An urn contains b black balls and r red balls. One of the balls is drawn at random, but when it is put back in the urn c additional balls of the same color are put in with it. Now suppose that we draw another ball.

Required:
Show that the probability that the first ball drawn was black given that the second ball drawn was red is b/(b r c).

Answer:

Please see the explanation below for proof

Step-by-step explanation:

Probability of choosing black ball from the bag

[tex]\frac{b}{b+r}[/tex]

Probability of choosing red ball from the bag

[tex]\frac{r}{b+r}[/tex]

After the addition of c balls of correspondent color –  

P (R2/B1) [tex]= \frac{r}{(r+b+c)}[/tex]

P (R2/B2) [tex]=\frac{r+c}{(r+b+c)}[/tex]

B1 and B2 are mutually exclusive event

P (B1/R2) = [tex]\frac{\frac{r}{(r+b+c)} *\frac{b}{b+r}}{\frac{r}{(r+b+c)}*\frac{b}{b+r} + \frac{r+c}{(r+b+c)}* \frac{b}{b+r}}[/tex]

P (B1/R2) [tex]=\frac{b}{(r+b+c)}[/tex]

Hence, Proved

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-3r - 13

r = 2