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A bag holds four balls. One of them is white. Each of the others is either black or white at equal chance. You randomly draw out two balls and discover they are both white. If you then randomly draw a third ball, what is the chance that it is white?

I've seen answers varying from 0.27 through 0.58 to 0.75. What is the correct answer?

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    $\begingroup$ Welcome to stackexchange. Can you show us your attempt? With just four balls you should be able to write the whole tree of possibilities. $\endgroup$
    – Ethan Bolker
    Feb 2, 2018 at 16:05
  • $\begingroup$ I feel this is a tough question, where interpretations can really cause a mix of answers, just my thoughts. $\endgroup$
    – imranfat
    Feb 2, 2018 at 20:13

1 Answer 1

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The probabilities that we have $(1,2,3,4)$ white balls in the bag are $$\left({1\over8},{3\over8},{3\over8},{1\over8}\right)\ .$$ Given that we have $(1,2,3,4)$ white balls in the bag the probabilities that we draw two white balls are $$\left(0,{1\over6}, {1\over2},1\right)\ .$$ The overall probability $p(2w)$ to draw two white balls then is given by $$p(2w)={1\over8}\cdot0+{3\over8}\cdot{1\over6}+{3\over8}\cdot{1\over2}+{1\over8}\cdot1={3\over8}\ .$$ Finally the probability that all three balls are white computes to $$p(3w)={1\over8}\cdot0+{3\over8}\cdot0+{3\over8}\cdot{1\over4}+{1\over8}\cdot1={7\over32}\ .$$ It follows that the conditional probability $p(3w|2w)$ is given by $$p(3w|2w)={p(3w\wedge 2w)\over p(2w)}={p(3w)\over p(2w)}={7\over12}=0.58333\ldots\ .$$

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  • $\begingroup$ While I certainly won't dispute your answer, I fail to recognize the conditional probability here. The probability that the third ball is white under the condition that the first two are white, means no balls have been drawn yet. But the OP states that the first 2 balls are already out. If I don't want to draw the third ball, but somebody else (coming out of nowhere and who doesn't know the history of the problem) will do it for me, how does that affect his chance for the third ball being white? Under that logic, the probability for the third ball would be 50-50 $\endgroup$
    – imranfat
    Feb 2, 2018 at 20:12
  • $\begingroup$ @imranfat: $p(3w)$ is the probability that all three drawn balls are white, and this is equal to the probability that the first two balls are white as well as all three. Check how I computed $p(3w)$. $\endgroup$
    – Christian Blatter
    Feb 2, 2018 at 20:16
  • $\begingroup$ hmm, something I have to think about. Thanks. $\endgroup$
    – imranfat
    Feb 2, 2018 at 20:27
  • $\begingroup$ @ChristianBlatter Maybe you can clarify something. I think I answered the question "What is the probability of the third ball being white, given the first two drawn were white?" This, I think, should exclude the case of one white ball out of four from the start (there are only then seven valid possibilities for the three "chance balls") . But your answer is different, and I tend to think correct, but I don't see why. $\endgroup$
    – John
    Feb 2, 2018 at 20:27
  • $\begingroup$ Note that the probability of drawing three white balls is not the same as the probability of drawing a third white ball after having drawn two white balls. $\endgroup$
    – Thomas Bitonti
    Feb 2, 2018 at 20:36

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