This question is a lot like your real estate agent problem. However, we're going to solve this one in a different way.
Let's let `p_1` be the probability that Truck 1 is available. Similarly, let's let `p_2` be the probability that Truck 2 is available.
Notice that if we just add the probabilities, we are grossly overcounting the chances of the two trucks being available!
`p_T = p_1 + p_2 = 0.96 + 0.96 = 1.92`
We certainly can't have a proability greater than 1!
However, let's consider the fact that we're forgetting one case, which, based on the probabilities, happens very often! Where both trucks are available at the same time!
We need to recognize that if one truck is available, we shouldn't be considering whether the other truck is also available! However, we should consider another alternate situation: where the first truck is unavailable AND the second truck is available.
This is a fairly simple conditional probability. The probability of two independent events occuring at the same time is the product of their respective probabilities. Let's let `p_(bar1,2)` be the probability that the first truck is not available and that the second truck is available.
`p_(bar1,2) = (1-p_1)p_2=0.04*0.96 = 0.0384`
So, now we consider combining the probabilities in a way that does not double-count the case in which both trucks are available:
`p_T = p_1 + p_(bar1,2) = 0.96 + 0.0384 = 0.9984`
Well, that looks like a more reasonable probability! In fact, it's the correct probability.
To confirm this probability, because it does seem a bit close to 1, we can think of it in a different way. We know that the only case in which no truck is available is the case where Truck 1 AND Truck 2 are unavailable! In math terms:
`p_(barT) = (1-p_1)(1-p_2) = 0.04*0.04 = 0.0016`
So, our total probability will be:
`p_T = 1-p_(barT) = 1-0.0016 = 0.9984`
And there it is confirmed! Our final probability of having at least one truck available is 0.9984.
I hope that helps!