A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ? A) 11/91 B) 4/11 C) 1/11 D) 90/91

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14

= $C414$ ways =  $\frac{14\times 13\times 12\times 11}{4\times 3\times 2\times 1}$= 1001

Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

   $\Rightarrow C15+ C44+ C15$ = 5+1+5 =11

 P(E) =  $\frac{n\left(E\right)}{n\left(S\right)}=\frac{11}{1001}=\frac{1}{91}$  

 Required probability =  $1-\frac{1}{91}=\frac{90}{91}$