Particle 1 has a characteristic decay rate of $\tau_1,$ particle 2 has a characteristic decay rate of $\tau_2.$

The decay rate is defined as: during a small time interval $\Delta t,$ the probability that the particle decays is $\Delta t/\tau.$ So the probability that it doesn't decay is $1-\Delta t/\tau.$ In a finite time $T$, we have to multiply the probabilities that the particle hasn't decayed during interval 1, nor during interval 2, nor during interval 3, etc. The probability that the particle hasn't decayed is:

$$P(\textrm{hasn't decayed})=\left(1-\frac{\Delta t}{\tau}\right)^{T/\Delta t}\approx e^{-T/\tau}$$

So the probability that particle 1 hasn't decayed and particle 2 hasn't decayed can be found by multiplying the two independent probabilities:

$$P(\textrm{no decays in time T})=e^{-T/\tau_1}e^{-T/\tau_2}$$

We see it makes sense to define $\tau_3$ as

$$\frac{1}{\tau_3}=\frac{1}{\tau_1}+\frac{1}{\tau_2}$$

I always have to use roundabout reasoning to think about these things; we have the probability that either particle decays is one minus the probability that neither particle decays:

$$P(\textrm{either decays in time T})=1-e^{-T/\tau_3}$$

and this should be interpreted as the cumulative distribution of either particle decaying at time $T$:

$$\rho(T)=\frac{1}{\tau_3}e^{-T/\tau_3}$$

The expectation value $\langle T\rangle=\int_0^\infty T\rho(T)dT$ is just $\tau_3.$ So the expected time of first decay of either particle is

$$\tau_3=\left(\frac{1}{\tau_1}+\frac{1}{\tau_2}\right)^{-1}$$

:~)