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Please help with me find the area to read up on to solve this type of problem:

I have two bags. Bag one has 7 black marbles and 4 white marbles. Bag two has 4 black marbles and 5 white marbles. I take two marbles from bag one and put them in bag two (with a blindfold). If I take one marble out of bag two, what is the probability that it's white?

While I would like to know the answer, I would most like to know how to solve similar questions on my own.

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This problem uses what is usually called the law of total probability.

For $i = 0, 1, 3$, let $A_i$ denote the event that $i$ white marbles and $2-i$ black marbles were transferred to bag #$2$. Note that exactly one of the three events $A_0, A_1, A_2$ must have occurred, and if event $A_i$ occurred, then bag #$2$ contains $5+i$ white marbles and $4+(2-i) = 6-i$ black marbles. If $W$ is the event that the marble drawn from bag #$2$, then the law of total probability gives us that $$P(W) = P(W\mid A_0)P(A_0) + P(W\mid A_1)P(A_1) + P(W\mid A_2)P(A_2)$$ where $P(W\mid A_i) = \frac{5+i}{5+i+6-i}=\frac{5+i}{11}$ (do you see why?). I will leave you to work out the values of the $P(A_i)$ and the final calculation to get the numerical value of $P(W)$.

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Yes, I understand exactly. This will give me the same answer as above. Thank you for introducing the law of total probability. – Alex Dec 3 '12 at 15:22

You can make a case-by-case analysis of the kind of marbles you take from the first bag. For example the probability to take two whites is ${4 \over 11} \cdot {3 \over 10}$. Now you have 4 black and 7 white marbles in the second bag, the probability to take a white one is ${7 \over 11}$. So for this case the probability is ${4 \over 11} \cdot {3 \over 10} \cdot {7 \over 11}$. Add up the probabilities of all three cases white/white, black/black and black/white (which is actual to two cases, black first or white first) to get the final result.

If you are familiar with probability tree diagrams, they are a good way to visualize the cases and calculate the probabilities of these kind of problems.

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Thanks, so the answer would be: 4/11 . 3/10 . 7/11 + 7/11 . 6/10 . 5/11 + 7/11 . 4/10 . 6/11 = 21/55 – Alex Dec 3 '12 at 14:33
Or does the last term, (7/11 . 4/10 . 6/11) for b/w need to be doubled because there are two options (b/w and w/b)? – Alex Dec 3 '12 at 14:43
Yes, it has to be doubled. This wasn't clear enough in my answer, fixed. – Simon S Dec 3 '12 at 14:46
Great, thanks for your help! – Alex Dec 3 '12 at 14:56

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