A bag contains 10 balls out of which k are red and (10-k) are black, where 0 k 10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
A. 14 55
Using Bayes' theorem: P(k=1| all black ) = P( all black |k=1) P(k=1) P( all black ) where P(k) = 1 11 for each k \ 0,1,...,10\ . When k=1: P( all 3 black |k=1) = 9 3 10 3 = 84 120 = 7 10 . Calculating P( all black ) = 1 11 _ k=0 ^ 10 10-k 3 120 . Numerator is 120 + 84 + 56 + 35 + 20 + 10 + 4 + 1 = 330. So P( all black ) = 1 11 330 120 = 1 4 . Therefore P(k=1| all black ) = 7 10 1 11 1 4 = 7 10 4 11 = 28 110 = 14 55
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