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?
Best Answer
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 $Pr = \dfrac{6}{12}$