A bag contains m blue and n yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with k other marbles of the same colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?

A bag contains m blue and n yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with k other marbles of the same colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?
13 Comments
Andrey Dyukmedzhiev
10/1/2016 09:36:04 am
Formula of Total Probability (B1 - blue at 1st try, etc.):
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10/1/2016 05:56:29 pm
Audrey's analysis is correct. But she could go one step further. By factoring, her result numerator can be rewritten as:
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ANDREY Dyukmedzhiev
10/1/2016 07:08:00 pm
My name is derivative of the Greek male name ANDREAS, not Audrey (like Audrey Hepburn for example). And I didn't simplify the expression because that is of no importance, the method is.
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tom
13/1/2016 03:28:05 am
I disagree about simplifying - if you don't simplify from where you reached you may not see that adding balls in this way afrter the first round has no effect whatoever on the second round, which (in my opinion, anyway, so maybe you will disagree) is the only interesting thing about this question. And I still thinks it's easier to get there by a quick look at the extremes instead of doing arithmetic, but maybe that's just different attitudes to these things.
tom
10/1/2016 06:47:03 pm
Another way of getting the same answer as Andrey is to say that the expected number of blue balls after the first round is m+mk/(m+n) so that the chance of getting blue on the second round is (m+mk/(m+n))/(m+n+k) which simplifies to the simple answer (as does Andrey's formula).
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While the conclusion is true, the argument lacks rigor. Let us modify the problem as follows: for positive k, if yellow add k yellow, if blue add k-1 blue and 1 yellow then the probability is (m+(k-1)(m/m+n))/(m+n+k). Following your argument, if we add none we have m/(m+n) and same if we take the limit. But a different value for every positive k.
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tom
22/2/2016 01:40:23 pm
No, your argument doesn't work because with your modified scheme if k = 0 the odds are not the same on the second round as on the first - since if a blue is drawn on the first round you add a yellow and remove a blue, leaving m-1 blue and n+1 yellow. Obviously if m and n are both 1, the chances of drawing a blue on the second round are 0 if a blue was drawn on the first round and 0.5 if a yellow was drawn on the first round, so overall probability of drawing a blue on the second round is 0.25, which isn't the same as the 0.5 probability of drawing blue on the first round. So your model doesn't tell us anything about the case where the probability of blue on the second round is the same as the probabilty of blue on the first round if k is 0 (as in the problem we are concerned with), only about some completely different problem.
I thought of the following alternative.
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Timothy
9/2/2016 12:17:55 pm
On a second read, apparently Tom made this observation in different words to which I failed to catch. Kudos.
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