We can only add probabilities when the events are alternatives. For example, let's say we want to calculate the probability that we achieve an A or a B or a C in a given exam. We cannot achieve an A and B for the same exam, so these events are indeed alternatives.

So let's say there's a 1/10 chance of getting an A, a 3/10 chance of getting a B and 4/10 chance of getting a C. The chance that we get either an A, B or C would be the sum of the individual probabilities: (1/10 + 3/10 + 4/10), so 8/10.

Think about the problem numerically. We know that all probabilities lie in the range from 0 to 1. Addition of such numbers will lead to an increased probability value. Multiplication of such numbers will lead to a decreased probability value. In this case, we'd expect the probability of attaining A or B or C to be greater than the probability of simply attaining an A. And yes, we have an 8/10 chance of attaining an A,B or C and only a 1/10 chance of attaining an A - this is what we'd expect.

We multiply probabilites when we want events to occur simultaneously, or consecutively. Let's say we're going to sit the same exam twice. We want to find the probability of attaining a C first time and an A on the re-sit. Before we dive in let's think about the expected probability value: getting a C AND THEN an A is a very unique scenario, and as such we'd expect quite a low probability in relation to the probabilities of individual events. This hints to us that multiplication could be the correct method.

So in this case, yes, we can say that the probability of getting a C and then an A is (4/10 * 1/10), so 1/25.

f is a function such that f(x)=2/(3x-3) Find the inverse function and ff^-1

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