Now, we discuss a trick that helps in some problems. Sometimes, instead of finding the probability of an event, it can be easier to find the probability that the event *does not occur*. Since we know that either an event will occur or not, we know that the probability of the event occurring plus the probability of the event not occurring must equal to 1. More formally, we write:

**Probabilities for Complements**

Let A be some event. Then, P(A) + P(A does not occur) = 1.

Here's an example where we apply this rule:

**Example 1**

You are about to roll a fair six-sided dice. What is the probability that you roll a prime number?

Answer
**Solution**

We know how to do this kind of problem in the traditional way: find the number of outcomes where we get a prime number and divide by the total number of outcomes. But we can also solve this problem by finding the number of outcomes where we *do not* get a prime number (1, 4, 6) and then calculating:

P(I get a prime number) + P(I do not get a prime number) = 1

Substituting, we get

P(I get a prime number) + 3/6 = 1

Subtracting, we get

P(I get a prime number) = 3/6 = 1/2

which is the same as our previous answer.

Why do we bother with this rule? Sometimes it is actually more convenient to calculate things this way. For example,

**Example 2**

You are going to roll a fair six-sided dice 10 times in a row. What is the probability that you get a six at least once?

Answer
**Solution**

Now, we are trying to find P(I get a 6 at least once in 10 rows). By our above rule, we know that

P(I get a 6 at least once in 10 rows) + P(I do not get a 6 at least once in 10 rows) = 1

Directly calculating P(I get a 6 at least once in 10 rows) is somewhat tricky. But P(I do not get a 6 at least once in 10 rows) is not nearly as hard to calculate. If I do not get a 6 at least once in 10 rows, then I must have gotten some number 1 through 5 on each roll. Now, the probability that I got some number 1 through 5 on the first roll is 5/6. And similarly, the probability that I got some number 1 through 5 on the second roll is 5/6. And so on for the third, fourth, …, tenth rolls.

Since each roll of the dice is independent of all the other rolls, we get:

P(I do not get a 6 at least once in 10 rows) =

Thus, we conclude that:

P(I get a 6 at least once in 10 rows) + P(I do not get a 6 at least once in 10 rows) = 1

P(I get a 6 at least once in 10 rows) +

P(I get a 6 at least once in 10 rows) =

which gives us the answer.

Now, it will often be easier to just directly calculate the probability of a given event. But if that calculation looks absurdly difficult or tedious, take a minute to step back and consider: can I calculate the probability that the event *doesn’t occur*? Would that be easier? Sometimes that can save you a lot of hassle.

**Practice Problems**

Question 1

You are rolling a fair spinner with seven sections, numbered 1 through 7. What is the probability that, in 10 spins, you get at least one prime number?

Answer
Solution 1

Let E = the event that you get at least one prime number in 10 spins. By our rule above, 1 = P(E) + P(not E). What is P(not E)? That's just the probability that E does not occur. What does it mean for E to not occur? That means that, in 10 spins, you never get a prime number. What is the probability of that? Well, on each spin, there is a 3/7 chance that you do not get a prime number (since 1, 4, 6 are not prime). Since each spin is independent of each other spin, we get: P(not E) = Thus, P(E) = 1 -

Question 2

One random employee will be selected to win a company-sponsored vacation. There are 500 employees, 300 of which are female. Also, 50 employees are managers, and half of the managers are female. What is the probability that the winner is not a female manager?

Answer
Solution 2

By our rule, we know P(Winner is not a female manager) + P(Winner is a female manager) = 1. What is the probability that the winner is a female manager? We know there are 25 female managers. So, P(Winner is a female manager) Thus, P(Winner is not a female manager) =

Question 3

The weather reporters say that there is a 60% chance of rain for each of the next seven days. If that is true, what is the probability that it rains at some point over the next seven days?

Answer
Solution 3

Let E = the event that it rains at some point over the next seven days. By our rule, P(E) + P(not E) = 1. It is easier to find P(not E), since that is just the probability that it does not rain at any point over the next seven days. Each day, there is a 40% chance that it does not rain. So over the seven days, P(not E) = Thus, P(E) =

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