Three balls are drawn from a box containing 3 red, 4 black and 5 white balls. Find the probability that 2 out of 3 balls will be of the same colour (event A) if the first one drawn is red (event B).
I tried to solve this question using combinations and conditional probability. First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $ and then I calculated $P(B) = \frac{3\choose1} {12\choose1} = 1/4$. Then I used the two results to find $P(A|B) = \frac{15/44}{1/4} = 1.36$, which is wrong (correct result is $34/55$). I know there are other ways of solving this exercise, but if possible, I would like to solve it using combinations.
Thanks in advance!
Best Answer
It is given that the first ball is red, so in effect you only have 2R,4B, 5W balls left, and only $2$ more to draw
$$ Pr= \frac{\binom21\binom{9}1+\binom42 +\binom 52}{\binom{11}2}= \frac{34}{55} $$