Probability that an independent exponential random variable is the least of three Let $Y_1, Y_2, Y_3$ be independent exponentially distributed random variables, with parameters $\lambda_1, \lambda_2, \lambda_3$ respectively. Why is it the case that:
$P(Y_1=min(Y_1,Y_2,Y_3))=\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$?
I just came across this fact but I'm not sure where it comes from. I do know that if we define $Y=min(Y_1,Y_2),$ then $Y$ is  an exponential random variable with parameter $\lambda_1+\lambda_2$. Is this fact related to the one above?
Thanks
 A: This yields the same result as @p.s. provided, but this is how I'd think to solve it.
Let $f_{Y_1}(t)$ denote the density of $Y_1$, and note that $$\{ Y_1 = min(Y_1,Y_2,Y_3) \} \iff \{ Y_2 \geq Y_1, Y_3 \geq Y_1\}.$$
Then, by independence,
\begin{align*}
P(Y_1 = min(Y_1,Y_2,Y_3)) &= \int_0^{\infty} P(Y_2 \geq Y_1, Y_3 \geq Y_1| Y_1 = t)f_{Y_1}(t)dt \\
&= \int_0^{\infty} P(Y_2 \geq t) P(Y_3 \geq t)f_{Y_1}(t)dt \\
&= \int_0^{\infty} e^{-\lambda_2t}e^{-\lambda_3t}\lambda_1e^{-\lambda_1t}dt\\
&= \int_0^{\infty} \lambda_1 e^{-(\lambda_1+\lambda_2+\lambda_3)t}dt \\
&= \frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}
\end{align*}
A: Not sure what the intuition is, but it's a straightforward integral to work out. 
$$
\begin{aligned}\int_0^\infty \int_{0}^\infty& \int_{0}^\infty [x_1 \le x_2] [x_1 \le x_3] \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\
&=
\int_0^\infty \int_{x_1}^\infty \int_{x_1}^\infty \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\
&=\int_{0}^\infty \lambda_1   e^{-(\lambda_1 +\lambda_2 +\lambda_3) x_1}dx_1 
\\&=\frac{\lambda_1}{\lambda_1 +\lambda_2 +\lambda_3}\end{aligned}$$
(This is using the fact that expectation of the indicator function of an event equals the probability of that event.)
