3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?

The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?


There are $9$ ways ($3\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.

  • $\begingroup$ the other thing i just realized is that the remaining unknown two balls aren't necessarily drawn after the 2 red and 2 green balls, the order is unspecified. Will this have any affect on the answer? For example, the orange and blue balls could be the first two balls drawn, followed by two red and two green balls. $\endgroup$ – sappgob Nov 2 '15 at 18:40
  • $\begingroup$ If you draw all six balls without looking at them, and you then look at four of them, and they are two red and two green, then the answer I gave is correct. If that's not what you mean, then you need to make your question more precise. $\endgroup$ – rogerl Nov 2 '15 at 18:44

Rogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.

$$\frac{\binom{3}{1}\binom{3}{1}\binom{2}{0}}{\binom{8}{2}}= \frac{9}{28}$$

If you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green:

$$P(O{=}1,B{=}1\mid R{=}2,G{=}2)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{5}$$

If you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:

$$P(O{=}1,B{=}1,R{=}2,G{=}2\mid R{\geq}2,G{\geq}2)=\dfrac{\binom{3}{1}\binom{3}{1}\binom{3}{2}\binom{3}{2}}{\binom{6}{2}\binom{3}{2}\binom{3}{2}+2\binom{6}{1}\binom{3}{3}\binom{3}{2}+\binom{3}{3}\binom{3}{3}}=\frac{81}{172}$$

This is why it is important to be clear about your specifications.

Similarly: I toss two coins behind a screen and tell you about one of them being a head.   What is the probability of the other being a tail when what I told you was ... ?:

  • The left coin is a head.
  • At least one coin is a head.
  • Exactly one of the coins is a head.

The probability space is $\{\rm HH, HT, TH, TT\}$

  • $\begingroup$ I actually like this answer better, Graham. $\endgroup$ – rogerl Nov 4 '15 at 22:49

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