Use the rate convention for the <exponential distribution>. By <convolution of independent random variables>, the <probability density function> of $S=X+Y$ is, for $s\geq0$,
$$
f_S(s)=\int_0^s\lambda e^{-\lambda x}\mu e^{-\mu(s-x)}\,dx
=\boxed{\frac{\lambda\mu}{\mu-\lambda}
\left(e^{-\lambda s}-e^{-\mu s}\right)}.
$$
It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage <hypoexponential distribution>. As a check, its limit as $\mu\to\lambda$ is $\lambda^2s e^{-\lambda s}$, the shape-two <gamma distribution>.

For $M=\min\{X,Y\}$, <independence> gives the <survival function> $\mathbb P(M>s)=e^{-\lambda s}e^{-\mu s}$ for $s\geq0$. Differentiating its complement gives
$$
\boxed{f_M(s)=(\lambda+\mu)e^{-(\lambda+\mu)s}\mathbf1_{\{s\geq0\}}.}
$$
Thus the minimum has an <exponential distribution> whose rate is the sum of the rates.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 10F / b / Solution

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A hypoexponential distribution is the distribution of a sum of <independent> variables with <exponential distributions>. For two positive unequal rates $\lambda,\mu$, <convolution of independent random variables> gives the <probability density function>
$$
f(s)=\frac{\lambda\mu}{\mu-\lambda}(e^{-\lambda s}-e^{-\mu s}),\qquad s\geq0.
$$
It is zero for negative $s$. Its equal-rate limit is the shape-two <gamma distribution> with rate $\lambda$.

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 Incoming links: Hypoexponential distribution