High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Statistics Textbook > Probability

Basic Rules of Probability

Tools for ProbabilityPrevious Section
Probability of Independent EventsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Interpretation of Probability

The probability for event A to happen is denoted as P(A) , where

  • For any given event, its probability of occurrence must be between 0 and 1, inclusive.
  • An event's probability cannot be negative and cannot be greater than 1.
  • P(A) = 0 means there is 0% probability that Event A will occur (no uncertainty).
  • P(A) = 1 means there is 100% probability that Event A will occur (no uncertainty).
  • The closer the probability is to 0, the less likely the event will occur.
  • The closer the probability is to 1, the more likely the event will occur.
0:00 / 0:00

Complement Rule

Two events are complementary if they are non-overlapping, and together cover all of the possible outcomes.

The sum of two complementary events is equal to 1.

Example: Two Possible Events (Win or Lose)

Event AA = the hockey team wins the Stanley Cup
Event AcA^c = the hockey team does not win the Stanley Cup

If the probability of the hockey team winning the Stanley Cup (Event AA) is P(A)=0.36P\left(A\right)=0.36 (or 36%), then what is the probability of the hockey team not winning the Stanley Cup P(Ac)P\left(A^c\right)?

P(Ac)=1−P(A)P\left(A^c\right)=1-P\left(A\right)
P(Ac)=1−0.36=0.64P\left(A^c\right)=1-0.36=0.64

Therefore, the probability that the hockey team doesn't win the Stanley Cup is 64%.

PAGE BREAK

Total Probability

The total probability of all possible, non-overlapping outcomes must be 1.

Example
There are 6 possible outcomes of rolling a six-sided die. Each outcome has a probability of 16\frac{1}{6}. The total probability of all of these possible outcomes is:
16+16+16+16+16+16=1\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=1

Bill has a meeting with the Human Resources manager. He will either be promoted, demoted, or fired. The probability of him being promoted is 26% and the probability of him being fired is 18%. What is the probability that he will be demoted?
0:00 / 0:00

The Addition Rule

If events A and B are defined on a sample space, then
P(A∪B)=P(A)+P(B)−P(A∩B)\boxed{P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)}


Wize Tip
If events A and B are mutually exclusive (don't overlap), then
P(A∩B)=0P(A\cap B)=0.

So, our probability formula becomes:
P(A∪B)=P(A)+P(B)\boxed{P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)}

PAGE BREAK
Example
The probability that Bill will buy a drink at Starbucks is 0.65, and the probability that he will buy a snack is 0.20. We also know that the probability Bill will buy a drink or a snack or both is 0.75.

a) What is the probability that he buys both a drink and a snack?

Here's what we know:
  • Probability that he buys a drink: P(D)=0.65P(D)=0.65
  • Probability that he buys a snack: P(S)=0.20P(S)=0.20
  • Probability that he buys a drink, a snack, or both: P(D∪S)=0.75P(D\cup S)=0.75
Using our formula:
P(D∪S)=P(D)+P(S)−P(D∩S)0.75=0.65+0.20−P(D∩S)0.75=0.85−P(D∩S)−0.10=−P(D∩S)0.10=P(D∩S)\begin{array}c P(D\cup S)&=&P(D)&+&P(S)&-&P(D\cap S)\\ 0.75&=&0.65&+&0.20&-&P(D\cap S)\\ 0.75&=&&0.85&&-&P(D\cap S)\\ -0.10&=&&&&-&P(D\cap S)\\ 0.10&=&&&&&P(D\cap S) \end{array}

Therefore, the probability that he buys both a drink and a snack is 0.10.

b) Are the events that Bill buys a drink and Bill buys a snack mutually exclusive?

Since P(D∩S)≠0P(D\cap S)\neq 0, the events that Bill buys a drink and Bill buys a snack are NOT mutually exclusive.