AP Statistics
6.3 General Probability Rules
General Probability Rules
Union: the event that AT LEAST ONE of the collection occurs (think OR)
Addition Rule for DISJOINT events: (If two events are disjoint, then the probability of them both
occurring at the same time is 0)
P(A or B or C) = P(A) + P(B) + P(C)
Example: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?
Possibilities: A two can be rolled, or a five can be rolled
These events are mutually exclusive since they cannot occur at the same time.
General Addition Rule for Unions of Two Events: (In events which aren't mutually exclusive, there is
some overlap. When P(A) and P(B) are added, the probability of the intersection (and) is added twice.
To compensate for that double addition, the intersection needs to be subtracted.)
P(A or B) = P(A) + P(B) – P(A and B)
Example: In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4 boys and 5 girls
made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl
or an A student?
Personal Probabilities
Probabilities or chance behavior based on ones personal opinion
Joint Events
Events that have a common outcome
Joint Probabilities: P(A and B)
Probabilities of joint events
Conditional Probability
 Written P(B|A) meaning “probability of B given A has occurred”
 Reduces the size of the Sample Space
 NOTE: it is easy to confuse probabilities and conditional probabilities of the same
event
General Multiplication Rule for any Two Events
P(A&B) = P(A) x P(B|A)
Conditional Probability Rule (dependent events)
𝑃(𝐴&𝐵)
P(B|A) =
𝑃(𝐴)
Example: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class
passed the first test. What percent of those who passed the first test also passed the second test?
Intersection
  two events occurring; all events are happening
(mainly two events A&B but not only in A or B alone)
More on Tree Diagrams
Make tree diagrams and assign probabilities to make problems easier
Bayes’s Rule


Used to calculate conditional probability
Shows how one conditional probability depends on its inverse
Example: In Orange County, 51% of the adults are males. (It doesn't take too much advanced mathematics
to deduce that the other 49% are females.) One adult is randomly selected for a survey involving credit card
usage.
a. Find the prior probability that the selected person is a male.
b. It is later learned that the selected survey subject was smoking a cigar. Also, 9.5% of males smoke
cigars, whereas 1.7% of females smoke cigars (based on data from the Substance Abuse and Mental Health
Services Administration). Use this additional information to find the probability that the selected subject is
a male.
Let's use the following notation:
M = male
M = female (or not male)
C = cigar smoker
C = not a cigar smoker.
Solution:
a. Before using the information given in part b, we know only that 51% of the adults in Orange County are
males, so the probability of randomly selecting an adult and getting a male is given by P(M) = 0.51.
b. Based on the additional given information, we have the following:
P(M) = 0.51 because 51% of the adults are males
P(M) = 0.49 because 49% of the adults are females (not males)
P(C|M) = 0.095 because 9.5% of the males smoke cigars (That is, the probability of getting
someone who smokes cigars, given that the person is a male, is 0.095.)
P(C|M) = 0.017 because 1.7% of the females smoke cigars (That is, the probability of getting
someone who smokes cigars, given that the person is a female, is 0.017.)
Let's now apply Bayes' theorem by using the
preceding formula with M in place of A, and C
in place of B. We get the following result:
Before we knew that the survey subject smoked a cigar, there is a 0.51 probability that the survey subject is
male (because 51% of the adults in Orange County are males). However, after learning that the subject
smoked a cigar, we revised the probability to 0.853. There is a 0.853 probability that the cigar−smoking
respondent is a male. This makes sense, because the likelihood of a male increases dramatically with the
additional information that the subject smokes cigars (because so many more males smoke cigars than
females).
Independent Events (Review)
P(B|A) = P(B)
P(A&B) = P(A) x P(B)
HW: pg. 440; 6.70, 6.71, 6.72 6.73, 6.82, 6.90