An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability =  $PA\cap B\cap C$ 

=  $PA P\frac{B}{A} P\frac{C}{A\cap B}$

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

  $\therefore P\frac{B}{A}=\frac{3}{17}$ 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

  $\therefore P\frac{C}{A\cap B} =\frac{2}{16}=\frac{1}{8}$

 Hence the required probability =  $\frac{2}{9}\times \frac{3}{17}\times \frac{1}{8}=\frac{1}{204}$