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
(iii) one is red
given: bag which contains 6 red, 8 blue and 4 white balls
formula:
two balls are drawn at random, therefore
total possible outcomes are 18C3
therefore n(S)=816
(i) let E be the event of getting one red and two white balls
E= {(W) (W) (R)}
n(E)= 6C14C2=36
(ii) let E be the event of getting two blue and one red
E= {(B) (B) (R)}
n(E)= 8C26C1=168
(iii) 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)= 6C14C18C1+6C14C2+6C18C2=396