The lifetime of an electronic device is a rv with exponential distribution ($\mu=1/10$). In a normal week, the hours that the device is used is a rv with Poisson distribution ($\lambda=12$). Calculate the probability that a new device, in a normal week, won't fail during that week.
I'm stuck with this type of problems. So far I have the following:
Let $X$: lifetime of an electronic device and $Y$: hours that the device is used. Then, I have to calculate $P[X>Y]$. So,
$$P[X>Y]=\sum_{y=0}^{\infty}P[X>Y|Y=y]P[Y=y].$$
I have no idea if I can suppose that X and Y are independent.