A bag contains “a” white and “b” black balls. Two players A and B alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. If A begins the game and the probability of A winning the game is three times that of B, then the value of   $\frac{a}{b}$ is 

Let  $E_1$  denote the event of drawing a white ball at any draw and $E_2$  that for a black ball and let  $E$  be the event for A winning the game
$\therefore P\left(E_1\right)=\frac{a}{a+b}$  and  $P\left(E_2\right)=\frac{b}{a+b}$
 $$\begin{gathered} \therefore P\left(E\right)=P\left(E_1\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}{E}_2{E}_2{E}_1\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}{E}_2{E}_2{E}_2{E}_2{E}_{1}or\mathrm{.....}\right) \\ =P\left(E_1\right)+P\left(E_2{E}_2{E}_1\right)+P\left(E_2{E}_2{E}_2{E}_2{E}_1\right)+\mathrm{...} \\ =P\left(E_1\right)+P\left(E_2\right)P\left(E_2\right)P\left(E_1\right)+P\left(E_2\right)P\left(E_2\right)P\left(E_2\right)P\left(E_2\right)P\left(E_1\right)+\mathrm{...} \\ [\because E_1\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{E}_2\text{\hspace{0.17em}\hspace{0.17em}are\hspace{0.17em}\hspace{0.17em}independent}] \end{gathered}$$                                          
   $=\frac{P\left(E_1\right)}{1-{\left(P\left(E_2\right)\right)}^2}$          [sum of infinite GP]
 $=\frac{\frac{a}{a+b}}{1-{\left(\frac{b}{a+b}\right)}^2}=\frac{a(a+b)}{a^2+2ab}\therefore P\left(E\right)=\frac{a+b}{a+2b}$ ,
Then,  $P\left(E^{\text{'}}\right)$  is the probability for B winning the game
 $\therefore P\left(E^{\text{'}}\right)=1-P\left(E\right)=1-\frac{a+b}{a+2b}=\frac{b}{a+2b}$
According to the problem,  $P\left(E\right)=3P\left(E^{\text{'}}\right)$
 $\Rightarrow \frac{a+b}{a+2b}=\frac{3b}{a+2b}\Rightarrow \alpha +\beta =3\beta \Rightarrow \alpha =2\beta$
 $\therefore \frac{a}{b}=\frac{2}{1}=2$
(A) α + β = 2, (B) δ - γ = 3, (C) δ + β = 4