# Find the probability that the first ball was red, knowing that the second ball was black?

In a box we have $3$ white balls, $4$ black balls and 6 red balls. We extract $4$ balls, without reintroducing them back. Find the probability that: $a.$ First ball is red, second is black and the next two are red $b.$ The first ball was red, knowing that the second ball was black

$a.$ I started with using hypergeometric distribution, and calculated that $p$ is the probability that if we extract $4$ balls, $1$ one them will be red, one black and $2$ red. But we have $12$ such permutations, and from them, only one will be good, so the actual probability will be $\frac{p}{12}$, $p=...$

$b.$ I'm stuck at this, I don't know even how should I start?

a.

$Pr = \dfrac {6}{13}\cdot \dfrac{4}{12} \cdot \dfrac{5}{11} \cdot \dfrac{4}{10}$

b

Be careful here.
We are given that the second ball is black, but the first could be red or black or white.

$RB$ can happen in $6\times 4$ ways, $BB$ in $4\times 3$ ways, and $WB$ in $3\times4$ ways,
thus $Pr = \dfrac{6\times4}{(6\times4) + (4\times3)+(3\times4)}$

Shortcut

Since we know that the second ball is black, we can take it that there are only $12$ balls from which the first red ball is chosen, thus we directly get $Pr = \dfrac{6}{12}$

• Thank you. Your answer seems right. Commented Feb 2, 2017 at 13:23
• You're welcome ! Commented Feb 2, 2017 at 13:38