Addition Rule of Probability
Example 2 (with non mutually exclusive events)
A study of patient records at a public health clinic showed that 22% of the patients had had a dental examination, 45% had had a general physical examination, and 7% had had both. If a patient's record is randomly selected, what is the probability that the patient received either a dental examination or a physical examination?
Analysis of the Problem
By the wording of the question asking for the probability of either a dental examination or a physical examination, you should immediately think of the Addition Rule. The next step is to analyse the problem into what is the statistical experiment, the sample space, the sample size, the two events, and their probabilities so that all the pieces of information can be placed into the Addition Rule to solve the problem.
The sample space S of this experiment is based upon the set consisting of the group of patients at the public health clinic. This statistical experiment consists of pulling patient records to determine what kinds of examinations the patient has already had. This is another case of categorizing a targeted population, but, as the question clearly indicates, the categories are not mutually exclusive because some patients can be placed into two distinct categories — namely having had both a dental and physical exam.
The kind of examination a patient has had is an "outcome" of the eperiment for that particular patient. In this case, a patient may be placed in more than one category. In statistical language, one patient may produce more than one outcome. The sample space S is, therefore, a list (set) of the examinations the patients have had, one or more for each of the patients in that clinic.
In contrast to the first example, this question does not give the sample size — that is, the total number of examinations undergone by the collective patient population — nor does it give how many patients fall into the various categories. What it does give are the probabilites themselves so that you do not have to calculate them.
Solution to 2
Now we proceed with the solution.
Let A be the event that the patient's record indicates he/she has had a dental examination. The event A consists of the outcome: examination is "dental".
Let B be the event that the patient's record indicates he/she has had a physical examination. The event B consists of the outcome: examination is "physical".
P(A) = 0.22
P(B) = 0.45
P(A and B) = 0.07
We substitute into the general form of the Addition Rule to get:
The probability that a randomly selected patient has received either a dental examination or a physical examination is 0.60.
Example 3
Last Updated by SAS October 07, 2026, 16:00:54
