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.

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And your question is? – Henning Makholm Sep 23 '11 at 23:04
I think (a.) should be 9/19. – Srivatsan Sep 23 '11 at 23:12
Question (c) seems to be missing some details? If we draw all of the balls then the expected number of white ones is 10. If we draw only one ball, then the expected number of white balls is 0.5. There must be a hidden explanation of when we stop drawing if the question is to be interesting. – Henning Makholm Sep 23 '11 at 23:19
As for (d), try red-white symmetry and the fact that P(last ball has some color | first two balls are different) must be 1. – Henning Makholm Sep 23 '11 at 23:21

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.

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This makes sense. For part d, I'm not sure whether the question is asking if the first two balls are white(different color from the last ball which is red) or the first two balls are(red or white). If the latter, then the probability of the last ball being red has to 9/19 since we know the first two balls are different colors(white,red). So these events are not independent? – lord12 Sep 24 '11 at 0:25
The wording I think is clear. We have two events; $X$ is the event the first two balls are of different colours, event $Y$ is the event last is red. We are asked whether $X$ and $Y$ are independent. The easiest approach is to show $P(Y|X)=P(Y)$. Clearly $P(Y)=1/2$, and as @Henning Makholm explained, by symmetry $P(Y|X)=1/2$. I don't understand your $9/19$, it would be $9/18$ ($1$ red ball gone, but two balls gone) which is $1/2$. – André Nicolas Sep 24 '11 at 0:37
One can reduce the arithmetic a little, though. The probability that the third ball is red given that the first two are white is $10/18=5/9$. The probability that the first two are white is $9/38$, so the probability of WWR is $5/38$. The probability that the third ball is red is $1/2$, so $P(X\cap Y)=1/2-5/38=14/38$, $P(Y)=1-9/38=29/38$, and $P(X|Y)=14/29$. – Brian M. Scott Sep 24 '11 at 0:54
@Brian M. Scott: And yes, when calculating I cheated and did not use my formulas, I am not fond of extra work. But I wanted to "lay out" things in maximally mechanical style. – André Nicolas Sep 24 '11 at 1:10