= Solution
Write $a=\beta^*=\beta N+\gamma$. If the first event is recovery, the epidemic has ended and a second event is impossible. If the first event is infection at time $s$, its density is $\beta N e^{-as}$, and the new state is $(N-1,2,0)$. The total next-event rate is
$$
b=2\beta(N-1)+2\gamma=2(a-\beta).
$$
Thus, by the <Markov property>, \b[the <two-event probability in a stochastic SIR model> by time one] is
$$
\int_0^1\beta N e^{-as}\left(1-e^{-b(1-s)}\right)\,ds.
$$
Evaluate the first term as $\beta N(1-e^{-a})/a$. For the second, $b-a=a-2\beta$, so
$$
\int_0^1\beta N e^{-as-b(1-s)}\,ds=\frac{\beta N e^{-a}}{a-2\beta}\left(1-e^{-(a-2\beta)}\right).
$$
Therefore
$$
\boxed{\mathbb P(\text{at least two events in }[0,1))=\frac{\beta N}{\beta^*}(1-e^{-\beta^*})-\frac{\beta N e^{-\beta^*}}{\beta^*-2\beta}(1-e^{-(\beta^*-2\beta)}).}
$$
For $N\ge2$ and $\gamma>0$, the denominator $\beta^*-2\beta=\beta(N-2)+\gamma$ is positive. Event times are continuous, so inclusion or exclusion of the endpoint one makes no difference.
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