tough probability question, please help

Question

three-fifth of balls from bag A, all the white balls from bag B and one-third of balls form bag c are placed in a new bag d. If 4 balls are picked in random, what is the possibility of three of them being black balls

Answer 1

As mentioned before, there is not enough information to solve this. One could make the following four assumptions and then be able to solve it.

1: there are an equal number of black and white balls in the bags.

2: Each bag has the same number of marbles.

3: Each bag has a large number of marbles so that we can ignore the issue of whether we have sampled with or without replacement.

4: Balls are selected from bag d.

With these 4 assumptions, one would get the following result.

Let each bag have n marbles; n/2 are white and n/2 are black.

Bag A: 3/5 of the marbles means 3/5 * n/2 = 3n/10 white and 3n/10 black marbles

Bag B: all white marbles = n/2 white marbles

Bag C: 1/3 of the marbles means n/6 white marbles and n/6 black marbles

Then, one selected 3n/10 + n/2 + n/6 = 29/30 n white marbles

One selected 3n/10 + n/6 = 14/30 n = 7/15 n black marbles

Then, probability of a black marble is 7/15 n/(29/30 n + 7/15 n) = 14/43

Finally, the probability of selecting 3 black marbles out of 4 will be, from the binomial distribution (or you can just use Excel / your calculator)

C(4,3) 14/43 ^3 (1-14/43)^1 = 4 * 14^3 * 29/43^4 = 318304/3418801 = .093104

Answer 2

How many balls are in bags A, B and C?