Probability that 2 out of 3 balls will be of the same colour if the first one drawn is red 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!


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 $

This is the probability that exactly two among the three are the same colour and at least one among the three is red.  That is not the required event.

Event $B$ specifies the first position's colour. So therefore, you must count ways to fill it separately from the later two in both the numerator and denominator.
The probability that the first one is red and exactly two among the first three are the same colour
$\qquad\begin{align}\mathsf P(A\cap B)&=\dfrac{\dbinom{3}{1}\left[\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}\right]}{\dbinom{12}{1}\dbinom{11}{2}}\end{align}$
And so:
$\qquad\begin{align}\mathsf P(A\mid B)&=\dfrac{\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}}{\dbinom{11}{2}}\end{align}$
