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Suppose we have three urns with black and white balls distributed as follows:

Urn A has $10$ black balls and $20$ white balls Urn B has $10$ black balls and $10$ white balls Urn C has $10$ black balls and $1$ white ball. Suppose we choose an urn (uniformly at random) and draw a ball (uniformly at random) from that urn. What is the probability that the ball is white?

I believe the answer is $$\frac{1}{3} \cdot \frac{2}{3} + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{11}$$ Is that correct?

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    $\begingroup$ Yes, your reasoning is correct! $\endgroup$
    – Remy
    Oct 12, 2018 at 9:59
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    $\begingroup$ Yes, the answer is correct. $\endgroup$
    – ArsenBerk
    Oct 12, 2018 at 10:00

2 Answers 2

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Correct.

Application of law of total probability is what they call this: $$P(E)=P(A)P(E\mid A)+P(B)P(E\mid B)+P(C)P(E\mid C)$$ where $A,B,C$ are mutually exclusive and covering.

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Your answer is indeed correct.

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