Urn 1 has 1 orange ball and 6 blue balls. Urn 2 has 2 orange balls and 5 blue balls. Suppose you draw 3 balls from one urn. To decide which urn to use you roll a fair 6-sided die. Draw from urn 1 if you roll an even number, urn 2 if you roll an odd number. What's the probability of drawing exactly one orange ball?

I understand that you have a $0.5$ chance of drawing from urn 1 and $0.5$ chance of drawing from urn 2. I initially drew out a tree diagram for this question which led me to the answer of $P(exactly\ 1\ orange) = 0.5(3/7 + 4/7)$. My issue is the other solution which involves combinations.

$$P(1\ orange | Urn_1) = \frac{6 \choose 2}{7 \choose 3} = 15 / 35 = \frac{3}{7}$$ and $$P(1\ orange | Urn_2) = 2 \left(\frac{5 \choose 2}{7 \choose 3} \right) = 20/35 = 4/7$$

My mind simply can't understand why the above works. I have a tree diagram in front of me where I manually calculate each of the options but I can't relate the two together.

I do know in the end you would just do $$P(1\ orange) = 0.5 \left(P(1\ orange | Urn_1) + P(1\ orange | Urn_2) \right) = 0.5$$

  • $\begingroup$ I do not follow... your question is why the probabilities in the "other" solution can be calculated using combinations? This is just a hypergeometric distribution. Your question is why a hypergeometric distribution works? Temporarily assume that the balls are all uniquely labeled. You recognize that there are $\binom{7}{3}$ distinct ways to select three balls where we didn't care about the order of the balls and that each of these ways is equally likely to have occurred. $\binom{6}{2}$ of them correspond to getting 1 orange. $\endgroup$
    – JMoravitz
    Oct 2, 2020 at 14:28
  • $\begingroup$ I guess it's the $ {6 \choose 2} $ that's giving me difficulty. Would it be correct in saying that $ {6 \choose 2} $ is a way of constraining my options such that I only draw 2 blue balls and my only other option would be an orange ball? $\endgroup$ Oct 2, 2020 at 14:39
  • 1
    $\begingroup$ Does it help to write it as $\binom{1}{1}\binom{6}{2}$? To describe some way of drawing balls such that one of them is orange and two of them are blue it suffices to describe it based on the orange ball you drew as well as the two blue balls you drew. $\endgroup$
    – JMoravitz
    Oct 2, 2020 at 14:53
  • $\begingroup$ That's exactly what I needed and was confused about. Thank you so much! $\endgroup$ Oct 2, 2020 at 14:59

1 Answer 1


I think the way you wrote the two cases makes it harder to see the rule. If you are drawing three balls, and want exactly $1$ of them to be orange, then the other two must be blue. When drawing from an urn with $N_o$ orange balls and $N_b$ blue balls, then, you can choose the single orange ball in $N_o$ ways and the blue balls in ${N_b}\choose{2}$ ways; to obtain the probability, the product of these should be divided by the total number of ways to choose $3$ balls, which is ${N_o+N_b}\choose{3}$: $$ P=\frac{N_o\cdot{{N_b}\choose{2}}}{{N_o+N_b}\choose{3}}. $$ More generally, if you wanted to find the probability of drawing exactly $n_o$ orange balls and $n_b$ blue balls when drawing $n_o+n_b$ without replacement, it would be $$ P=\frac{{{N_o}\choose{n_o}}{{N_b}\choose{n_b}}}{{N_o+N_b}\choose{n_o+n_b}}. $$


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