A bag contains a mixed lot of red and blue balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls and the probability of drawing one ball of each colour is six times the probability of drawing two blue balls. Then, the total number of red and blue balls in the bag is

Let, a = number of red balls.
b = number of blue balls,
$p_1$= probability of drawing two red balls

$=\frac{{ }^a{C}_2}{{ }^{a+b}{C}_2}=\frac{a(a-1)}{(a+b)(a+b-1)}$

$p_2=$probability of drawing two blue balls $=\frac{{ }^b{C}_2}{{ }^{a+b}{C}_2}=\frac{b(b-1)}{(a+b)(a+b-1)}$

$p_3=$probability of drawing one red and one blue ball

$=\frac{{ }^a{C}_1.{}^b{C}_1}{{ }^{a+b}{C}_2}=\frac{2ab}{(a+b)(a+b-1)}$

Given that $p_1=5p_2\&p_3=6p_2$

$\iff a(a-1)=5b(b-1)$ and 2ab = 6b (b – 1)

$\Rightarrow a=6,b=3\Rightarrow$ Total number of balls = 9