# Equivalence between exponential clocks that ring in different ways

Could someone explain to me why the following facts are equivalent.

1) I have a clock that rings with exponential distribution of parameter $$1$$. When the click rings a transition to the state $$s_1$$ occurs with probability $$p$$ and a transition to $$s_2$$ occurs with probability $$1-p$$.

2) I have two independent clocks that ring with exponential distribution of parameter $$p$$ and $$1-p$$ respectively. If the clock with parameter $$p$$ rings first then a transition to $$s_1$$ occurs, if the clock with parameter $$1-p$$ rings first then a transition to the state $$s_2$$ occurs.

Why the two systems are equivalent?

This equivalence comes down to the following two facts (which you can try proving as an exercise). Suppose that $$X \sim \mathrm{Exponential}(a)$$ and $$Y\sim \mathrm{Exponential}(b)$$ (where $$a$$ and $$b$$ are rate parameters) and $$X$$ and $$Y$$ are independent. Then:
1. $$\min(X,Y)\sim \mathrm{Exponential}(a + b)$$
2. $$\mathbb{P}\left(X \leq Y \right) = \dfrac{a}{a + b}$$.
Using these, you can see that in both your formulations, the time to transition is $$\mathrm{Exponential}(1)$$ and the probability of transitioning to $$s_{1}$$ is $$p$$.