You have 2 boxes that contains 100 balls each, and the balls can be white, blue, red, or half red and blue.

Box A contains:

  • 90 white balls
  • 10 red balls

Box B contains:

  • 86 white balls
  • 4 red balls
  • 9 blue balls
  • 1 red and blue ball

If you don't know which box you pick a ball from (the probability is equal for what box you chose), what is the probability for pulling exactly 2 white balls and 1 red ball, if you don't put it back and don't take a look at the balls until all 3 are pulled from the box you chose? (You only pick 3 balls).

I thought this would be pretty straight forward by using combinatorics:

P(1 red and 2 white balls) = $\frac{176 \choose 2}{200 \choose 3}$$14\choose 1$

But this gives me the wrong answer.

Please help me solve this problem, any suggestions would be appreciated.

  • $\begingroup$ There is a contradicition. Box B contains 110 balls, not 100 balls. This contradiction has to be resolved first. $\endgroup$ – callculus Sep 14 '15 at 13:44
  • $\begingroup$ Box $B$ contains $110$ balls (in stead of $100$)? How many balls do you pick? $3$ Maybe? If you have picked a ball then is this ball replaced? Quite some things are unclear. $\endgroup$ – drhab Sep 14 '15 at 13:45
  • $\begingroup$ @calculus Thank you, it has not been resolved. $\endgroup$ – martin Sep 14 '15 at 13:53
  • $\begingroup$ @drhab Yes you pick 3 balls, I hope the rest is explained well now. $\endgroup$ – martin Sep 14 '15 at 13:54
  • $\begingroup$ Are all balls picked from the same box (after choosing one)? If so then the answer of 5xgum works with $P(A)=P(B)=\frac12$. To be found are $P(X|A)$ and $P(X|B)$ separately. $\endgroup$ – drhab Sep 14 '15 at 13:58

$X$ be the event "you draw $2$ white balls and one red ball" Also, let $A$ be the event "you draw from box $A$" and $B$ be the event "you draw from box $B$.

Then, as a certain Bayes might tell you, because $A$ and $B$ represent a partition (their union is the universe set, their intersection is empty), that

$$P(X) = P(X|A)\cdot P(A) + P(X|B)\cdot P(B)$$

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