A box contains 12 red and 6 white balls. Balls are drawn from the bag one at a time without replacement. If in 6 draws, there are at least 4 white balls, find the probability that exactly one white ball is drawn in the next two draws. (Binomial coefficients can be left as such).

Answer

`(""^(10)C_(1)xx""^(2)C_(1))/(""^(12)C_(2))xx(""^(12)C_(2)xx""^(6)C_(4))/(""^(18)C_(6))+(""^(11)C_(1)xx""^(1)C_(1))/(""^(12)C_(6))xx(""^(12)C_(1)xx""^(6)C_(5))/(""^(18)C_(6))`

Explanation

Let us define the following events
`{:(A:4,"white balls are drawn in first six draws"),(B:5,"white balls are drwn in fiest six draws"),(C:6,"white balls are drawn in first six draws"),(E:, "exactly one white ball is drawn in next two draws"):}`
`" "("i.e., one white and one red")`
Then `P(E)=P(E//A)P(A)+P(E//B)P(B)+P(E//C)P(C)`
But `P(E//C)=0" "["as there are only 6 white balls in the bas"]`
`thereforeP(E)=P(E//A)P(A)+P(E//B)P(B)`
`=(""^(10)C_(1)xx""^(2)C_(1))/(""^(12)C_(2))(""^(12)C_(2)xx""^(6)C_(4))/(""^(18)C_(6))+(""^(11)C_(1)xx""^(1)C_(1))/(""^(12)C_(1))(""^(12)C_(1)xx""^(6)C_(5))/(""^(18)C_(6))`