Sometimes we are interested in computing probabilities of
more complicated events. Here we introduce two basic probability rules. The
first rule is useful for finding the probability of one event *or *another
event. The second rule tells us how to compute the probability that an event *does
not *occur.

We will illustrate this rule with an example. Suppose

{00, 01, 02, 03, ..., 97, 98, 99}.

There are 100 possible winning numbers and since each has the same chance of being chosen, we assign a probability of 1/100 = .01 to each number.

Suppose we want to find the probability that the winning number has the same
two digits *or *the winning number is between 89 and 96 inclusive. If
these two events ("same two digits" and "between 89 and 96")
are nonoverlapping, then we can find the probability of "same two digits"
or "between 89 and 96" by adding:

Prob("same two digits" or "between 89 and 96") = Prob("same two digits")+Prob("between 89 and 96")

Are these two events nonoverlapping? Nonoverlapping means that it is impossible for the two events to occur at the same time. Here "same two digits" means the winning number is from the set {00, 11, 22, 33, 44, 55, 66, 77, 88, 99}. "Between 89 and 96" means that the number is in the set {89, 90, 91, 92, 93, 94, 95, 96}. Note that these two sets have nothing in common; in other words, it is impossible for the winning number to have the same two digits and be between 89 and 96. So we can add the probabilities to find the probability of the "or" event. The probability of "same two digits" is 10/100 and the probability of "between 89 and 96" is 8/100. Therefore the probability of interest is

Prob("same two digits" or "between 89 and 96") = 10/100 + 8/100 = 18/100 = .18

What if we wanted to find the probability of "same two digits" or "an even second digit"? Here we can't use this addition rule, since these two events are overlapping. It is possible for the winning to have the same two digits and have an even second digit -- the number 44 (and other numbers) is in both events. So this rule cannot be used in this case.

This rule is also applicable in the case where you want to find the probability of a collection of different outcomes. Suppose you toss a coin five times and you wish to find the probability that the number of heads is 2 or fewer. You can think of the event "2 or fewer heads" as an "or" event:

{2 or fewer heads} = {0 heads} or {1 head} or {2 heads}

By definition, the three outcomes {0 heads}, {1 heads} and {2 heads}, since you can only observe at most one of these outcomes when you toss the coin three times. So the addition rule can be used and

Prob(2 or fewer heads) = Prob(0 heads) + Prob(1 head) + Prob(2 heads)

Let's return to our lottery example. What if you're
interested in the probability that the winning number *does not *have
the same two digits. The rule for "not" events is called the *complement
*rule:

Probability("not" an event) = 1 - Probability(event)

In this case, we can write

Probability(not same digits) = 1 - Probability(same digits)

We have already found the probability that the winning number has the same two digits, so the probability of interest is

Probability(not same digits) = 1 - 10/100 = 90/100

The complement rule is especially useful in the case where it hard to compute the probability of an event, but it is relatively easy to compute the probability of "not" the event. For example, suppose we wish to compute the probability of tossing at least one head in 10 tosses of a coin. In this case, it would make sense to first perform the easier computation, the probability of "not at least one head" or "no heads". Then we apply the complement rule to find the probability of the event of interest.

Probability(at least one head) = 1 - Probability(no heads)

Return to AN INTRODUCTION TO PROBABILITY

albert@bayes.bgsu.edu

Document: http://www-math.bgsu.edu/~albert/m115/probability/add_probs.html

Last Modified: February 18, 2005