Question

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

Answer 1

Let us define the following events

A ≡ 4 white balls are drawn in first six draws

B ≡ 5 white balls are drawn in first 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 bag.] 

P (E) = P (E| A) P (A) + P (E| B) P (B)