Statistics Unit
Lesson: General Probability
Learning Target: To calculate general and conditional probabilities.
Success Criteria: By the end of the lesson you should be able to:
a. Calculate probabilities using the General Addition Rule
b. Calculate conditional probabilities
1. Addition Rule: If A and B are disjoint events, then the probability of A or B is:
P(A or B) = P(A) + P(B)
2. General Addition Rule: For any two events when an outcome is shared, the
probability of A or B is:
P(A or B) = P(A) + P(B) - P(A and B)
(P(A and B) = Probability of A and B at the same time, shared)
3. In a survey of high school students: 38% take biology, 22% take statistics, and 12%
take both. Find the probability that a randomly selected student:
a. Takes biology, but not statistics?
P(biology, but not statistics) = 0.38 – 0.12 = 0.26
b. Takes statistics, but not biology?
P(statistics, but not biology) = 0.22 – 0.12 = 0.10
c. Takes either biology or statistics, but not both?
P(biology or statistics) = 0.38 + 0.22 – 0.12 = 0.48
d. Takes neither biology nor statistics?
P(neither biology nor statistics) = 1 – 0.48 = 0.52
4. Example: The probability that a student owns a car is 0.65, and the probability that a
student owns a computer is 0.82. If the probability that a student owns both is 0.55,
what is theprobability that a randomly selected student owns a car or computer?
a. What is the probability that a randomly selected student does not own a car or
computer?
b. What is the probability that a randomly selected student owns a computer, but
not a car?
c. What is the probability that a randomly selected student owns a car or a
computer, but not both?
d. What is the probability that a randomly selected student owns a car, but not a
computer?
5. Example: At Athens Country Club, 73% of the members swim, and 82% play bridge,
and 65% do both. If a member is selected at random, find the probability that the
member:
a. Swims, but doesn’t play bridge?
b. Plays bridge or swims, but not both?
c. Plays neither bridge nor swims?
6. Conditional Probability: When we want the probability of an event from a conditional
distribution, we write P(B│A) pronounced “ the probability of B given that A happens”
or “ the probability of A given that B happens”. It is calculated as such:
P(B│A) = P(A and B) or P(A│B) = P(A and B)
P(A)
P(B)
The probability that A and B happens, given that A happens, or
The probability that A and B happens, given that B happens
7. Example: Look at the previous problem about cars and computers
a. What is the probability that a randomly selected student has a car given that they
have a computer?
P(car│computer) = P(both)/P(computer) = 0.55/0.82 = 0.671
b. What is the probability that a selected student had a computer given that they
have a car?
P(computer│car) = P(both)/P(car) = 0.55/0.65 = 0.846
8. Example: Look at the previous problem about club members swimming are playing
bridge:
a. What is the probability that a randomly selected club member is playing bridge
given that they are swimming?
P(bridge│swimming) = 0.65/0.73 = 0.890
b. What is the probability that a randomly selected club member is swimming given
that they are playing bridge?
P(swimming│bridge) = 0.65/0.82 = 0.793
9. Probabilities and Contingency Tables:
A Gallup poll asked 1008 Americans age 18 and over whether they planned to watch
the upcoming Super Bowl. The pollster also asked those who planned to watch
whither they were looking forward more to seeing the football game or the
commercials. The results are summarized in the table:
a. What is the probability that a randomly selected person from the poll is a male
who won’t watch the game?
P(male who won’t the game) = 132/1008 = 0.131
b. What is the probability that the person is a male given that he won’t watch the
game?
P(male│won’t watch the game) = 132/292 = 0.452
c. What is the probability that the person who won’t watch the game given that he is
a male?
P(won’t watch the game│male) = 132/492 = 0.268