A number is selected at random from 1 through 100, inclusive.  What is the probability that the number is a multiple of 3?
So the probability of an event occurring is the number of favorable outcomes divided by the total number of possible outcomes. Right, in this case the total number of possible outcomes is 100 since any number from 1 to 100 could be selected. Now we need to determine the number of favorable outcomes. That would be the number of numbers from 1 to 100 that are multiples of 3. So it's the number of multiples of 3 from 1 to 100. The multiples of 3 from 1 to 100 are 3, 6, 9, ..., 99. That's correct. Now we need to determine how many numbers are in that list. We could count them but there's a better way. Right. Since the first term is 3 and the last term is 99 and the common difference is 3, we can use the formula for the number of terms in an arithmetic sequence: $n = \dfrac{last\:term - \:first\:term}{common\:difference} + 1$. So there are $\dfrac{99 - 3}{3} + 1 = 33$ multiples of 3 from 1 to 100. Now that we have the total number of possible outcomes and the number of favorable outcomes, we can compute the probability. That is $33 \div 100 = \dfrac{33}{100}$. Right, so the probability is $\dfrac{33}{100}$.

