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- Jan 26, 2012

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__Basic problem:__

Urn I has 4 white and 4 black balls. Urn II has 2 white and 6 black balls. Flip a fair coin. If the outcome is heads, then a ball from urn I is selected, whereas if the outcome is tails, then a ball from urn II is selected. Suppose a white ball is selected and the replaced. Denote this event by $W_1$. Now another ball is withdrawn at random from the same urn.

(a) What is the probability that the first ball is from urn I (given that it is white)?

__My solution to part (a):__

This can be solved by Bayes' Theorem in a pretty straightforward way. Let the notation $W_1$ mean that the first ball is white and let $U_1$ mean that it was chosen from urn I.

\(\displaystyle P \left( U_1|W_1 \right) = \frac{ P \left( W_1 \cap U_1 \right)}{P \left(W_1 \right)} = \frac{P \left( W_1|U_1 \right)P(U_1)}{P \left( W_1|U_1 \right)P(U_1)+P \left( W_1|U_2 \right)P(U_2)}\).

Now plugging in the information I have into the last expression I get:

\(\displaystyle P \left( U_1|W_1 \right)=\frac{\frac{1}{2}\frac{1}{2}}{\frac{1}{2} \frac{1}{2}+\frac{1}{4}\frac{1}{2}}=\frac{2}{3}\)

So my question is how does that look?

(b) What is the probability that the second ball is also white?

__My solution to part (b):__

This one I'm not 100% sure on. I think it should be similar to part (a) except the values of \(\displaystyle P(U_1)\) and \(\displaystyle P(U_2)\) will be different. Instead of $\frac{1}{2}$ and $\frac{1}{2}$ they will be $\frac{2}{3}$ and $\frac{1}{3}$ respectively. So the solution should be found using the same method just replacing those values, correct?