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"Urn 1 contains two white chips and one red chip. Urn 2 has one white chip and two red chips. One chip is drawn at random from urn 1 and transferred into urn 2. Then one chip is drawn from urn 2. Suppose that a red chip is selected from urn 2. What is the probability that the transferred chip was white?"

Here is my Tree-Diagram: Tree

The question wants us to find the probability that we draw a red chip from urn 2 given a white chip is drawn from urn 1.

I have some confusion about conditional probability and total probability.

This question is asking for $P(W|R)$. So we must look 'backwards' in the tree-diagram.

Backwards Tree

$P(W|R) = \frac{P(W \cap R)}{P(R)} = \frac{P(W_{1})P(R|W_{1})}{P(R_{1})P(R|R_{1}) + P(W_{1})P(R|W_{1})}$

But there are values that I am missing. Can anyone see what I am doing wrong?

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You have done everything correctly. The missing values need to be evaluated. 1. $P(R|W_1) = 1/2$ since you have 2 red and 2 white chips in urn2. 2. $P(R|R_1) = 3/4$ since you have 3 red and 1 white chips in urn2.

solving you get P=4/7.

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  • $\begingroup$ Thank you for answering. So we assume that we do indeed have an extra white chip in Urn 2? I didn't know we were allowed to assume that $\endgroup$ – krlo Sep 22 '15 at 18:55
  • $\begingroup$ $p(R|W_1)$ implies probability that a red chip was picked from urn 2 given that a white chip was transferred from urn 1. $\endgroup$ – Sriharsha Madala Sep 22 '15 at 19:18
  • $\begingroup$ Ah, I understand. Thank you. $\endgroup$ – krlo Sep 22 '15 at 19:48

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