Probability

Outline

Addition and Multiplication Rules for ProbabilityLecture 10, STAT 2246

Julien Dompierre

Departement de mathematiques et d’informatique

Universite Laurentienne

30 janvier 2007, Sudbury

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OutlineAddition RulesMultiplication Rules

Outline

1 Addition and Multiplication Rules for ProbabilityAddition RulesMultiplication Rules

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Outline


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Mutually Exclusive Events (p. 195)

Two events of the same experiment are mutually exclusive

events if they cannot occur at the same time (i.e., they have nooutcomes in common).

U

AB

In this case, the intersection of the sets A and B is empty.

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Mutually Exclusive Events

If A and B are mutually exclusive events of the same experiment,then the probability that A and B will occur is

P(A ∩ B) =n(A ∩ B)

n(S)=

0

n(S)= 0.

For example. The experiment is to roll a die. The sample space isthe set of all possible outcomes is S = {1, 2, 3, 4, 5, 6}.The event A is to get an odd number, A = {1, 3, 5} ⊆ S .The event B is to get a 6, B = {6} ⊆ S .In probability theory, we say that the events A and B are mutuallyexclusive because they have no outcomes in common.In set theory, we say that the sets A and B are mutually exclusivebecause their intersection is empty.

P(A ∩ B) =n(A ∩ B)

n(S)=

n(∅)

n(S)=

0

6= 0.

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Addition Rule 1 (p. 196)

When two events A and B of the same experiment are mutuallyexclusive, the probability that A or B will occur is

P(A ∪ B) =n(A ∪ B)

n(S)=

n(A)

n(S)+

n(B)

n(S)= P(A) + P(B).

For example. The experiment is to roll a die. The sample space isthe set of all possible outcomes is S = {1, 2, 3, 4, 5, 6}.The event A is to get an odd number, A = {1, 3, 5} ⊆ S .The event B is to get a 6, B = {6} ⊆ S .

P(A ∪ B) =n(A ∪ B)

n(S)=

n(A)

n(S)+

n(B)

n(S)=

3

6+

1

6= 4/6.

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Principle of Inclusion-Exclusion (p. 197)

When two events are not mutually exclusive, we must subtract oneof the two probabilities of the outcomes that are common to bothevents, since they have been counted twice.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

BA

A ∩ B

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Addition Rule 2 (p. 197)

When two events A and B of the same experiment are not

mutually exclusive, the probability that A or B will occur is

P(A ∪ B) =n(A ∪ B)

n(S)=

n(A) + n(B) − n(A ∩ B)

n(S)

= P(A) + P(B) − P(A ∩ B).

Note: This rule can also be used when the events are mutuallyexclusive, since (A ∩ B) will always equal 0. However, it isimportant to make a distinction between the two situations.

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Example of Addition Rule 2

For example. The experiment is to roll a die. The sample space isthe set of all possible outcomes is S = {1, 2, 3, 4, 5, 6}.The event A is to get an odd number, A = {1, 3, 5} ⊆ S .The event B is to get a number greater than 4, B = {5, 6} ⊆ S .As A ∩ B = {1, 3, 5} ∩ {5, 6} = {5} 6= ∅, the events A and B arenot mutually exclusive.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) =3

6+

2

6−

1

6=

4

6

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Principe of Inclusion-Exclusion for Three Sets

n(A ∪ B ∪ C ) = n(A) + n(B) + n(C )

− n(A ∩ B) − n(A ∩ C ) − n(B ∩ C )

+ n(A ∩ B ∩ C ).

BAA ∩ B

C

A ∩ C B ∩ C

A ∩ B ∩ C

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Principe of Inclusion-Exclusion for Four Sets

n(A1 ∪ A2 ∪ A3 ∪ A4)

= n(A1) + n(A2) + n(A3) + n(A4)

− n(A1 ∩ A2) − n(A1 ∩ A3) − n(A1 ∩ A4) − n(A2 ∩ A3)

− n(A2 ∩ A4) − n(A3 ∩ A4)

+ n(A1 ∩ A2 ∩ A3) + n(A1 ∩ A2 ∩ A4) + n(A1 ∩ A3 ∩ A4)

+ n(A2 ∩ A3 ∩ A4)

− n(A1 ∩ A2 ∩ A3 ∩ A4)

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Principe of Inclusion-Exclusion for n Sets

Let A1, A2, ..., An be n finite sets. Then

n(A1 ∪ A2 ∪ · · · ∪ An) =∑

1≤i≤n

n(Ai )

−∑

1≤i<j≤n

n(Ai ∩ Aj)

+∑

1≤i<j<k≤n

n(Ai ∩ Aj ∩ Ak)

− · · · + · · · − · · ·

+ (−1)n+1n(A1 ∩ A2 ∩ · · · ∩ An)

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Outline


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Independent Events (p. 205)

The multiplication rules can be used to find the probability of two

or more events that occur in sequence. For example, if a coinis tossed and then a die is rolled, one can find the probability ofgetting a head on the coin and a 4 on the die. These two eventsare said to be independent since the outcome of the first event(tossing a coin) does not affect the probability outcome of thesecond event (rolling a die).

Two events A and B are independent events if the fact that A

occurs does not affect the probability of B occurring.

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Multiplication Rule 1 (p. 206)

When two events are independent, the probability of bothoccurring is

P(A ∩ B) = P(A) · P(B)

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Remarks on the Multiplication Rule 1

1. Multiplication rule 1 can be extended to three or moreindependent events by using the formula

P(A ∩ B ∩ C ∩ · · · ∩ K ) = P(A) · P(B) · P(C ) · · ·P(K )

2. In this sequence, the experiments may or may not be the same.If the experiments are the same, the events may or may not be thesame.

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Dependent Events (p. 208)

When the outcome or occurrence of the first event A affects theoutcome or occurrence of the second event B in such a way thatthe probability is changed, the events A and B are said to bedependent events.

The conditional probability of an event B in relationship to anevent A is the probability that event B occurs given that the

event A has already occurred. The notation for conditionalprobability is P(B|A). This notation does not mean that B isdivided by A; rather, it means the probability that event B occursgiven that event A has already occurred.

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Multiplication Rule 2 (p. 208)

When two events are dependant, the probability of both occurringis

P(A ∩ B) = P(A) · P(B|A)

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Formula for Conditional Probability (p. 210)

The probability that the second event B occurs given that the firstevent A has occurred can be found by dividing the probability thatboth events occurred by the probability that the first event hasoccurred. The formula is

P(B|A) =P(A ∩ B)

P(A)

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Conditional Probability and Independent Events

Two events A and B are independent if P(B|A) = P(B) and aredependent otherwise.

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Probabilities for “At Least” or “At Most” (P. 213)

In some case, it is easier to compute the probability of thecomplement of an event than the probability of the event itself.This is still true for a sequence of events.Example: A coin is tossed 5 times. Find the probability of gettingat least one tail. This is equal to 1 minus the probability ofgetting no tail at all, which is all heads.Find the probability of getting at most four tails. This is equal to1 minus the probability of getting five tails.

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