I have this exercise:

In a box there are $3$ green balls, $1$ blue and $1$ yellow. If $3$ balls are removed from the urn at the same time. What is the probability that only one of them is yellow?

My attempt was:

Let $Y$ the prob. of the yellow ball, $G$ the prob. of the green ball and $B$ of the blue.

The unique cases are: $YGG, YGB$. Since the balls are removed from the urn at the same time, the order doesn't matter.

Therefore, the probability is equal to:

$YGG$ Case:$\frac{1\cdot 3\cdot 2}{5\cdot 4\cdot 3} = \frac{1}{10}$

$YGB$ Case: $\frac{1\cdot3\cdot1}{5\cdot4\cdot3} = \frac{1}{20}$

And adding them $\frac{3}{20}$ is the prob.

But according to the guide the correct answer must be $\frac{3}{10}$ and i don't know what is wrong. Thanks in advance.

  • $\begingroup$ I think the question was meant to ask for the probability that only one of the drawn balls is green. $\endgroup$ Oct 21 '19 at 17:01

Are you sure the correct answer is 3/10? I am getting 3/5 in two different ways.

First, consider explicitly labeling the green balls as $G_1, G_2, G_3$. Then there are $\binom{5}{3} = 10$ ways to pick the balls from the urn, and 6 include yellow: $YG_1G_2, YG_2G_3, YG_1G_3$ and $YBG_i$ for each $G_i$. So this gives 6/10, or 3/5.

Alternatively, consider the probability you do not pick the yellow ball. There is a 4/5 chance to not do so on the first pick, then a 3/4 chance, then a 2/3 chance, and multiplied together you get a 2/5 chance to not pick yellow, and thus a 3/5 chance to pick yellow.

As for where you went wrong, you correctly said order does not matter, but then multiplied as if order mattered. When you multiply in this fashion, you are implicitly assuming you pick Y first, then G, then B (eg for your second example). So to make sure order truly doesn't matter, you should actually include cases YBG, BYG, BGY, GBY, and GYB. All of these together get you a 6/20 probability for this particular combination, which matches our calculation from the first attempt. (Similarly, including GYG and GGY gets you up to 6/20 there as well, for our total 3/5.) In general, you can take the probability of one occurrence and multiply by the number of possible rearrangements of the sequence.


I agree with the answer above. I initially thought of this as yellow and not yellow (i.e. there is 1 ball that is yellow and 4 balls that are not.

Now we can model this using a hypergeometric distribution. In the below equation, $n$ is the number of balls, $r$ is the number of yellow balls, $k$ is the number of yellow balls that we draw, and $m$ is the number of balls we are drawing in total. Then $n = 5$, $r = 1$, $k = 1$ and $m = 3$.

$$ \frac{{r \choose k}{n - r \choose m - k}}{n \choose m} = \frac{{1 \choose 1}{5 - 1 \choose 3 - 1}}{5 \choose 3} = \frac{6}{10} = \frac{3}{5} $$


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