Given: A bag containing 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = \( \dfrac{favourable\; outcomes }{total\; possible\; outcomes}\)
Three balls are drawn so, we have to find the probability that one is red, one is white and one is blue.
Total number of outcomes for drawing 3 balls is 18C3
n (S) = 18C3 = 816
Let E be the event that one red, one white and one blue ball is drawn.
n (E) = 6C14C18C1 = 192
P (E) =\( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{192}{816}\)
= \( \dfrac{4}{17}\)
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