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Chapter 3: Probability
3.1 Probabilities of Events (Sections 2.1 and 2.3)
Basic terms in Probability:
Probability Experiment - Random process which generates distinct outcomes
Outcome (Element, Sample Point) - A result of a probability experiment
Sample Space, S - The set of all possible outcomes of a probability
experiment
Event - A subset of outcomes from the sample space
Notation for events: Use capital letters near the beginning of the alphabet.
Examples:
1. Experiment - Toss a regular die
Outcome - The number observed on the top of the die
Event - Roll a number greater than or equal to 3
Sample Space?
Event in set notation?
2. Experiment - Randomly select a Miami University student and ask
"Should there be stricter controls on firearms?"
Outcome - One of Yes, No, Undecided
Event - Student has an opinion
Sample Space?
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Event in set notation?
3. Experiment - Toss a coin three times
Outcome - result of first toss, result of second toss, result of third toss
(For example: HTH)
Event - Exactly one head is tossed
Sample Space?
Event in set notation?
Probability of an event, P(A) - relative frequency with which the event A will
occur
Special Types of Events (Each event is illustrated using the probability
experiment: Toss a regular die):
Null Event,  - Event with no outcomes that satisfy it
Example - E: Toss a number greater than 8
Simple Event - An event which contains exactly one outcome
Example - E 1 = {1}, E 2 = {2}, ..., E 6 = {6}
Union Event, A  B - The event which contains all outcomes which satisfy
event A or event B or both. (Note: A  B is read as ‘A or B’)
Example - A: Toss an even number
B: Toss a number greater than or equal to 4
A  B in set notation?
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Intersection event, A  B - The event which contains all outcomes which
satisfy both event A and event B. (Note: A  B is read as ‘A and B’)
Example - A: Toss an even number
B: Toss a number greater than or equal to 4
A  B in set notation?
Complement of an Event, A c - Event which contains all outcomes in the
sample space which do not satisfy event A. (Note: A c is read as ‘not A’)
Example - A: Toss an even number
B: Toss a number greater than or equal to 4
A c , B c , A c  B c , and (A  B) c in set notation?
Mutually Exclusive Events (Disjoint Events) - Two events which have no
outcomes in common.
Examples:
1. A: Toss an even number
B: Toss a number less than or equal to 1
2. A and A c .
3. Any two simple events.
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Independent Events - Two events which are such that knowing whether one
event occurs does not change the probability that the other event occurs.
Example - A: Toss an even number
B: Toss a number greater than or equal to 5
Note: Ways of checking whether two events are independent will be
given later.
Properties of probabilities:
1. P(S) = 1
2. 0  P(A)  1
3. If A 1 , A 2 , ... are mutually exclusive events, then P(A 1  A 2  ...) =
P(A 1 ) + P(A 2 ) + ...
Note: As a consequence of properties 1 and 3, we deduce that the sum of
the probabilities of all simple events is 1 (Since S = E 1  E 2  ... and
consequently 1 = P(S) = P(E 1  E 2  ...) = P(E 1 ) + P(E 2 ) + ... =  P(E i ).)
Examples
1. Which of the following assignments of probabilities are legitimate for
the tossing of a regular die? (Let E 1 = {1}, E 2 = {2}, ..., E 6 = {6}.)
a. P(E 1 ) = .3, P(E 2 ) = .1, P(E 3 ) = .2, P(E 4 ) = .1, P(E 5 ) = .2,
P(E 6 ) = .1.
b. P(E 1 ) = P(E 2 ) = P(E 3 ) = P(E 4 ) = P(E 5 ) = P(E 6 ) = .2.
c. P(E 1 ) = .3, P(E 2 ) = P(E 3 ) = P(E 4 ) = P(E 5 ) = .2, P(E 6 ) = -.1.
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2. The candy in a bag of M&M’s comes in six colors: brown, red,
yellow, green, orange, and blue. The manufacturer says that 30% are
brown, 20% are red, 20% are yellow, 10% are green, 10% are orange.
a. What is the probability that a randomly selected M&M is blue?
b. What is the probability that a randomly selected M&M is red,
yellow, or green?
Experiment - Randomly select an M&M
Outcomes – color of the M&M
S = {br, r, y, g, o, bl}
Simple events: Let E 1 = {br}, E 2 = {r}, E 3 = {y}, E 4 = {g},
E 5 = {o}, E 6 = {bl}
Assignment of probabilities: P(E 1 ) = .3, P(E 2 ) = .2, P(E 3 ) = .2,
P(E 4 ) = .1, P(E 5 ) = .1
6
a. Since 1 =

P(E i ),
i 1
5
P(E 6 ) = 1 -

i 1
P(E i ) = 1 – (.3 + .2 + .2 + .1 + .1) = .1
b. A: M&M is red, yellow, or green
A = E2 E3 E4
P(A) = P(E 2 ) + P(E 3 ) + P(E 4 )
= .2 + .2 + .1 = .5
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(since mutually exclusive)
3. Consider the following:
Experiment - Toss a fair coin three times
Outcomes - result of first, result of second, result of third toss
So, S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
a. Assign appropriate probabilities to each of the simple events.
Let E 1 = {HHH}, E 2 = {HHT}, ..., E 8 = {TTT}
Now, it is reasonable to assume that each of these simple events
are equally likely.
So, P(E 1 ) = P(E 2 ) = ... = P(E 8 ) = a. What is a?
Since 1 =  P(E i ) =  a = 8a,
a=
1
.
8
b. What is the probability that exactly one head is tossed?
A: exactly one head
A = {HTT, THT, TTH} = E 5  E 6  E 7
P(A) = P(E 5 ) + P(E 6 ) + P(E 7 )
=
1
1
1
3
+ + =
8
8
8
8
Other properties of probabilities:
4. If all outcomes in S are equally likely, then
P(A) =
k
# of outcomes in A
= .
# of outcomes in S
N
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(since mutually exclusive)
Example: See part b of the previous example.
A = {HTT, THT, TTH}.
P(A) =
# of outcomes in A
3
= (Since the outcomes are equally likely)
8
# of outcomes in S
5. Addition Rule for Mutually Exclusive Events: If events A and B are
mutually exclusive, then
P(A  B) = P(A) + P(B)
6. Complement Rule: P( A c ) = 1 - P(A)
(Likewise, P(A) = 1 - P( A c ).)
Why? Since 1 = P(S) = P(A  A c ) = P(A) + P( A c )
7. P(  ) = 0
Why? Since  = Sc , P(  ) = P( Sc ) = 1 - P(S) = 1 - 1 = 0
8. Multiplication Rule for Independent Events: Events A and B are
independent if and only if P(A  B) = P(A) P(B)
Note: This formula can be used to check if two events are independent.
Examples:
1. Experiment - Toss a balanced die
Events - A: Toss an even number
B: Toss a number greater than or equal to 5
Find the probabilities for A, B, and A  B.
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A = {2, 4, 6}
P(A) =
P(B) =
B = {5, 6}
A  B = {6}
# of outcomes in A
3
1
= =
6
2
# of outcomes in S
# of outcomes in B
2
1
= =
6
3
# of outcomes in S
P(A  B) =
Since the
outcomes in S
are equally
likely.
# of outcomes in A  B
1
=
6
# of outcomes in S
Are events A and B independent?
1
1 1
P(A) P(B) =     = = P(A  B)
 2 3
6
Yes, A and B are independent.
2. Experiment: Toss a balanced die twice.
Find the probability that the first toss is even and the second toss is
less than or equal to 2.
Events - A: First toss is even
1
2
1
P(B) =
3
P(A) =
B: second toss is less than or equal to 2
Since the two tosses of a die are considered to be independent, these
two events are independent.
So, P(A  B) = P(A) P(B)
(since A and B are independent)
1
1 1
=     =
 2 3
6
Check this?
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Conditional Probability of B given A, P(B|A) P(B|A) =
P(A  B)
P( A)
Justification: Suppose the outcomes in S are equally likely, then
P(A  B) =
# of outcomes in A  B
# of outcomes in S
and P(A) =
# of outcomes in A
# of outcomes in S
Therefore, P(B|A) =
# of outcomes in A  B
P(A  B)
=
# of outcomes in A
P( A)
Consequently, if A is known to have occurred, the sample space is
reduced from S to A.
Example:
The table below shows the number of jobs lost (in thousands) in the
United States over a three year period. (There were 5,584,000 jobs lost.)
Reason for job loss
Workplace
moved/closed
Slack Work
Male
Female
1,703
1,210
1,196
564
548
363
3,447
2,137
Total
2,913
1,760
911
5,584
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Position
abolished Total
Suppose that a person is randomly selected from the group of all persons
who lost their jobs over the three year period.
a. What is the probability that the person is male?
Experiment: Randomly select a person
Outcome: Person selected (Outcomes are equally likely.)
A: person is male
P(A) =
# of outcomes in A
3447
=
= .617
5584
# of outcomes in S
b. Given that the person selected lost the job due to the fact that the
position was abolished, what is the probability that the person is male?
B: position was abolished
P(A|B) =
# of outcomes in A  B
P(A  B)
548
=
=
= .602
911
# of outcomes in B
P(B)
c. Given that the person selected is male, what is the probability that he
lost the job because the position was abolished?
P(B|A) =
# of outcomes in A  B
P(A  B)
548
=
=
= .159
3447
# of outcomes in A
P( A)
9. Multiplication Rule: P(A  B) = P(A) P(B|A) = P(B) P(A|B)
Note: From properties 8 and 9 we get that A and B are independent if
and only if P(B|A) = P(B). (In the above example, are A and B
independent?)
Example: A plant manufacturing transistors has two shifts: I and II. It is
known that shift I produces 60% of the transistors and 3% of its
production will not meet customer specifications. Five percent of
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shift II’s production will not meet customer specifications. If a
transistor is selected at random, what is the probability that it will not
meet customer specifications?
A: Transistor is from shift I
B: Transistor does not meet specifications
P(A) = .6
P(B|A) = .03
P(B| A c ) = .05
P(B) = P((A  B)  ( A c  B))
= P(A  B) + P( A c  B)
(since mutually exclusive)
= P(A) P(B|A) + P( A c ) P(B| A c )
= P(A) P(B|A) + (1 - P(A)) P(B| A c )
= .6 (.03) + (1 - .6) (.05) = .018 + .02 = .038
An alternate (simpler) solution.
A: Shift I
B: Does not meet Specs
Create a tree diagram as follows:
Event
Probability
P(B|A)
B
AB
P(A) P(B|A)
P( Bc |A)
Bc
A  Bc
P(A) P( B c |A)
P(B| A c )
B
Ac  B
P( A c ) P(B| A c )
P( Bc | A c )
Bc
A c  Bc
P( A c ) P( B c | A c )
A
P(A)
P( A c )
Ac
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For our problem,
Event
Probability
.03
B
I, no specs .018
.97
Bc
I, specs
.05
B
II, no specs .02
.95
Bc
II, specs
A
.6
.582
.4
Ac
.38
P(B) = .018 + .02 = .038
(What happens in the tree diagram when the events are independent?)
10. Addition Rule: P(A  B) = P(A) + P(B) - P(A  B)
proof: A  B = (A  B)  (A  B c )  ( A c  B)
P(A  B ) = P((A  B)  (A  B c )  ( A c  B))
= P(A  B) + P(A  B c ) + P( A c  B)
(since m.e.)
= P(A  B) + P(A  B c ) + P( A c  B) + P(A  B) - P(A  B)
= (P(A  B) + P(A  B c )) + (P( A c  B) + P(A  B)) - P(A  B)
= P(A) + P(B) - P(A  B)
Example: A producer of automobile batteries knows that 3% of the batteries
have defective terminals, 4% have defective plates, and 1% have both
defective terminals and defective plates. What is the probability that a
randomly selected battery has neither of the defects?
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A: defective terminal
B: defective plate
P(A) = .03
P(B) = .04
P(A  B) = .01
P( A c  Bc ) = P( (A  B) c ) = 1 – P( A  B ) = 1 – (P(A) + P(B) - P(A  B))
= 1 – (.03 + .04 - .01) = .94
Example illustrating many of the previous properties:
A house needs to be re-roofed in the spring. To do this a dry, windless
day is needed. The probability of getting a dry day is 0.7, a windy day is
0.4, and a wet and windy day is 0.2. What is the probability of getting
a. a wet day?
b. a day that is wet or windy?
c. a day when the house can be re-roofed?
d. a calm day given that the day is dry?
Probabilities given by the problem:
A: Dry
B: Windy
P(A) = .7
P(B) = .4
A c  B: not dry and windy
P( A c  B) = .2
a. P( A c ) = 1 - P(A) = 1 - .7 = .3
b. A c  B: not dry or windy
P( A c  B) = P( A c ) + P(B) - P( A c  B) = .3 + .4 - .2 = .5
c. A  B c : dry and not windy
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Note that this event is the complement of the one in part b. That is,
A  B c = (Ac  B) c and
P(A  B c ) = P( (Ac  B) c ) = 1 - P( A c  B) = 1 - .5 = .5.
d. P( Bc | A ) =
P(Bc  A)
.5
= = .714
.7
P(A)
An alternate (simpler) solution to parts b and c.
A: Dry
P(A) = .7
B: Windy
P(B) = .4
P( A c  B) = .2
Create a two dimensional table as follows:
B
Bc
Total
A
P(A  B)
P(A  B c )
P(A)
Ac
P( A c  B)
P( A c  B c )
P( A c )
Total
P(B)
P( B c )
1
Not Windy
Total
For this problem, we get:
Windy
Dry
Not Dry
Total
1
b. P( A c  B) = .2 + .2 + .1 = .5
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c. P(A  B c ) = .5