A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one without replacement, find the probability of getting all white balls.

Let A, B, C, D denote events of getting a white ball in first, second, third and fourth draw  respectively. Then,

Required probability = P(A ∩ B ∩ C ∩ D) = P(A) P (B/A) P (C/A ∩ B) P(D/A ∩ B ∩ C)….(i)

Now, P(A) = Probability of drawing a white ball in first draw = 5/20 =1/4

When a white ball is drawn in the first draw there are 19 balls left in the bag, out of which are white.

∴ P (B/A) = 4/19 

Since the ball drawn is not replaced, therefore after drawing a white ball in second draw there are 18 balls left in the bag, out of which 3 are white.

∴ P (C/A ∩ B) = 3/18 = 1/6

After drawing a white ball in third draw there are 17 balls left in the bag, out of which 2 are white.

∴ P (D/A ∩ B ∩ C) = 2/17

Hence, required probability =P (A ∩ B ∩ C ∩ D)

= P(A) P (B/A) P (C/A ∩ B) P (D/A ∩ B ∩ C) = 1/4 × 4/19 × 1/6 × 2/17 = 1/969