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I suck at those but here is my try :
You could draw 2 red balls in (15 2) = 105 ways (every pair with same probabiilty).
2 red balls from first bag in (10 2) = 45 ways
So its 45/105 = 9/21.
Explanation :
I think its like that : there is 190 possibilites of drawing (including green balls). We are only interested in draws which has 2 red balls , they are 105 of them (also there are 75 draws with one green and on red and 10 with both green).
So now we calculate probability of drawing 2 red balls from 1st bag assuming one OUT OF THOSE 105 draws happened because we know drawing 2 red balls already happened (bayes theorem).
So its 45/105.
EDIT : Sry for my english I dont have time to correct it now I hope reasoning is clear; would love to have somebody smart to comment on it.
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This solution is invalid because the 105 ways to draw 2 red balls are not equally likely. Huskerfan has the right answer assuming that the 1st red ball is not replaced before the 2nd spin. The easiest way to solve this type of Bayes' theorem problem is to compute the probability of each case as follows:
The probability of selecting the bag with all reds both times is (1/2)*(1/2) = 1/4, and if we select this bag both times, then the probability of both balls being red is 1, so the probability of picking this bag both times and getting both reds is (1/4)*1 = 1/4. Now the probability of selecting the bag with both red and green both times is also (1/2)*(1/2) = 1/4, and if we select this bag both times, then the probability of both balls being red is (5/10)*(4/9) = 2/9, assuming that the 1st ball is not replaced before the 2nd spin, so the probability of picking this bag both times and getting both reds is (1/4)*(2/9) = 1/18. The remaining possibility is that we pick one of each bag, and this has probability 1/2. In this case, the probability of both balls being red is just the probability that we draw a red ball from the bag with half reds, or 1/2, so the probability of picking one of each bag and getting both reds is (1/2)*(1/2) = 1/4. So the probability of NOT picking the bag with all reds both times and picking 2 red balls is 1/18 + 1/4 = 11/36, while the probability of picking the bag with all reds both times and picking both reds is 1/4 = 9/36, so the odds are 11:9 against having picked the bag with all reds both times, or a probability of 9/20 = 0.45.
Note that if the 1st ball is replaced before the 2nd spin, this would change the probability of the 2nd case from (1/4)*(2/9) to (1/4)*(5/10)*(5/10) = 1/16, and this would change the probability of NOT picking the bag with all reds both times and picking 2 red balls to 1/16 + 1/4 = 5/16. Then the odds against having picked the bag with all reds both times would be (5/16)
1/4) = 5:4 or a probability of 4/9.