There are ‘a’ white and ‘b black balls in an Urn. A ball is drawn and its colour is noted then the ball is not replaced. If a second ball is drawn, the probability that both the balls are white is

Let E₁, E₂ be the events that the 1^st, 2^nd balls drawn are both white. Since then first ball drawn is replaced before drawing the 2nd ball, the events E₁, E₂ are independent 
$\begin{array}{l}\therefore \mathrm{P}\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\right)=\mathrm{P}\left({\mathrm{E}}_1\right)\cdot \mathrm{P}\left({\mathrm{E}}_2\right)\\ =\left(\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}\right)\left(\frac{\mathrm{a}-1}{\mathrm{a}+\mathrm{b}-1}\right)\\ =\frac{a\left(a-1\right)}{\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}-1\right)}\end{array}$