Probability addition rule

The addition rule for probability

1. 1
MARIO F. TRIOLA
Essentials ofSTATISTICS
Section 3-3 Addition Rule

2. 2
Example: Let event A represent a woman and B represent
blue eyes.
A or B is a compound event representing the set of
people who are women or who have blue eyes.
A and B is a compound event that represents the set of
people who are women AND have blue eyes (i.e. blue
eyed women)
A compound event is any event that
combines two or more simple events.

3. 3
Venn Diagrams
Women blue
eyes
Let the circle on the left represent the
characteristic “subject is a woman”
Let the circle on the right represent the
characteristic “subject has blue eyes”

4. 4
Venn Diagrams– “or”
Women blue
eyes
Together, the blue and yellow circles represent
the compound event “is a woman or has blue
eyes.”

5. 5
Venn Diagrams
Women blue
eyes
The intersection of the two circles
represents the event “subject is a woman
and has blue eyes.”

6. 6
A={accountant} B = {is a downhill skier}
A or B ?
A and B?
Interpret these Events:

7. 7
A={accountant} B = {is a downhill skier}
A or B ?
Is an accountant or is a downhill skier
A and B?
Is an accountant and a downhill skier
Interpret these Events:

8. 8
When finding the probability that event A
occurs or event B occurs, find
1. the total number of ways A can occur and
2. the number of ways B can occur,
3. but find the total in such a way that no
outcome is counted more than once.
General Addition Rule

9. 9
P(A or B) = P(A) + P(B) - P(A and B)
where P(A and B) denotes the probability that A and B
both occur at the same time.
Formal Addition Rule
If P(A) = .5 and P(B) = .2 and P(A and B) = .1,
then,
P(A or B) = 0.5 + 0.2 – 0.1 = 0.6

10. 10
Example: Addition Rule
In a study of 82 young drivers, 39 were men who at some
point in time were ticketed for traffic violations. 11 were
men who were never ticketed. 8 were women who were
ticketed, and 24 were women who were never ticketed.
If one of these subjects is randomly selected, what is the
probability of:
(a)Getting a man or someone who was ticketed?
(b)Getting a man or someone who was NOT ticketed?
(c)Getting a woman or someone who was ticketed?

11. 11
Example: Addition Rule
In a study of 82 young drivers, 39 were men who at some point in time
were ticketed for traffic violations. 11 were men who were never
ticketed. 8 were women who were ticketed, and 24 were women
who were never ticketed.
P(man) = 50 men / 82 surveyed = 0.6098
P(woman) = 32 women / 82 surveyed = 0.3902
P(ticketed) = 47 ticketed / 82 surveyed = 0.5732
P(never ticketed) = 35 never ticketed / 82 surveyed = 0.4268

12. 12
Example: Addition Rule
If one of these subjects is randomly selected, what is the probability of:
(a) Getting a man or someone who was ticketed?
P(man or ticketed) = P(man) + P(ticketed) – P(man and ticketed)
= 50/82 + 47/82 - 39/82 = 58/82
(b) Getting a man or someone who was NOT ticketed?
P(man or not ticketed) = P(man) + P(not ticketed) –
P(man and not ticketed)
= 50/82 + 35/82 – 11/82 = 74/82
(c) Getting a woman or someone who was ticketed?
P(woman or ticketed) = P(woman) + P(ticketed) –
P(woman and ticketed)
= 32/82 + 47/82 - 8/82 = 71/82

13. 13
Mutually Exclusive
Events A and B are mutually exclusive if they
cannot occur simultaneously.

14. 14
Mutually Exclusive
Events A and B are mutually exclusive if they
cannot occur simultaneously.
P(A) P(B)
P(A and B)
NOT mutually exclusive!

15. 15
Definition
Events A and B are mutually exclusive if they
cannot occur simultaneously.
P(A) P(B)
Events A and B
are
Mutually exclusive
Nothing in common

16. 16
Mutually Exclusive Events
If two events A and B are mutually exclusive, they cannot
occur at the same time.
The implication is that it is impossible for these two events
A and B to occur together, so
P(A and B) = 0

17. 17
Mutually Exclusive Events
If two events are mutually exclusive, this simplifies the
addition rule.
P(A or B) = P(A) + P(B) - P(A and B)
= P(A) + P(B) - 0
= P(A) + P(B)

18. 18
Applying the Addition Rule
P(A or B)
Addition Rule
Are
A and B
mutually
exclusive
?
P(A or B) = P(A)+ P(B) - P(A and B)
P(A or B) = P(A) + P(B)
Yes
No

19. 19
Find the probability of randomly selecting a man or a boy.
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table

20. 20
Find the probability of randomly selecting a man or a boy.
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table

21. 21
Find the probability of randomly selecting a man or a boy.
P(man or boy) = 1692 + 64 = 1756 = 0.790
2223 2223 2223
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table

22. 22
Find the probability of randomly selecting a man or a boy.
P(man or boy) = 1692 + 64 = 1756 = 0.790
2223 2223 2223
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table
* Mutually Exclusive *

23. 23
Find the probability of randomly selecting a man or
someone who survived.
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table

24. 24
Find the probability of randomly selecting a man or
someone who survived.
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table

25. 25
Find the probability of randomly selecting a man or
someone who survived.
P(man or survivor) = 1692 + 706 - 332 = 1756
2223 2223 2223 2223
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table
= 0.929

26. 26
Find the probability of randomly selecting a man or
someone who survived.
P(man or survivor) = 1692 + 706 - 332 = 1756
2223 2223 2223 2223
Men Women Boys Girls Totals
Survived 332 318 29 27 706
Died 1360 104 35 18 1517
Total 1692 422 64 56 2223
Contingency Table
* NOT Mutually Exclusive *
= 0.929

27. 27
Complementary Events

28. 28
Complementary Events
P(A) and P(A)
are
mutually exclusive

29. 29
Complementary Events
P(A) and P(A)
are
mutually exclusive
All simple events are either in A or A.

30. 30
Complementary Events
If two events are mutually
exclusive, then they cannot
occur at the same time.
The probabilities of these two
events will make 1.
P(A) + P(A) = 1

31. 31
Rules of Complementary Events
P(A) + P(A) = 1

32. 32
P(A)
Rules of Complementary Events
P(A) + P(A) = 1
= 1 - P(A)

33. 33
P(A) + P(A) = 1
= 1 - P(A)
P(A) = 1 - P(A)
P(A)
Rules of Complementary Events

34. 34
Example:
When a baby is born in the United States,
the true probability that the child is a boy
is 0.512. What is the true probability that
the child is a girl?
The events are complementary. One
cannot be a boy and a girl at the same
time.
P(girl) = 1 – P(boy) = 1 – 0.512 = 0.488

35. 35
Example:
A poll shows that 61% of Americans say
they believe that life exists elsewhere in
the galaxy. What is the probability that
someone does not have that belief?
The events are complementary. One
cannot believe in life and not believe in
life at the same time.
P(not) = 1 – P(believes) = 1 – 0.61 = 0.39

36. 36
Example:
Your alarm clock has a 97.5% probability
of working on any given morning. What
is the probability that it will NOT work?
The events are complementary. The alarm
cannot “work” and “not work” at the same
time.
P(not) = 1 – P(works) = 1 – 0.975 = 0.025

37. 37
Example:
Suppose your use the principle of redundancy
to wake you up in the morning. You set
TWO alarms, both with 97.5% probability
of working on any given morning. What
is the probability that your system works?
The events “works” and “fails” are two
complementary events.
P(works) = 1 – P(both fail)
= 1 – (0.025)(0.025) = 0.999375
The redundant alarm system is almost
certainly going to work every morning.