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Suppose there are 3 boxes: X, Y and Z, and in box X there are 3 black balls and 1 white ball, box Y has 2 white balls and 2 black balls, box Z has 3 white balls and 1 black ball. The probability of choosing box X is 1/6, box Y is 1/3, and box Z is 1/2.

The question is what is the probability of getting a black ball?

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    $\begingroup$ Imagine running the experiment 24 times. How many times would you choose box X? Y? Z? how many times would you get black? $\endgroup$
    – Gerry Myerson
    Oct 8, 2012 at 23:02
  • $\begingroup$ Hint:Use the Law of total probability. $\endgroup$
    – Aaron
    Oct 8, 2012 at 23:03

2 Answers 2

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Are you familiar with the Law of Total Probability? Denote $S$ the probability of sampling a black ball. Obviously we need to somehow account for the fact that the exactly one black ball has to be chosen from any of the 3 boxes in a single experiment: $$ P(S)=P(S|X)P(X)+P(S|Y)P(Y)+P(S|Z)P(Z) $$ where $P(X), P(Y), P(Z)$ are the probabilities to select the corresponding box. Hence, $$ P(S)=\frac{3}{4} \cdot \frac{1}{6}+\frac{2}{4} \cdot \frac{1}{3}+\frac{1}{4} \cdot \frac{1}{2}=\frac{5}{12} $$

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You may want to break down the probabilities.

The probability of choosing a black ball is the same as the probability of

  1. Choosing a black ball from box X
  2. or Choosing a black ball from box Y
  3. or Choosing a black ball from box Z

If we choose a black ball from box X, that's the same as first choosing box X and then choosing a black ball given that we have chosen box X. Can you take it from here?

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  • $\begingroup$ I think so. But just to double check: the entire sample space (the denominator) will consist of the Prob(Box X)*Prob(Black Ball| Box X) + Prob(Box X)*Prob(White Ball| Box X) + ...? While the events we are interested in is namely the probability of "choosing box X then choosing a black ball given that we have chosen box X" which are in the numerator. $\endgroup$
    – kuantumbro
    Oct 9, 2012 at 1:34
  • $\begingroup$ what a quantum bro. $\endgroup$
    – pneumatics
    Oct 9, 2012 at 2:10
  • $\begingroup$ @kuantumbro Not quite. I'm not sure separating it into numerator and denominator is the best way of looking at it. But I think you have the overall idea. The expression you wrote "Prob(Box X)*Prob(Black Ball| Box X) + Prob(Box X)*Prob(White Ball| Box X) + ..." would just sum to $1$, since it is the entire range of possibilities. The one we're interested in would be something like "Prob(Box X)*Prob(Black|Box X) + Prob(Box Y)*Prob(Black|Box Y) + ..." $\endgroup$
    – EuYu
    Oct 9, 2012 at 2:48
  • $\begingroup$ Ok I see that you mean. The probability of a black ball is just the sum ("or") of the individual events like you mentioned in your first post. I think I was getting confused between finding the Prob(Black ball) with the Prob(Box Y | Black ball). If you don't mind me asking: what if you're told a black ball was drawn and you want to know what the prob is that it came from Box Y. It is just the Prob(Black Ball | Box Y)*Prob(Box Y) / Prob(Black Ball) ? $\endgroup$
    – kuantumbro
    Oct 9, 2012 at 3:31
  • $\begingroup$ You would have to break it down using conditional probabilities. You will have $$P(\text{Box Y}|\text{Black Ball}) = \frac{P(\text{Box Y and Black Ball})}{P(\text{Black Ball})}$$ So yes, your expression is correct. $\endgroup$
    – EuYu
    Oct 9, 2012 at 3:40

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