A box contains 2 white balls, 2 red balls and a black ball. Balls are chosen without replacement from the box. What is the probability that red ball is chosen before the black ball?

I am quite confused about the question because this is a exercise arranged in "combination" section, however I intuitively think it as a "permutation" problem. The red balls and the white balls have to be different, and as a red is chosen before the black, then its order should be accounted with. Furthermore, the question is that "the red", thus the red ball labeled 1 and the red ball labeled 2 can be chosen without considering their orders. Can anyone give me a clue to solve this kind of problem?


4 Answers 4


It seems to me that first you can ignore the white balls: if you configure the black and red ball order first, then adding the white balls in any combination cannot affect your answer because the while ball positions are all equally possible and equally likely regardless of the situation [*].

This would suggest that the probability is 2/3.

Consider the three nonwhite balls. The black ball can either be first, second, or third. These possibilities are all equally likely. If the black ball is first, you lose because you've drawn it before a red ball. In the other two outcomes, you win because you draw a red ball before a black ball.

[*] To see this, you can imagine all of the balls arranged in a row in order of how you drew them out. You can change the positions of the white balls without (a) changing the probability of that configuration, and (b) without changing whether you drew a red ball before a black ball.

  • $\begingroup$ Thank you, but can you give me a clue or the solution with the concepts of combinatorics. Actually, I am more interested in understanding the concepts by this example. Thank u again! $\endgroup$ Jul 28, 2017 at 3:40
  • 1
    $\begingroup$ Good answer! Well explained! $\endgroup$
    – Bram28
    Jul 28, 2017 at 3:47
  • $\begingroup$ If a box contains $b$ black balls, and $r$ red balls, the probability of choosing "black before red" without replacement, no matter what else is in the box, is $\dfrac{b}{{b+r\choose 1}}$ which is of course $\dfrac{b}{b+r}$ $\endgroup$
    – FormerMath
    Jul 28, 2017 at 3:58
  • $\begingroup$ Thank you all guys very much. I have a further question, that whether the event $\{R_1, R_2, B, W_1, W_2\}$ and event $\{R_2, R_1, B, W_2, W_1\}$ are the same events or not? \As I am new to the combinatorics world, I may misunderstand the concept of combination. $\endgroup$ Jul 28, 2017 at 7:18
  • $\begingroup$ There are five balls and there are 5! ways to draw and that the two are different in those 120. and both the cases count towards red balls are picked before the black ball is picked. $\endgroup$ Jul 28, 2017 at 7:34

@user326210 has given a most simple answer. If you nevertheless insist on an answer using "combinatorial concepts" you can argue as follows:

Number the white balls $1$ and $2$, the red balls $3$ and $4$, and the black ball $5$. The result of the drawing then is a random permutation of $[5]$, like $(41352)$. By symmetry one fifth of these permutations begin with $5$ and two fifths with $3$ or $4$. From the permutations beginning with $1$ or $2$ one fourth have $5$ at second place and two fourths have $3$ or $4$ at second place. Finally from the permutations beginning with $12$ or $21$ one third have $5$ at third place and two thirds have $3$ or $4$ at third place.

The conclusion is that $5$ appears before $3$ or $4$ in exactly a third of all permutations. Have you obtained any insight by going through all these motions?


One way is to think from opposite: the probability of red before black implies one minus the probability of nonred before black. Hence: $$1-P(B)-P(W\cap B)-P(W\cap W\cap B)=$$ $$1-\frac{1}{5}-\frac{2}{5}\cdot \frac{1}{4}-\frac{2}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}=\frac{2}{3}.$$

  • $\begingroup$ Very clear solution, thank you, but I still concentrate on the combinatorics way. $\endgroup$ Jul 28, 2017 at 7:10

The possible configurations are {RWB}, {WRB}, {RBW}.

Probability that the last ball to become extinct is black =$\frac{b}{r+w+b}$. Given that the last ball to become extinct is black, the White ball is the second to become extinct is $\frac{w}{r+w}$. Thus for the first configuration the

P(RWB) = $\frac{w}{r+w}.\frac{b}{r+w+b} =\frac{2}{4}.\frac{1}{5}$


P(WRB) = $\frac{r}{r+w}.\frac{b}{r+w+b} = \frac{2}{4}.\frac{1}{5}$

and lastly

P(RBW) = $\frac{b}{r+b}.\frac{w}{r+w+b} =\frac{1}{3}.\frac{2}{5}$

Adding all you get $\frac{1}{3}$,


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