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




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