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

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

$$P(T_{1}

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?

• I think it's easiest to observe $P(T_1 < T_2) = E[P(T_1 < T_2 | T_2 = t)]$ and compute the latter term. – David Kraemer Feb 18 at 16:14

Compute the following integral. \begin{align} P(T_1