A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) one is red and two are white
(ii) two are blue and one is red

Given is a bag containing 6 red, 4 white and 8 blue balls.

By using the formula of probability we get,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random, so the total possible outcomes will be ${}^{18}C_3$

n (S) = 816

(i) Let E be the event of getting one red and two white balls

E = {(W) (W) (R)}

$n (E) = {}^6C_1{}^4C_2 = 36$

P (E) = n (E) / n (S)

= 36 / 816

= 3/68

(ii) Let E be the event of getting two blue and one red

E = {(B) (B) (R)}

$n (E) = {}^8C_2{}^6C_1 = 168$

P (E) = n (E) / n (S)

= 168 / 816

= 7/34