Question -
Answer -
Given: A bagcontaining 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Two balls are drawn atrandom.
Total possibleoutcomes are 18C3
n (S) = 816
(i) Let E be theevent of getting one red and two white balls
E = {(W) (W) (R)}
n (E) = 6C14C2 =36
P (E) = n (E) / n (S)
= 36 / 816
= 3/68
(ii) Let E be theevent of getting two blue and one red
E = {(B) (B) (R)}
n (E) = 8C26C1 =168
P (E) = n (E) / n (S)
= 168 / 816
= 7/34
(iii) Let E be theevent 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
P (E) = n (E) / n (S)
= 396 / 816
= 33/68