Need to calculate $\Pr \left( {\frac{{X + c}}{Y} \le t} \right)$ where $c>0$
My attemp of solving this probability was $\Pr \left( {\frac{{X + c}}{Y} \le t} \right) = \Pr \left( {X \le tY - c} \right) = \int_{y = 0}^{y = \infty } {\Pr \left( {{{\left. {X \le ty - c} \right|}_{Y = y}}} \right)} {f_Y}(y)dy$.
This would lead to [\begin{array}{l} \int_{y = 0}^{y = \infty } {\Pr \left( {{{\left. {X \le ty - c} \right|}_{Y = y}}} \right)} {f_Y}(y)dy = \int_{y = 0}^{y = \infty } {\left( {1 - {e^{ - {\lambda _1}\left( {ty - c} \right)}}} \right) \times } {\lambda _2}{e^{ - {\lambda _2}y}}dy\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \frac{{{{\rm{e}}^{c{\lambda _1}}}{\lambda _2}}}{{t \times {\lambda _1} + {\lambda _2}}} = \frac{{t \times {\lambda _1} + {\lambda _2} - {{\rm{e}}^{c{\lambda _1}}}{\lambda _2}}}{{t \times {\lambda _1} + {\lambda _2}}} \end{array}]
However while I was doing this with Mathematica, there was some inconsistency
Mathematica said that my answer would only valid for $c < 0$, denoted by the red arrow, while for c>0 it must be $\frac{{{{\rm{e}}^{ - \frac{{c{\lambda _2}}}{t}}}t{\lambda _1}}}{{t{\lambda _1} + {\lambda _2}}}$
Is there any mistake in my derivation ? Any help would be greatly appreciated, thank you !
