## Probability Rules

**Overview:**

The following are a list of rules you may use when dealing with different probability. These are used when trying to find the chance that an event or events may occur.

**Addition Rule**

For two

**disjoint**events A and B, the probability that one or the other occurs is the sum of the probabilities of two events. Disjoint events have no outcomes in common.

- P(A B) = P(A) + P(B)

**General Addition Rule**

There are times when you’re going to run into probabilities that are not disjoint and share a common outcome. If that does not happen you will be using this rule.

- P(A B) = P(A) + P(B) – P(AB)

By doing this rule you’ll be adding the probability of the two events and then subtracting out the common areas/ intersections.

**Multiplication Rule**

When solving the probability of two independent events A and B, the probability that both A and B will occur is the product of the probabilities of the two events.

- P (AB) = P(A) x P(B)

**General Multiplication Rule**

This rule is used for compound events that do not necessarily have to be independent.

- P (AB) = P(A) x P(B | A)

**Complement Rule**

The probability of an event occurring is 1 minus the probability it doesn’t occur. The set of outcomes that are not in the event A is called the complement of A and is denoted Ac.

- P(A) = 1 – P(Ac)

**Good Rules of Thumb:**

Though solving probability problems might be troublesome by looking out for certain words you’ll know what to do

*Not -*subtract from 1*At least-*subtract from 1*O**r*- add (must be disjoint)*And-*multiply (must be independent)

**Example 1:**

Lewis is to choose one ball from a sack. The sack contains six red balls, four green balls, two yellow balls, and three purple balls. What is the probability of Lewis getting a green or yellow ball?

*Solution:*

*4/15: number of green balls/total balls*

*2/15: number of yellow balls/total balls*

*4/15 + 2/15 = 6/ 15 = 2/ 5 = 40%*

**Example 2:**

A survey of college students found that 64% live in a campus residence hall, 72% have a car, and 45% live on campus and have a car. What’s the probability that a randomly selected student lives on campus or has a car?

*Solution:*

*64% ->*

*.64: percentage of student living in a campus residence hall*

*72% ->*

*.72: percentage of students who own a car*

*45% ->*

*.45: percentage of students who live on campus and have a car*

*.64 + .72 - .45 = .91*