Final answer:
In probability functions, P(0 < x < 12) would be 1 if f(x) is uniform over that interval. Describing P(x > 3) for a function between 1 and 4 would result in 1/4, assuming a uniform distribution. The probability of an event's complement, P(A'), is calculated as 1 minus the probability of the event, P(A).
Step-by-step explanation:
The question regards understanding probability functions and calculating specific probabilities given the conditions of a continuous probability function. For instance, if f(x) represents a continuous probability function that is equal to 1/12 over the interval from 0 to 12, to find P(0 < x < 12), we would recognize that this is a uniform probability distribution. Since the total area under the function from 0 to 12 must equal 1 (the probability of all outcomes), the probability of any subset of that interval is simply the length of the interval divided by 12, which would be 1. This mean P(0 < x < 12) = 1, because it constitutes the entire probability space.
For the question asking to describe P(x > 3) when f(x) is a probability function between 1 and 4, we can infer that the probability is the area under the probability function from x equals 3 to 4. Assuming a uniform distribution, this would be 1/4 since the interval from 3 to 4 is one-fourth of the total interval from 1 to 4.
Finding complements of events such as P(A') is another common task in probability. The complement A' includes all outcomes not in A, and by definition, P(A) + P(A') = 1. For a discrete sample space like {1, 2, 3, ..., 19}, each outcome has an equal chance, so P(A) and P(A') can be calculated directly from the number of outcomes in each set.