# [Math] Conditional Probability with red and blue balls

conditional probabilityprobability

A jar contains 7 red balls and 9 blue balls. We choose, uniformly at random and without
replacement, 3 balls. Define the following two events:

A = "exactly 2 of the balls are red"

B = "the number of red balls is even"

What is the conditional probability $Pr(A \vert B)$?

My attempt to obtain the answer to this begins by showing the formula used:
$$Pr(A \vert B) = \frac{Pr(A\cap B)}{Pr(B)}$$

$Pr(B) = 9 \cdot {7 \choose 2}$ since we choose a combination of two balls from 3 and multiply it by 9, which is the selecting any of the blue balls to accommodate our reds.

$Pr(A \cap B) = {9 \choose 3} + 9\cdot {7 \choose 2}$ because this is the intersection of $Pr(A)$ and $Pr(B)$, we must take an even amount of reds. This means that ${9 \choose 3}$ is one of the cases where we choose zero reds (zero is even) plus the case where we choose two reds $9 \cdot {7 \choose 2}$

Therefore:
$$Pr(A \vert B) = \frac{ {9\choose 3} + 9\cdot {7 \choose 2}}{9\cdot {7 \choose 2}}$$

Correct?

$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
$$= \frac{P(2 \,red \,balls\, AND \,even\, number\, of\, red\, balls\,)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$= \frac{P(2\,red\,balls)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$=\frac{\frac{\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}{\frac{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}$$
$$=\frac{\binom{7}{2}\binom{9}{1}}{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}$$