1. The Addition Rule

2. In order to use the Addition Rule, you must first remember what “Mutually Exclusive” is…
Now, if the problem involves items which are mutually exclusive, you will use a formula which is different than a formula you will use if the items are not mutually exclusive…

3. If they are MUTUALLY EXCLUSIVE… 1
If they are MUTUALLY EXCLUSIVE… 1. You will add the probabilities of the events together. This will give you the probability that one or the other has occurred.

4. Two Problems
Fruit
Probability
Apple
.21
Orange
.14
Strawberry
Pear
.13
Banana
.33
Cherry
.05
If he wants either a pear or an apple, what is the probability that he will get it?
What if he wants anything except a banana?

5. The Answers:
The probability he wants a pear or an apple: = % probability he wants a pear or an apple.
The probability he wants anything but a banana:
1 – 0.33 = 0.67
67% probability he wants anything but a banana.

6. Alternate Ways to Solve the Banana Problem
The way demonstrated was using “complements”:
1 - the probability of a banana =
probability of everything else
You could also add all the probabilities, except for bananas together.

7. What if it’s not “mutually exclusive”?
Example: Use a standard deck of 52 cards:
4 suits – hearts, diamonds, clubs, spades (hearts & diamonds are red, clubs and spades are black)
13 cards per suit – 2 through 10, J, Q, K, A
Face Cards – only the Jacks, Queens, and Kings (Aces don’t have faces ☺)

8. The Formula
If it is mutually exclusive: P(A) + P(B) Now, since it is not mutually exclusive: P(A) + P(B) – P(A and B) Where P(A and B) is where they overlap; that is where the same item has been counted twice.

9. An Example What is the probability you draw a red card or an ace?
Since there are 26 red cards and 4 aces, but 2 of the aces are red…
26/52 + 4/52 – 2/52
= 28/52 = 7/13

10. Another Example
Find the probability of drawing a face card or a spade:
12/ /52 – 3/52
= 22/52 = 11/26
12 face cards
13 spades
3 face cards which are spades

11. How About You… Write down and solve 1 each mutually exclusive and not
addition problems related
to each activity (total of 8).