Question

An urn contains 11 amber balls and 17 black balls. Two balls are selected at random, without replacement, and removed from the urn. The colors of the two balls removed are not observed. A third ball is then drawn from the urn at random. Find the probability that the third ball removed from the urn is amber.

Answer 1

iii) (17/28)*(11/27)*(10/26) = 0.095136

iv) (17/28)*(16/27)*(11/26) = 0.152218

i) (11/28)*(10/27)*(9/26) = 0.050366

ii) (11/28)*(17/27)*(10/26) = 0.095136

Next, we need to add all these four probabilities, we get

Required probability = 0.050366 + 0.095136  + 0.095136  + 0.152218 = 0.392857 ( This is the final answer).