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 co-efficients can be left as such). 

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

$P\left(E\right)=P(E/A)P\left(A\right)+P(E/B)P\left(B\right)+P(E/C)P\left(C\right)$

But

$\mathrm{P}(\mathrm{E}/\mathrm{C})=0$ [as there are only 6 white balls in the bag]

$\begin{array}{l}\therefore P\left(E\right)=P(E/A)P\left(A\right)+P(E/B)P\left(B\right)\end{array}$

$\frac{{ }^{10}{\mathrm{C}}_1{\times }^2{\mathrm{C}}_1}{{ }^{12}{\mathrm{C}}_2}\frac{{ }^{12}{\mathrm{C}}_2{\times }^6{\mathrm{C}}_4}{{ }^{18}{\mathrm{C}}_6}+\frac{{ }^{11}{\mathrm{C}}_1{\times }^1{\mathrm{C}}_1}{{ }^{12}{\mathrm{C}}_1}\frac{{ }^{12}{\mathrm{C}}_1{\times }^6{\mathrm{C}}_5}{{ }^{18}{\mathrm{C}}_6}$