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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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    Hint:Use the [Law of total probability](http://en.wikipedia.org/wiki/Law_of_total_probability).2012-10-08

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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    Ok tha$n$ks for your help!2012-10-09