Download - Addition Rule for Probability
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!