**Basic Probability Rules**

**General Addition Rule**

The following addition rule for probability is always valid if you wish to find the probability that event A or B occurs.

**P(A or B) = P(A) + P(B) - P(A and B)**

**Mutually Exclusive or Disjoint Events**

If two events cannot occur at the same time, they are labeled as *mutually exclusive* or *disjoint* events. Furthermore, If two events are disjoint, then the probability of them both occurring at the same time is 0.

**If A and B are disjoint events: P(A and B) = 0**

**Addition Rule for Disjoint/Mutually Exclusive Events**

If two events are mutually exclusive, then the following formula applies.

**If A and B are disjoint events: P(A or B) = P(A) + P(B)**

**Complement Rule**

In probability, we may find ourselves wanting to find the probability that an event does not occur. This leads us to the Complement Rule. Specifically, this rule states that the probability that an event does not occur is equal to 1 minus the probability that the event does occur. Regarding notation, one usually denotes the complement of event A as either A' or A^{c}, depending on the book used. Therefore the probability of not observing event A is typically denoted as P(A') or P(A^{c} ).

**Complement Rule: P(A**

^{c}) = 1 - P(A)Another way to think of complementary events is that they are mutually exclusive, but when combined make up the entire sample space. One may also express the above formula as:

**P(A) + P(A**

^{c}) = 1

EXAMPLES

In rolling a fair die, what is the chance of rolling a {2 or lower} or an even number?

Solution:

A = Getting a {2 or lower} number = {1,2}

P(A) = 2/6

B= Getting an even number = {2,4,6}

P(B) = 3/6

A ∩ B = {2}

P(A ∩ B) = 1/6

Therefore, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 2/6 + 3/6 - 1/6 = 4/6 = 2/3

**2. In Chesterfield High School, the probability a randomly chosen student is a freshman is 0.29, and the probability a randomly chosen student is a senior is 0.21. Calculate the probability that a randomly chosen student at Chesterfield High school is either a sophomore or a senior.**

P(freshman or senior) = P(freshman) + P(senior) - P(freshman and senior)

However, because a student can not be both a freshman and senior, these two events are disjoint. Therefore, P(freshman and senior) = 0. Simplifying:

P(freshman or senior) = P(freshman) + P(senior) = 0.29 + 0.21 = 0.50

**3. John has an extra ticket to tonight's concert. He randomly texts two friends, Jane and Molly. Neither knows each other. He estimates the probability that Jane will say yes to buying the ticket and 0.4 and Molly will say yes with a probability of 0.1.**

**a. John is nervous that both will respond back and say they want to go. Using John's estimates, determine the probability that this will happen?**

P(J and M) = P(J)P(M) = (0.4)(0.1) = 0.04

**b. Calculate the probability that either Jane or Molly will go.**

P(J or M) = P(J) + P(M) - P(J and M) = 0.4 + 0.1 - 0.04 = 0.46

**c. Calculate the probability that neither Jane or Molly decide to go?**

P(J^{c} and M^{c}) = P(J^{c})P(M^{c} ) = (0.6)(0.9) = 0.54

**4. The following is a data collected by a survey of 200 random people.**

Favorite Pet / Favorite Sport |
Basketball |
Football |
Baseball |
Sum |

Dog |
47 |
54 |
10 |
111 |

Cat |
25 |
22 |
31 |
78 |

Rabbit |
2 |
3 |
6 |
11 |

Sum |
74 |
79 |
47 |
200 |

**a. A random person's survey results were viewed. What is the probability that this individual chose football as their favorite sport or a cat as their favorite pet.**

P(Football or Cat) = P(Football) + P(Cat) - P(Football and Cat)

P(Football or Cat) = (79 / 200) + (78 / 200) - (22 / 200) = .395 + .39 - .11 = .675

**b. What is the probability that another randomly chosen individual chose cat or rabbit as their favorite pet?**

Since there an individual can't choose both cat and rabbit as their favorite pet, they are considered disjoint, or mutually exclusive events.

P(Cat or Rabbit) = P(Cat) + P(Rabbit) - 0

P(Cat or Rabbit) = (78 / 200) + (11 / 200) = .39 + .055 = .445