this case, the prior probability would be the probability of selecting 5 random chips with at least 1 blue, which is 41/42 as you have correctly calculated. The likelihood would be the probability of selecting 2 red and 3 blue chips from bowl 1, given that 1 blue chip was selected from bowl 2. This can be calculated by multiplying the probabilities of each event, which would be (6/10)*(5/9)*(4/8)*(3/7)*(7/14) = 1/42.

Using Bayes' Theorem, the conditional probability can be calculated as follows:

P(2r3b | 1b) = (P(1b | 2r3b) * P(2r3b)) / P(A)

= (7/14 * 1/42) / (41/42)

= 5/14

This method is much quicker and more efficient than calculating each possible combination separately. It also allows you to easily adjust for different scenarios, such as changing the number of chips or bowls involved. I hope this helps you in your future calculations!