Let $Y$ and $W$ are independent random variable with PDF $$f_Y(y)=\begin{cases} \frac13~e^{\frac {-y}{3}}& y\geq0\\~\\ 0&\text{else} \end{cases}$$ $$f_W(w)=\begin{cases} \frac14~e^{\frac {-w}{4}}& w\geq0\\~\\ 0&\text{else} \end{cases}$$ Find $P(Y>W)$.
Define $Z=Y+(-W)$ and we find $P(Z>0)$. Using substitution I find density of $-W$ as $$f_{(-W)}(w)=\begin{cases} \frac14~e^{\frac {w}{4}}& w\leq0\\~\\ 0&\text{else} \end{cases}$$ And using convolution I find density of $Z$ as $$f_Z(z)=\begin{cases} \frac17~e^{\frac {z}{4}}& z\leq0\\~\\ \frac17~e^{\frac {-z}{3}}& z\geq0 \end{cases}$$Now $$\begin{align} P(Z>0)&=\int\limits_{(0,\infty)} f_Z(z)~dz\\~\\ &=\int\limits_{(0,\infty)}\frac17~e^{\frac {-z}{3}}~dz\\~\\ &=\frac37 \end{align}$$
This answer is given in options ,I don't have answer key though. Is my calculation correct? Even if it is I think there is another easy way to find $P(Y>W)$.
Please share other approaches to solve this question.
