Minimum of exponential distributions

Question

Let's say we have three independent random variables $X, Y$ and $Z$ with Exponential distributions of parameter 1.

What is the probability $$ P( X = \min(X,Y,Z)) \quad ?$$

I obtained that it is $1/3.$ Anyone could confirm me that it is correct?

Answer 1

$$F_{min}(x) = P(W_{min} \leq x)$$ $$= 1- P(W_{min} > x)$$$$= 1-P(X > x; Y > x; Z> x)$$ $$= 1 - [1 - F_X(x)][1 - F_Y(x)][1 - F_Z(x)]$$

$$ = 1-(1-1+e^{-x})^3 = 1-e^{-3x}$$

$$f_{min}(x) = 3e^{-3x}$$

$$E(X_{min}) = \frac{1}{3}$$

Answer 2

Average $P(Y,\,Z>x)=\exp -2X$, viz. $\int_0^\infty \exp -3xdx=\frac{1}{3}$.