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:

Step-by-step solution

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