All Seasons Plumbing has two service trucks which frequently break
down. If the probability the first truck is available is .75, the
probability the second truck is available is .50, and the probability
that both trucks are available is .30, what is the probability neither
truck is available?

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Interesting problem--if indeed related to business.
Assuming that it is, I believe an answer might be as follows:
First, let the first truck's availability be event A and the
availability of the second truck be event B. From your info, then, we
are provided with P(A) =0.75 & P(B)= 0.5. Now, if events A & B were
mutually exclusive(i.e., independent from each other), then the
probability of both event A and event B occuring together, P(AB),
would be
P(AB)=P(A)P(B)
And conversely, P(A'B')=P(A')P(B')
Where A' = Event of Nonavailability of First Truck
      B' = Event of Nonavailability of Second Truck
      A'B' = Event of neither the First and Second Truck being available
And P(A') = 1 - P(A)
    P(B') = 1 - P(B)
However, since P(AB) = 0.3 (as given), then we know that these events
are not independent from one another (perhaps they share the same
maintenance resources,etc.) and we must pursue another path.
Consider the following:
P(A) = P(AB) + P(AB')
P(A') = P(A'B) + P(A'B')
P(B) = P(AB) + P(A'B)
P(B') = P(AB') + P(A'B')
Solving for P(AB') using the first equation yields:
0.75 = 0.3 + P(AB')
P(AB') = 0.45 (Note: This is the probability of the First truck being
available and the Second truck not being available concurrently)
Further, as P(B') = 1 - P(B), then P(B') = 1 - 0.5 = 0.5
Solving for P(A'B') yields:
0.5 = 0.45 + P(A'B')
Therefore,

P(A'B') = 0.05

Thus, the probability that neither the first and second truck will not
be available is 5%.