A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 =  $15C3$  = $\frac{15*14*13}{3*2*1}$= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = $5C3$ = $\frac{5*4}{2*1}$ = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.