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.
I understand that we can use Bayes theorem so we should find $P(E|A)+P(E|B)+P(E|C)$ isn't it? where $A=$drawing $4$ white balls
$B=$drawing $5$ white balls
$C=$drawing $6$ white balls ( can be neglected, no white balls will be left for the next draws)
$E=$exactly one white ball is drawn in the next two draws
What's wrong with my solution?