Thread: Probability Problem

If one bag contains 4 white and 2 black balls; another contains 3 white and 5 black balls. If one ball is drawn from each bag
find the probability that
i) both are white
ii) one is white and one is black.

Originally Posted by zorro

if one bag contains 4 white and 2 black balls; another contains 3 white and 5 black balls. If one ball is drawn from each bag
find the probability that
i) both are white
ii) one is white and one is black.

(1) (4/6)(3/8)

(2) (4/6)(5/8)+(2/6)(3/8)

(1)
Since 1 ball should be white from first bag (4W,2B;Total=6) and 1 ball should be white from second bag (3W,5B;Total=8)...

Therefore ((4C1)/(6C1))*((3C1)/(8C1)) = 1/4

(2)
There are 2 cases...

1.When First bag produce white and second bag produce black, then

((4C1)/(6C1))*((5C1)/(8C1)) = 5/12

2.When First bag produce black and second bag produce white, then

((3C1)/(8C1))*((2C1)/(6C1)) = 1/8

Add both the results and the answer is 13/24

P.S. It can be wrong... I am a learner too...

Originally Posted by zorro

If one bag contains 4 white and 2 black balls; another contains 3 white and 5 black balls. If one ball is drawn from each bag
find the probability that
i) both are white
ii) one is white and one is black.

Drawing a tree diagram would be an excellent way of answering these questions ... (I suspect that's what mathaddict did).