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Assume that you randomly pick one of the urns: X, Y, or Z. You then randomly draw one ball out of the selected urn and note its color. The urns contain the following colored balls:

Urn X: 4 red, 2 white, and 2 blue.
Urn Y: 2 red, 1 white, and 3 blue.
Urn Z: 2 red, 2 white, and 4 blue.

  1. If the ball is blue, what is the probability that it was drawn from urn X?

  2. If the ball is either blue or white, what is the probability that it was drawn from urn Y?

Progress

I keep getting ${2/27 \over 2/27+1/9+4/27}$ which turns out to be $2/9$. Which is the wrong answer according to the computer system. Any ideas where I'm going wrong?

I have been stuck on this problem for a while. Thanks in advance!

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  • $\begingroup$ I keep getting 2/27 over 2/27+1/9+4/27 which turns out to be 2/9. Which is the wrong answer according to the computer system. Any ideas where I'm going wrong? $\endgroup$ – Eric Oct 3 '14 at 9:03
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Essential:

$$P\left(X\mid B\right)=\frac{P\left(X\cap B\right)}{P\left(B\right)}=\frac{P\left(B\mid X\right)P\left(X\right)}{P\left(B\mid X\right)P\left(X\right)+P\left(B\mid Y\right)P\left(Y\right)+P\left(B\mid Z\right)P\left(Z\right)}$$

Same method for second question. Look at "blue or white" as one color: "blite".

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Think of it like this:

Of the 6 blue balls, 2 are in urn A, 3 are in urn B, and 4 are in urn C. If you pick a blue ball, the probability of the said ball being from urn A is 2/6 or 1/3.

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