Suppose $X_1$, $X_2$ are i.i.d. exponential with rate $\lambda = 3$ and $Y_1, Y_2, Y_3, Y_4$ are i.i.d. exponential with $\lambda = 4$. Find $P(X1+2X2+Y1>3)$.
This was only part of the question, but this was a part that I had no idea how to do. Obviously, $X_1 + 2X_2 + Y_1$ are all independent so this ends up being an $\exp(3)+\exp(3/2)+\exp(4)$, and we are trying to find the probability that the sum of three independent exponential variables is greater than $3$. My initial thought was to use some kind of order statistics, where I find that the minimum of the three variables is greater than $3$. However, I then realized that was completely wrong since it is the sum of three exponential random variables. Does anybody know how to proceed with such a question?
If $X\sim\mathsf{Expo}(\lambda)$ and $Y\sim\mathsf{Expo}(\mu)$ are independent where $\lambda\ne\mu$, then the density of $Z:=X+Y$ can be found by convolution: \begin{align} f_Z(t) &= \int_{\mathbb R}f_X(s)f_Y(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\mu e^{-\mu(t-s)}\ \mathsf ds\\ &= \lambda\mu e^{-\mu t} \int_0^t e^{-(\lambda +\mu)s}\ \mathsf ds\\ &= \frac{\lambda\mu}{\mu-\lambda} (e^{-\lambda t}-e^{-\mu t}). \end{align} Moreover, if $c>0$ then for all $t>0$ we have $$ \{cX\leqslant t\} = \left\{X\leqslant \frac tc\right\}, $$ and a quick computation shows that $cX$ is exponentially distributed with rate $\frac\lambda c$. So we can use convolution to find the distribution of $Z:=X_1+2X_2+Y_1$; assuming $\mathbb E[X_1] = \frac1\lambda$ and $\mathbb E[Y_1]=\frac1\mu$, we find that $$ f_Z(z) = \frac{\lambda \mu e^{-(\lambda+\mu)t} \left(\lambda e^{\mu t}-2 \lambda e^{\frac{\lambda t}{2}+\mu t}+2 \mu e^{\frac{\lambda t}{2}+\mu t}+\lambda e^{\lambda t}-2 \mu e^{\mu t}\right)}{(\lambda -2 \mu ) (\lambda -\mu )}. $$ From this we find $$ \mathbb P(Z>3) = \int_3^\infty f_Z(t)\ \mathsf dt = \frac{e^{-3 (\lambda +\mu )} \left(e^{3 \lambda } \lambda ^2+e^{3 \mu } \mu \left(-4 e^{\frac{3 \lambda }{2}} (\lambda -\mu )+\lambda -2 \mu \right)\right)}{(\lambda -2 \mu ) (\lambda -\mu )}. $$ Substituting $\lambda=3$ and $\mu=4$, the above is $$ \frac15 e^{-12}\left(4 e^3 \left(4 e^{9/2}-5\right)+9\right)\approx 0.0350662. $$ I sincerely hope your instructor does not expect you to do computations like these by hand...