An urn contains 5 black, 5 red, and 5 white balls. 3 balls are chosen without replacement at random. We are asked to find the probability of getting 2 black and 1 red ball.

One way to solve it: $\frac{\binom{5}{2}*\binom{5}{1}}{\binom{15}{3}} \approx 0.109$

Another way is to use conditional probability. We have:

P(2 black $\cap$ 1 red) = P(2 black | 1 red) * P(1 red) = $\frac{\binom{5}{2}}{\binom{14}{2}} * \frac{5}{15} \approx 0.036$. Where did I forget to multiply by 3? What did I do wrong? I feel stupid

  • 3
    $\begingroup$ You correctly computed the probability that the first ball is red and the next two black, but that is only $\frac 13$ of the cases. $\endgroup$
    – lulu
    Sep 26, 2022 at 18:12

2 Answers 2


In drawing without replacement, the question could ask for

(i) drawing in a specific order, eg $BBR$
(ii) drawing in all possible orders, viz $BBR,\;BRB,\; RBB$

For (i) it is easier to use the general multiplication rule (with implied conditional probabilities) as $\frac5{15}\frac4{14}\frac5{13}$

For (ii) it is easier to use the choose route, viz $\large\frac{\binom52\binom51}{\binom{15}3}$

Since a particular order has not been specified, it is implied that all possible orders need to be considered, and the choose route is easiest.

If instead you decide to use the general multiplication route, you need to have a multiplier, thus $\binom31\times \frac5{15}\frac4{14}\frac5{13}$
because the single red ball could be at any of the three positions.

On the other hand, if a specific $BBR$ order had been specified and you want to use the choose route, you will need to divide by $\binom31$

Beginners often forget this, so be careful,
and as a general practice, use the simpler route without multiplier/divider depending on what has been asked.


Suppose $X$ is the event 2 black balls and 1 red ball in a sample of 3 balls from the specified urn without replacement. You want $\text{Pr}(X)$.

You might think in 2 ways: (i) sample 3 balls at the same time or (ii) sample each one of the 3 balls in order. From the stand point of the experiment the probability won't change.

Let us think using method (ii): $X=(R,B,B)\cup (B,R,B) \cup (B,B,R)$. Remember that in the sequence notation $(R,B,B)$ order is relevant, this is the sequence in which $R$ is the first ball, $B$ is the second and $B$ the third.

As these events are disjoint,

$$\Pr(X)=\text{Pr}((R,B,B))+\text{Pr}((B,R,B))+\text{Pr}((B,B,R))\ \ \ (1)$$

When you computed $$\text{Pr}(2\ blacks\cap 1\ red)=\text{Pr}(2\ blacks|1\ red)\text{Pr}(1\ red)=\frac{10}{21\cdot 13}\approx 0.037$$ you've found only $\text{Pr}((R,B,B))$ in expression $(1)$ instead of $$\text{Pr}(X)=\frac{10}{21\cdot 13}+\frac{10}{21\cdot 13}+\frac{10}{21\cdot 13}\approx 0.110$$


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