Probability Question

Question

Bowl A=2 bb & 4 rb

Bowl B=8 bb & 4 rb

Bowl C=1 bb & 3 rb

we draw 1 ball from each bowl, what is the probability to draw a bb from bowlA if we know that we drew exactly a total of 2 bb

bb= Blue Balls

rb= Red Balls

Answer 1

bowl A has 2 bb and 4 rb

bowl B has 8 bb and 4 rb

bowl C has 1 bb and 3 rb

P(b/A) = 2/6 = 1/3 P(r/A) = 2/3

P(b/B) = 8/12=2/3 P(r/B)= 1/3

P(b/C) = 1/4 P(r/C) =3/4

two ways to get a bb from A with exactly 2bbs with one ball drawn from bowls A, B and C

either bb from A and B, bb from A and C

3rd outcome is bb from C and B with rb from A

1st way has probability P= (1/3(2/3)(3/4) = 1/6

2nd way has probability P = (1/3(1/4)(1/3)= 1/36

3rd outcome has probability P=(2/3(1/4(2/3) = 1/9

given 2 bb, probability 1 bb came from A

= (1/6+1/36)/(1/6+1/36+1/9)

= (7/36)/11/36

= 7/11

Answer 2

This is an example of conditional probability and we can use Bayes's Law.

P(bb from bowl A | 2 blue balls) = P(bb from bowl A and 2 blue balls) / P(2 blue balls)

To find the probability of picking 2 blue balls, consider all the ways that 2 blue balls can be picked: One from A and one from B, one from A and one from C, one from B and one from C. Find their individual probabilities and add them up. This will be your denominator.

In finding the probabilities above, you found the probability of picking 2 blue balls with at least one of them coming from bowl A. That will be you numerator.

Once you have the numerator and denominator, divide and you will get your answer.