probability that the die will NOT yield a 6. Thus, the probability that on all three rolls the die will not yield a 6 is:
.As shown in the previous section, you can solve “OR” problems (explicit or disguised) by combining the probabilities of individual events. If there are many individual events, though, such calculation may be tedious and time-consuming. The good news is that you may not have to perform these calculations. In certain types of “OR” problems, the probability of the desired event not happening may be much easier to calculate.
For example, in the previous section, you could have calculated the probability of getting at least one head on two flips by considering how you would NOT get at least one head. However, it would not be too much work to compute the probability directly, using the slightly more complicated “OR” formula.
Say, however, that a salesperson makes five sales calls, and you want to find the likelihood that he or she makes at least one sale. If you try to calculate this probability directly, you will have to confront five separate possibilities that constitute “success”: exactly one sale, exactly two sales, exactly three sales, exactly four sales, or exactly five sales. It would seem that you would have no choice but to calculate each of those probabilities separately and then add them together. This will be far too much work, especially under timed conditions.
However, consider the probability of failure—that is, the salesperson does not make at least one sale. Now you have only one possibility to consider: 0 sales. You can now calculate the probability in which you are interested, because for any event, the following relationship is true:
Probability of SUCCESS + Probability of FAILURE = 1
(the event happens) (it does not happen)
If, on a GRE problem, “success” contains multiple possibilities—especially if the wording contains phrases such as “at least” and “at most”—then consider finding the probability that success does not happen. If you can find this “failure” probability more easily (call it x), then the probability you really want to find will be 1 − x. For example:
What is the probability that, on three rolls of a single fair die, AT LEAST ONE of the rolls will be a 6?
You could list all the possible outcomes of three rolls of a die (1–1–1, 1–1–2, 1–1–3, etc.), and then determine how many of them have at least one 6, but this would be very time-consuming. Instead, it is easier to think of this problem in reverse before solving:
Failure: What is the probability that NONE of the rolls will yield a 6?
On each roll, there is a
probability that the die will NOT yield a 6. Thus, the probability that on all three rolls the die will not yield a 6 is:
.
Now, success was originally defined as rolling at least one 6. Because you have found the probability of failure, you can answer the original question by subtracting this probability from 1:
The probability that at least one 6 will be rolled is 
.
If a die is rolled twice, what is the probability that it will land on an even number at least once?