Draw balls from bin and add more balls after draws

Question

A bin contains $r$-many red balls and $s$-many black balls. We draw randomly a ball and afterwards we return the ball into the bin and add $c$-many balls of the color we have previously drawn.

We draw two times, what is the probability that the first ball was black if the second one has been red?

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Let be $\Omega=\{r,s\}^2$ the sample space and the probability measure:

$P(\{(r,r)\})=\frac{r}{r+s}\frac{r+c}{r+s+c},$ $P(\{(r,s)\})=\frac{r}{r+s}\frac{s}{r+s+c},$ $P(\{(s,r)\})=\frac{s}{r+s}\frac{r}{r+s+c},$ $P(\{(s,s)\})=\frac{s}{r+s}\frac{s+c}{r+s+c}.$

Now we simply collect all the elements that correspond to the event we are looking for:

$P($first ball black $\mid$ second ball red $)=\frac{P(\{(s,r)\})}{P(\{(r,r)\})+P(\{(s,r)\})}=\dots=\frac{s}{r+c+s}.$

Is this correct?

Answer 1

What is being asked is

P(first ball black | second ball red)

= P(first black $\cap$ second red)/ P(second red)

= $\dfrac{\text{P(first black $\cap$ second red)}}{\text{P(first black $\cap$ second red) + P(first red $\cap$ second red)}}$

$ = {\dfrac{s}{s+r}}{\dfrac{r}{s+r+c}}\over{{\dfrac{s}{s+r}}{\dfrac{r}{s+r+c}}+{\dfrac{r}{s+r}}{\dfrac{r+c}{s+r+c}}}$ $= \dfrac{s}{s+r+c}$

Answer 2

The "restricted sample space" is

$$ \Omega' = \{s,r\}^2/\{(r,s),(s,s)\} $$

We removed all the sets corresponding to the second ball having been black, since we are given that the second ball is red.

The measure of $\Omega'$ is $\frac{r}{r+s}$ so the probability that the first ball was black is the measure of $\{(s,r)\}$ divided by the measure of $\Omega'$ which gives us

$$P(\{(s,r)\})|\{(*,r)\}) = \frac{s}{r+s}\frac{r}{r+s+c}\times\frac{r+s}{r}$$ $$ =\frac{s}{r+s+c}$$

Which is indeed the same answer as what you got, but you had a $P(\{(s,s)\})$ in the denominator, which was probably a typo and should have been $P(\{(r,r)\})$?