Let $X_1, X_2$ and $X_3$ exponential random variables with the same parameter $\beta$?
The PDF and CDF of $X_i$ are $$ f_{X_i}(x)=\beta e^{-x \beta}, $$ $$ F_{X_i}(x)=1-e^{-x \beta}. $$
The PDF and CDF of $Y=\min\{X_1,X_2\}$ are $$ f_{Y}(y)=2\beta e^{-2\beta y}, $$ $$ F_{Y}(y)=1-e^{-2\beta y}. $$ I would lieke to find the probability that $$ P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}, $$ and the probability $$ P\Big\{\min\{X_1,X_2\}> X_3\Big\}. $$ I use the following steps \begin{align} P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}&=P\Big\{Y\leq X_3\Big\}\\ &=\int_{x_3=0}^{\infty}\left(\int_{y=0}^{x_3}f_Y(y)dy\right)f_{X_3}(x_3)dx_3\\ &=\int_{x_3=0}^{\infty}F_y(x_3)f_{X_3}(x_3)dx_3\\ &=1-\int_{x_3=0}^{\infty}e^{-2\beta x_3}\beta e^{-\beta x_3}dx_3\\ &=1-\beta\int_{x_3=0}^{\infty}e^{-3\beta x_3}dx_3\\ &=1-\frac{\beta}{3\beta}\\ &=2/3. \end{align} For $$ P\Big\{\min\{X_1,X_2\}> X_3\Big\}=1-P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}=1/3. $$
IS my derivation true?
Thanks.