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


  • $\begingroup$ If you have three balls in your hand, and the number of red ones is even, how many of the balls in your hand can be red? $\endgroup$ – Steve Kass Apr 18 '14 at 0:36

$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$

$$= \frac{P(2 \,red \,balls\, AND \,even\, number\, of\, red\, balls\,)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$

$$= \frac{P(2\,red\,balls)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$



  • $\begingroup$ I think you have properly defined $Pr(B)$ so your answer is correct. $\endgroup$ – GivenPie Apr 18 '14 at 1:05

Notice that, in your solution you have that $\Pr(B)=189$, which impossible. The same goes for your computation of $\Pr(A\cap B)$. If you want to use counting methods to calculate probabilities, don't forget to divide the "number of ways to satisfy the event" by the total number of possible outcomes!

As for a hint to solve your problem, think about how the events $A$ and $B$ are related to the others. For instance, if $A$ is satisfied, does that automatically mean that $B$ is also satisfied? Is the reverse also true? Do the answers to these questions simplify your task?

  • $\begingroup$ I understand. Would $Pr(B) = {16 \choose 3}$ satisfy the clause? $\endgroup$ – GivenPie Apr 18 '14 at 0:23

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