$P(T_{1}<T_{2})=\frac{\omega_{1}}{\omega_{1}+\omega_{2}}$

Question

Supposing that $ T_{1} $ and $T_{2}$ are independent and exponentially distributed, with parameters $\omega_{1} $ and $\omega_{2}$ respectively. Then,

$$P(T_{1}<T_{2})=\frac{\omega_{1}}{\omega_{1}+\omega_{2}}$$

I have recently been looking into exponential and gamma distributions and came across this statement, but am unsure on how to prove it. I was thinking of using the memory-less property. $$P(T>t+s|T>s)=P(T>t)$$ Would this be a correct way to start?

Answer 1

Compute the following integral. $$ \begin{align} P(T_1<T_2)&=\int_{0}^\infty \int_{0}^y \omega_1e^{-w_1x}\omega_2e^{-w_2y}\, dx\, dy\\ \end{align} $$