Question

A bag contains 5 white and 3 black balls. Another bag contains 4 white and 5 black balls. From any one of these two bags, two balls are drawn at random at one time. What is the probability that the two drawn balls are white and black?

Answer 1

hmm...identical question of another user
http://openstudy.com/updates/4f97b4fde4b000ae9ecd7601

since it doesn't matter which bag they are drawn from, essentially we can think of them as all being part of 1 big bag with 9 white and 8 black marbles

There are 3 possible outcomes: both white, both black, one of each
\[P(both black) = \frac{8C2}{17C2}\]
\[P(both white) = \frac{9C2}{17C2}\]
\[P(white/black) = 1-P(bothblack)-P(bothwhite)\]

Answer 2

The probability is 9/34

Answer 3

the question states - "From any one of these two bags, two balls are drawn at random at one time" - so i think it's either 1 bag or the other.

from bag 1, to get a white and then a black, or a black then a white, the prob is (5/8 x 3/7) + (3/8 x 5/7) {NB NO replacement, ball is not put back in bag}

for bag 2, the same prob is: (4/9 x 5/8) + (5/9 x 4/8)

totalling this and assuming it's 50:50 that one bag will be chosen to the other, the odds are

1/2( (5/8 x 3/7) + (3/8 x 5/7)) + 1/2 ( (4/9 x 5/8) + (5/9 x 4/8)) = 0.5456

when i use the same method to calculate the probability of both being white or black, i get these numbers:
P{both white} = 0.2619
P{both Black} = 0.1924

when i add these 3 numbers together, just to check, i get 1, which is what i would expect.