Conditional probabilities regarding urns An urn contains 10 white balls and 10 red balls. Ten balls are drawn from the urn without replacement. 
a. Find the probability that the fourth ball is red given that the first ball is red.
b. Find the probability the third ball is red given at least one of the first two balls is red.
c. Find the expected number of white balls drawn
d. Is the event that the last ball is red independent of the event that the first two balls are of different colors?
Attempt:
a.Pr(fourth ball is red| first ball is red) = 9/19
b. Pr(third ball is red|at least one of first two are red) = 
   Pr(third ball is red|both red) + Pr(third red|first red) + Pr(third red|second red)
   =   9/19 + 9/19 -7/17
c. 10*0.5 = 5
d. Stuck here. 
 A: All but (b) have answers by now, and (essentially) solutions.  We deal with (b), which is more technical, and less interesting.
Let $X$ be the event that the third ball drawn is red.  Let $Y$ be the event that at least one of the first two is red.  We want $P(X|Y)$.  Now we calculate, with unfortunately minimal thinking.
We have
$$P(X|Y)P(Y)=P(X \cap Y).$$
It remains to calculate $P(Y)$ and $P(X\cap Y)$.
The event $Y$ happens precisely if (under the obvious notation) RR or RW or WR. Thus
$$P(Y)=\frac{10}{20}\frac{9}{19} + \frac{10}{20}\frac{10}{19}+\frac{10}{20}\frac{10}{19}.$$
The event $X\cap Y$, or better, $Y\cap X$ happens if RRR or RWR or WRR. Thus
$$P(Y\cap X)=\frac{10}{20}\frac{9}{19}\frac{8}{18} + \frac{10}{20}\frac{10}{19}\frac{9}{18}+\frac{10}{20}\frac{10}{19}\frac{9}{18}.$$ 
Now it only remains to calculate. After not much work, we get $P(X|Y)=14/29$, I think. There may be a clever way to "see," without calculation, that this must be the answer, but I have not thought of one.  
