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:
- 7110
- 711
- 755
- 1455
Answer
(4) 1455
Solution
Let Ek: bag has k red balls, k = 0, 1, …, 10 (11 equally likely cases) P(Ek) = 111
A: the three balls drawn are black P(A Ek) = 10-kC310C3
By Bayes' theorem:
P(E1 A) = 9C3 k=010 10-kC3
k=010 10-kC3 = 3C3 + 4C3 + + 10C3 = 11C4 = 330
P(E1 A) = 84330 = 1455
Answer: (4)