There are 8 blue and 4 white balls in a bag . A ball is drawn at random. Without replacing it another ball is drawn. Find the probability that both the balls drawn are blue?

Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E₁) = ⁸C₁ / ¹²C₁
= 8/12 = 2/3
After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.
Probability of drawing one blue ball in the second drawn = P(E₂) = ⁷C₁ /¹¹C₁ = 7/11
∴ Probability that both are blue P(E₁ ∩ E₂) = P(E₁) x P(E₂)
= 2/3 x 7/11 = 14 / 33