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

Let E be the event that one of the balls must be red

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

$n (E) = {}^6C_1{}^4C_1{}^8C_1+{}^6C_1{}^4C_2+{}^6C_1{}^8C_2 = 396$

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

= 396 / 816

= 33/68