Combined probability distribution for two hazard functions

Question

I have two items with two different failure rates. I need to find the combined hazard function corresponding to both items failing, from which I can find the combined probability distribution.

I have a Monte Carlo based example solution which should show what I am trying to figure out:

"Item1" and "item2" have independent failure rates of 0.1 and 0.05 respectively.

The hazard function for either failing is simply the sum of their failure rates, $h_{either}(t) = h_1(t) + h_2(t) = 0.15$.

The hazard function for both failing is what have not been able to find. Intuition tells me the equation is $h_{both}(t) = h_1(t) \cdot F_2(t) + h_2(t) \cdot F_1(t)$, which is what I have shown in the plot (orange line), but that is wrong based on its comparison to the Monte Carlo results (intuition does not seem to work with statistics).

Just to be complete, here is psudo code for the Monte Carlo approach:

M = 1000
N = 500000
end_time = 100
h1(t) = 0.1
h2(t) = 0.05
sample_weight = M / (end_time * N)
dt = end_time / (N - 1)
For N samples
    failed1 = 0
    failed2 = 0
    failed_either = 0
    for M time steps
        if (not failed1) and (random() < h1(t) * dt)
            f1(t) = f1(t) + sample_weight
            failed1 = 1
        if (not failed2) and (random() < h2(t) * dt)
            f2(t) = f2(t) + sample_weight
            failed2 = 1
        if (not failed_either) and (failed1 or failed2)
            f_either(t) = f_either(t) + sample_weight
            failed_either = 1
        if failed1 and failed2
            f_both(t) = f_both(t) + sample_weight
            break

And the "(hazard function)" lines in the plot are just numeric solutions for $f(t)$ given $h(t) = \frac{f(t)}{R(t)}$.

So I am looking for the equation which combines two hazard functions into one, e.g. a computer with redundant power supplies such that both have to fail in order for the computer to fail. Note that although the given example has constant failure rates, I am looking for a general solution.

Answer 1

I figured this out, my mistake was in confusing failure rates for probabilities. Rather than $h_{both}(t) = h_1(t) \cdot F_2(t) + h_2(t) \cdot F_1(t)$ the correct answer is $f_{both}(t) = f_1(t) \cdot F_2(t) + f_2(t) \cdot F_1(t)$, from which you can find the combined hazard rate with $h(t) = \frac{f(t)}{R(t)}$. Here are updated Monte Carlo results to show veracity: And the equation holds for non-constant failure rates ($f_1$ is normal distrib with $\mu = 12, \sigma = 3$, $f_2$ with $\mu=16, \sigma = 7$):