Bag A contains 2 white, 1 black and 3 red balls, Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls. One Bag is chosen at random and 2 balls are drawn at random from that Bag. Of the randomly drawn balls happen to be red and black, what is the probability that both balls come from Bag B?

What makes this question messy is the fact that the bags contain different number of balls, so we have to take all cases

BagA:
WW - 2/6 * 1/5 = 2/30 = 1/15
WB - 2/6 * 1/5 * 2 = 4/30 = 2/15 , times 2 since it could be BW
WR - 2/6 * 3/5 * 2 = 12/30 =6/15
BR - 1/6 * 3/5 * 2 = 6/30 = 3/15
RR - 3/6 *2/5 = 6/30 = 3/15
BB - not possible, there is only 1 Black
notice the sum is 15/15 , all accounted for

Bag B:
WW - 3/9 * 2/8 = 6/72 = 1/12 = 3/36
WB - 3/9 * 2/8 * 2 = 12/72 = 1/6 = 6/36
WR - 3/9 * 4/8 * 2 = 24/72 = 1/3 = 12/36
BR - 2/9 * 4/8 * 2 = 16/72 = 2/9 = 8/36 = 40/180
RR - 4/9 * 3/8 = 1/6 = 6/36
BB - 2/9 * 1/8 = 1/36
Again, note the sum is 36/36

Bag C
WW = 4/9 * 3/8 = 6/36
WB - 4/9 * 3/8 * 2 = 12/36
WR - 4/9 * 2/8 * 2 = 8/36
BR - 3/9 * 2/8 * 2 = 6/36
RR - 3/9 * 2/8 = 3/36
BB - 2/9 * 1/8 * 2 = 1/36

It would be nice if they all had the same common denominator
which would be 180
So if we change all the fractions above out of 180, we would have 540 in total,
the one we need is BagB,RB = 8/36 = 40/180

so the prob of your event is 40/540 = 2/27

To find the probability that both balls come from Bag B given that they are red and black, we can use Bayes' theorem.

Let's denote the events:
A - Both balls come from Bag B
B - Both balls are red and black
We want to find P(A|B).

According to Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)

To find P(B|A), the probability of drawing both red and black balls given that they come from Bag B, we need to consider the number of ways we can choose a red and black ball from Bag B divided by the total number of ways we can choose two balls from Bag B.

In Bag B, there are 3 red balls and 2 black balls. Therefore, the probability of drawing a red and black ball from Bag B is (3/9) * (2/8).

Next, we need to find P(A), the probability of choosing Bag B. Since we are choosing a bag at random, the probability of choosing Bag B is (1/3).

To find P(B), the probability of drawing both red and black balls, we need to consider the probability of drawing both red and black balls from any of the bags.

In Bag A, there are 3 red balls and 1 black ball. Therefore, the probability of drawing a red and black ball from Bag A is (3/6) * (1/5).
In Bag B, we already calculated the probability as (3/9) * (2/8).
In Bag C, there are 2 red balls and 3 black balls. Therefore, the probability of drawing a red and black ball from Bag C is (2/7) * (3/6).

Since we are choosing a bag at random, the probability of choosing any of the bags is equal. Therefore, P(B) = (1/3) * [(3/6) * (1/5) + (3/9) * (2/8) + (2/7) * (3/6)].

Finally, we can plug these values into Bayes' theorem to find P(A|B):
P(A|B) = [(3/9) * (2/8) * (1/3)] / [(1/3) * [(3/6) * (1/5) + (3/9) * (2/8) + (2/7) * (3/6)]]

Simplifying this expression will give you the answer - the probability that both balls come from Bag B given that they are red and black.