A bag contains 10 balls of which 2 are red and the remaining are either blue or black. If the probability drawing 3 balls of the same colour is $\frac{11}{20}$ and if the  number of blue balls exceeds the number of black balls, the number of blue balls is

$\text{ Let the number of blue balls be }n\text{. }$

$\therefore \text{ the number of black balls }=10-(n+2)=8-n\text{. }$

$\text{ If all the three balls drawn are of the same colour, they must be either blue or black, }$

$\text{ the probability for which is }\frac{{ }^n{C}_3}{{ }^{10}{C}_3}+\frac{{ }^{8-n}{C}_3}{{ }^{10}{C}_3}=\frac{11}{120}$

$\begin{array}{l}\Rightarrow n(n-1)(n-2)+(8-n)(7-n)(6-n)\\ =10\times 9\times 8\cdot \frac{11}{120}\\ \Rightarrow 18n^2-144n+336=66\\ \Rightarrow n^2-8n+15=0\\ \Rightarrow (n-5)(n-3)=0\\ \Rightarrow n=5\text{ (as }n>\text{ number of black balls) }\end{array}$