A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :
(a) three are red;
(b) the first is blue and the other two are red;
(c) two are white and one is blue.
Explanation
n(R) = number of red balls = 12;
n(W) = number of white balls = 16;
n(B) = number of blue balls = 8
Total number of balls = 36
(a) P(three are red) = (frac{12}{36} times frac{11}{35} times frac{10}{34})
= (frac{11}{357})
(b) P(first is blue and other two red) = (frac{8}{36} times frac{12}{35} times frac{11}{34})
= (frac{44}{1785})
(c) P(two are white and one is blue) = (frac{16}{36} times frac{15}{35} times frac{8}{34})
= (frac{16}{357})