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An urn contains $r$ Red and $b$ Blue marbles. A fair coin is flipped. If the flip is Heads then $h$ Red marbles are added to the urn. If the flip is Tails then $t$ Blue marbles are add to the urn. Now a random marble $M$ is drawn from the urn.

(a) What is the probability that $M$ is Red?
(b) What is the probability that the flip was Heads given that $M$ is Blue?


My attempt: Original marbles: $r$, $b$

After flip:

$H: r+h = R$, $b$.

$T: b+t = B$, $r$.

a) $$P(M_{red}) = \frac{1}{2}\left [ \frac{R}{R+b}+\frac{r}{B+r} \right ]$$

b) $$\frac{1}{2} \cdot \dfrac{\dfrac{b}{R+b}}{\dfrac{b}{R+b}+\dfrac{B}{B+r}}$$

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    $\begingroup$ All seems right. Probably it would look nicer if all $\binom x1$ would be replaced simply by its value, $x$. $\endgroup$ – Berci Mar 11 '13 at 0:31
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You may have gone a little crazy with the chooses there. Let $R$ be the event that the flip is heads (and so $h$ red marbles are added), and let $B$ be the event that the flip is tails (and so $t$ blue marbles are added). Then

$$ P(M \text{ is red}) = \frac12P(M \text{ is red } | R) + \frac12P(M \text{ is red } | B) $$

In the first case there are $r+h+b$ total marbles and $r+h$ red ones, in the second case there are $r+b+t$ total marbles and $r$ red ones, so the previous line is $$ = \frac12 \left(\frac{r+h}{r+h+b}+ \frac{r}{r+b+t}\right). $$

Then $$ P(R| M \text{ is blue }) = \frac{P(R \cap \{ M \text{ is blue}\})}{P( M \text{ is blue})} = \frac{P(\{ M \text{ is blue}\}|R)P(R)}{P( M \text{ is blue})} $$ $$ = \frac{(1-P(M \text{ is red }|R))\frac12}{1-P(M \text{ is red})} $$ and all the remaining quantities have been already calculated.

Short answer: Yes your answers look correct, but can be simplified.

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