Addition Rulefor

Probability

Vicki Borlaug

Walters State Community College

Morristown, Tennessee

Spring 2006

Are the statements TRUE or FALSE?

Rita is playing the violin and soccer.

Rita is playing the violin or soccer.

This is Rita.

“or”means one or the other

(or both) are true

“and”means both must be true

FALSE

TRUE

Elm

St.

Maple St.

Elm

St.

Maple St.

Which one is “Elm and Maple”?

Which one is “Elm or Maple”?

Elm and Maple Elm or Maple

This is called

UNION.

This is called

INTERSECTION.

Like when two streets cross.Like when you put the North and the South together.

Next we will look at Venn Diagrams.

In a Venn Diagram the box represents the entire sample space.

A B

Members that fit

Event A go in this

circle.

Members that fit

Event B go in this

circle.

A B A B

Event A and B Event A or B

Which is “A and B”?

Which is “A or B”?

This is called

INTERSECTION.

This is called

UNION.

A BA BA BA B

+ _=

The Addition Rule for Probability

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

A B

But we have added this

piece twice! That is one extra time!

We need to

subtract off the extra time!

Example #1)

Given the following probabilities:

P(A)=0.8 P(B)=0.3 P(A and B)=0.2

Find the P(A or B).

This can be solved two ways.

1. Using Venn Diagrams

2. Using the formula

We will solve it both ways.

Example #1 (continued)

P(A)=0.8 P(B)=0.3 P(A and B)=0.2

Find the P(A or B).

Solution using Venn Diagrams:

A B In this example we

will fill up the Venn Diagram

with probabilities.

Solution using Venn Diagrams:

A B

First fill in where the

events overlap.The

probability that a student fits the event A and B is 0.2.

That means the entire A circle must

add up to 0.8.

0.20.6 0.1

0.1

The probability

that a student fits the event

B is 0.3.

The box represents the entire

sample space and must add

up to 1.

0.2

0.1

0.10.6

The probability

that a student fits the event

A is 0.8. That means the entire B circle must

add up to 0.3.

Example #1 (continued)

P(A)=0.8 P(B)=0.3 P(A and B)=0.2

Find the P(A or B).

Then find the probability of A or B.

A B

0.20.6 0.10.2

0.1

0.10.6

P(A or B) = 0.6 + 0.2 + 0.1

I will start by shading A or B.

Then I will add up the

probabilities in the shaded

area.

= 0.9 Answer

Solution using the formula:

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

= 0.8 + 0.3 - 0.2

= 0.9

Example #1 (continued)

P(A)=0.8 P(B)=0.3 P(A and B)=0.2

Find the P(A or B).

Answer

Example #2.) There are 50 students. 18 are taking

English. 23 are taking Math. 10 are taking English and Math.

If one is selected at random, find the probability that the student is taking English or Math.

E = taking English

M = taking Math

Solution using Venn Diagrams:

E M In this

example we will fill up the

Venn Diagram with the number of students.

Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math.If one is selected at random, find the probability that the student is taking English or Math.

Solution using Venn Diagrams:

E M

First fill in where the

events overlap.The number of students

taking English and Math is 10.

That means the number of

students taking English must add up

to 18.

108 13

19

The number of students

taking Math is 23.

The box represents the entire

sample space and must add

up to 50.

10

19

138

The number of students

taking English is 18. That means

the number of students

taking Math must add up

to 23.

Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math.If one is selected at random, find the probability that the student is taking English or Math.

Then find the probability of English or Math.

E M

108 1310

19

138

P(E or M) =

I will start by shading E or

M. Then I will

find the probability in the shaded

area.

= 0.62

8 10 1350

Answer

Solution using the formula:

P(E or M) = P(E) + P(M) - P(E and M)

= 0.62

Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math.If one is selected at random, find the probability that the student is taking English or Math.

18 23 1050 50 50

Answer

Class Activity #1) There are 1580 people in an amusement park. 570 of these people ride the rollercoaster. 700 of these people ride the merry-go-round. 220 of these people ride the roller coaster and merry-go-round.

If one person is selected at random, find the probability that that person rides the roller coaster or the merry-go-round.

a.) Solve using Venn Diagrams.

b.) Solve using the formula for the Addition Rule for Probability.

Example #3) Population of apples and pears.

Each member of this population can be described in two ways.

1. Type of fruit2. Whether it has a worm or not

We will make a table to organize this data.

Example #3) Population of apples and pears.

5

59

4

8

62

3apple

pear

no worm worm

grand total 14? ?

?

??

?

?

?

Ex. #3 (continued)

5

59

4

8

62

3apple

pear

no worm worm

grand total 14

Experiment: One is selected at random.Find the probability that . . .

a.) . . . it is a pear and has a worm.

b.) . . . it is a pear or has a worm.

Ex. #3 (continued)

5

59

4

8

62

3apple

pear

no worm worm

grand total 14

Solution to #3a.)

P(pear and worm) = 214 0.1429

Answer

Ex. #3 (continued)

5

59

4

8

62

3apple

pear

no worm worm

grand total 14

Solution to #3b.)

P(pear or worm) = 4 2 314 0.6429

Answer

Ex. #3 (continued)

5

59

4

8

62

3apple

pear

no worm worm

grand total 14

Alternate Solution to #3b.)

P(pear or worm)=214

0.6429

P(pear) + P(worm) – P (pear and worm)614

514

Answer

Class Activity #2)

There are our modes of transportation – horse, bike, & canoe. Each has a person or does not have a person.

1.) Make a table to represent this data.2.) If one is selected at random find the following:

b.) P( horse and has a person)a.) P( horse or has a person)

c.) P( bike or does not have a person)

The end!