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)
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, thenP.S. It can be wrong... I am a learner too...
((4C1)/(6C1))*((5C1)/(8C1)) = 5/122.When First bag produce black and second bag produce white, then
((3C1)/(8C1))*((2C1)/(6C1)) = 1/8Add both the results and the answer is 13/24