A box contains 4 red balls and 6 white balls. A sample of size 3 is drawn without replacement from the box, what is the probability of obtaining 1 red ball and 2 white ball, given at least 2 of the balls in the sample size are white?
I know the numerator must be $\binom{4}{1}$$\binom{6}{2}$, but I don't know what is the denominator. How do you deal with the conditional probability?
I know that $P(A\cap{B})=P(B\vert{A})*P(A)$ . How do I use this formula in this case?
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$\begingroup$If you have at least two white balls in the sample, either they are the only such white balls (the number you just computed, $\binom41\binom62$ ways) or there is a third white ball and zero red balls: $\binom40\binom63$ ways. The denominator of the conditional probability is thus these two counts added together.
(addendum by Graham Kemp)
I know that $P(A\cap{B})=P(B\vert{A})*P(A)$ . How do I use this formula in this case?
$\mathsf P(A\cap B)$ is the probability that at one ball is red and two are white, that is: $\left.\binom 41\binom 62 / \binom{10}3\right.$
$\mathsf P(A)$ is the probability that at least two of the balls are white, which is: $\left.\left(\binom 63+\binom 41\binom 62\right) \middle/ \binom {10}3\right.$
$\mathsf P(B\mid A)$ is what you seek; the probability that one ball is red when given that at least two are white. $$\mathsf P(B\mid A)~=~\dfrac{\mathsf P(A\cap B)}{\mathsf P(A)}$$
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