Math 1300: Section 8-1 Sample Spaces, Events, and Probability

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ExperimentsSample Spaces and Events

Probability of an EventEqually Likely Assumption

Math 1300 Finite MathematicsSection 8-1: Sample Spaces, Events, and Probability

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics




Probability theory is a branch of mathematics that has beendeveloped do deal with outcomes of random experiments. Arandom experiment (or just experiment) is a situationinvolving chance or probability that leads to results calledoutcomes.





DefinitionThe set S of all possible outcomes of an experiment a waythat in each trial of the experiment one and only one of theoutcomes (events) in the set will occur, we call the set S asample space for the experiment. Each element in S iscalled a simple outcome, or simple event.

An event E is defined to be any subset of S (including theempty set and the sample space S). Event E is a simpleevent if it contains only one element and a compoundevent if it contains more than one element.We say that an event E occurs if any of the simple eventsin E occurs.





DefinitionThe set S of all possible outcomes of an experiment a waythat in each trial of the experiment one and only one of theoutcomes (events) in the set will occur, we call the set S asample space for the experiment. Each element in S iscalled a simple outcome, or simple event.An event E is defined to be any subset of S (including theempty set and the sample space S). Event E is a simpleevent if it contains only one element and a compoundevent if it contains more than one element.

We say that an event E occurs if any of the simple eventsin E occurs.





DefinitionThe set S of all possible outcomes of an experiment a waythat in each trial of the experiment one and only one of theoutcomes (events) in the set will occur, we call the set S asample space for the experiment. Each element in S iscalled a simple outcome, or simple event.An event E is defined to be any subset of S (including theempty set and the sample space S). Event E is a simpleevent if it contains only one element and a compoundevent if it contains more than one element.We say that an event E occurs if any of the simple eventsin E occurs.





Example: Suppose an experiment consists of simultaneouslyrolling two fair six-sided dice (say, one red die and one greendie) and recording the values on each die.

Some possibilitiesare (Red=1, Green=5) or (Red=2, Green=2).

(1,5) (2,2)

What is the sample space S for this experiment?





Example: Suppose an experiment consists of simultaneouslyrolling two fair six-sided dice (say, one red die and one greendie) and recording the values on each die. Some possibilitiesare (Red=1, Green=5)

or (Red=2, Green=2).

(1,5)

(2,2)






Example: Suppose an experiment consists of simultaneouslyrolling two fair six-sided dice (say, one red die and one greendie) and recording the values on each die. Some possibilitiesare (Red=1, Green=5) or (Red=2, Green=2).

(1,5) (2,2)






(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)





To clarify, the sample space is always a set of objects. In thiscase,

S =

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

We can often use the counting techniques we learned in thelast chapter to determine the size of a sample space. In thiscase, by the multiplication principle:

n(S) = 6× 6 = 36





To clarify, the sample space is always a set of objects. In thiscase,

S =

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

We can often use the counting techniques we learned in thelast chapter to determine the size of a sample space. In thiscase, by the multiplication principle:

n(S) = 6× 6 = 36





Events are subsets of the sample space:

A simple event is an event (subset) containing only oneoutcome. For example,

E = {(3,2)}

is a simple event.A compound event is an event (subset) containing morethan one outcome. For example,

E = {(3,2), (4,1), (5,2)}

is a compound event.





Events are subsets of the sample space:A simple event is an event (subset) containing only oneoutcome. For example,

E = {(3,2)}

is a simple event.

A compound event is an event (subset) containing morethan one outcome. For example,

E = {(3,2), (4,1), (5,2)}






Events are subsets of the sample space:A simple event is an event (subset) containing only oneoutcome. For example,

E = {(3,2)}

is a simple event.A compound event is an event (subset) containing morethan one outcome. For example,

E = {(3,2), (4,1), (5,2)}






Events will often be described in words, and the first step willbe to determine the correct subset of the sample space.

Forexample

“A sum of 11 turns up” corresponds to the event

E = {(5,6), (6,5)}.

Notice that n(E) = 2.“The numbers on the two dice are equal” corresponds tothe event

F = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Here we have n(F ) = 6.“A sum less than or equal to 3” corresponds to the event:

G = {(1,1), (1,2), (2,1)}

Here n(G) = 3





Events will often be described in words, and the first step willbe to determine the correct subset of the sample space. Forexample


E = {(5,6), (6,5)}.

Notice that n(E) = 2.

“The numbers on the two dice are equal” corresponds tothe event

F = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.


G = {(1,1), (1,2), (2,1)}

Here n(G) = 3







E = {(5,6), (6,5)}.


F = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Here we have n(F ) = 6.

“A sum less than or equal to 3” corresponds to the event:

G = {(1,1), (1,2), (2,1)}

Here n(G) = 3







E = {(5,6), (6,5)}.


F = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.


G = {(1,1), (1,2), (2,1)}

Here n(G) = 3Jason Aubrey Math 1300 Finite Mathematics




Example: A wheel with 18 numbers on the perimeter is spunand allowed to come to rest so that a pointer points within anumbered sector.

(a) What is the sample space for this experiment?

S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}

(b) Identify the event “the outcome is a number greater than15”?

E = {16,17,18}

(c) Identify the event “the outcome is a number divisible by 12”?

E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}







S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}


E = {16,17,18}


E = {12}





The first step in defining the probability of an event is to assignprobabilities to each of the outcomes (simple events) in thesample space.

Suppose we flip a fair coin twice. The sample space is

S = {HH,HT ,TH,TT}

Since the coin is fair, each of the four outcomes is equallylikely, so P(HH) = P(HT ) = P(TH) = P(TT ) = 1

4 .







S = {HH,HT ,TH,TT}


4 .







S = {HH,HT ,TH,TT}


4 .





Suppose the local meteorologist determines that thechance of rain is 15%. As an experiment, we go out toobserve the weather. The sample space is

S = {Rain,No Rain}

The two outcomes here are not equally likely. We haveP(Rain) = 0.15 and P(No Rain) = 0.85.

Notice that in both cases, each probability was between zeroand one, and the sum of all of the probabilities was one.






S = {Rain,No Rain}








S = {Rain,No Rain}







Definition (Probabilities of Simple Events)Given a sample space

S = {e1,e2, . . . ,en}

with n simple events, to each simple event ei we assign a realnumber, denoted by P(ei), called the probability of the eventei .

1 The probability of a simple event is a number between 0and 1, inclusive. That is, 0 ≤ P(ei) ≤ 1.

2 The sum of the probabilities of all simple events in thesample space is 1. That is,

P(e1) + P(e2) + · · ·+ P(en) = 1.






S = {e1,e2, . . . ,en}




P(e1) + P(e2) + · · ·+ P(en) = 1.






S = {e1,e2, . . . ,en}




P(e1) + P(e2) + · · ·+ P(en) = 1.





Two coin flips. . .

S = {HH,HT ,TH,TT}

e P(e)HH 1

4HT 1

4TH 1

4TT 1

4

A possibly rainy day. . .

S = {Rain,No Rain}

e P(e)Rain 0.15

No Rain 0.85





Example: Suppose a fair coin is flipped twice. What is theprobability that exactly one head turns up.

The event “exactly one head turns up” is the set

E = {HT ,TH}

We know that P(HT ) = 14 and P(TH) = 1

4 .

To determine P(E), just add the probabilities of the simpleevents in E .

P(E) = P(HT ) + P(TH) =14+

14=

12







E = {HT ,TH}


4 .


P(E) = P(HT ) + P(TH) =14+

14=

12







E = {HT ,TH}


4 .


P(E) = P(HT ) + P(TH) =14+

14=

12







E = {HT ,TH}


4 .


P(E) = P(HT ) + P(TH) =14+

14=

12





Definition (Probability of an Event E)Given a probability assignment for the simple events in asample space S, we define the probability of an arbitraryevent E , denoted by P(E), as follows:

1 If E is the empty set, then P(E) = 0.2 If E is a simple event, i.e. E = {ei}, then P(E) = P(ei) as

defined previously.3 If E is a compound event, then P(E) is the sum of the

probabilities of all the simple events in E .4 If E is the sample space S, then P(E) = P(S) = 1.






1 If E is the empty set, then P(E) = 0.

2 If E is a simple event, i.e. E = {ei}, then P(E) = P(ei) asdefined previously.

3 If E is a compound event, then P(E) is the sum of theprobabilities of all the simple events in E .

4 If E is the sample space S, then P(E) = P(S) = 1.







defined previously.

3 If E is a compound event, then P(E) is the sum of theprobabilities of all the simple events in E .









probabilities of all the simple events in E .









probabilities of all the simple events in E .4 If E is the sample space S, then P(E) = P(S) = 1.





Example: In a family with 3 children, excluding multiple births,what is the probability of having exactly 2 girls? Assume that aboy is as likely as a girl at each birth.

First we determine the sample space S:

S = {GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB}

Since a boy is as likely as a girl at each birth, each of the 8outcomes in S is equally likely; so each outcome hasprobability 1

8 .









8 .









8 .





Now we identify the event we wish to find the probability of:

E = {GGB,GBG,BGG}

Therefore,

P(E) = P(GGB) + P(GBG) + P(BGG)

=18+

18+

18=

38





Now we identify the event we wish to find the probability of:

E = {GGB,GBG,BGG}

Therefore,

P(E) = P(GGB) + P(GBG) + P(BGG)

=18+

18+

18=

38





Procedure: Steps for Finding the Probability of an Event E

1 Set up an appropriate sample space S for the experiment.2 Assign acceptable probabilities to the simple events in S.3 To obtain the probability of an arbitrary event E , add the

probabilities of the simple events in E .





Procedure: Steps for Finding the Probability of an Event E1 Set up an appropriate sample space S for the experiment.

2 Assign acceptable probabilities to the simple events in S.3 To obtain the probability of an arbitrary event E , add the






Procedure: Steps for Finding the Probability of an Event E1 Set up an appropriate sample space S for the experiment.2 Assign acceptable probabilities to the simple events in S.

3 To obtain the probability of an arbitrary event E , add theprobabilities of the simple events in E .





Procedure: Steps for Finding the Probability of an Event E1 Set up an appropriate sample space S for the experiment.2 Assign acceptable probabilities to the simple events in S.3 To obtain the probability of an arbitrary event E , add the






Recall two past examples. . .

Two coin flips. . .

S = {HH,HT ,TH,TT}

e P(e)HH 1

4HT 1

4TH 1

4TT 1

4

The outcomes are equallylikely.


S = {Rain,No Rain}

e P(e)Rain 0.15

No Rain 0.85

The outcomes are not equallylikely.






Two coin flips. . .

S = {HH,HT ,TH,TT}

e P(e)HH 1

4HT 1

4TH 1

4TT 1

4



S = {Rain,No Rain}

e P(e)Rain 0.15

No Rain 0.85







Two coin flips. . .

S = {HH,HT ,TH,TT}

e P(e)HH 1

4HT 1

4TH 1

4TT 1

4



S = {Rain,No Rain}

e P(e)Rain 0.15

No Rain 0.85






Sometimes we can assume that all outcomes in a samplespace are equally likely.

If S = {e1,e2, . . . ,en} is a sample space in which alloutcomes are equally likely, then we assign the probability1n to each outcome. That is

P(ei) =1n

and we have. . .





Sometimes we can assume that all outcomes in a samplespace are equally likely.If S = {e1,e2, . . . ,en} is a sample space in which alloutcomes are equally likely, then we assign the probability1n to each outcome. That is

P(ei) =1n

and we have. . .





Theorem (Probability of an Arbitrary Event under an EquallyLikely Assumption)If we assume that each simple event in a sample space S is aslikely to occur as any other, then the probability of an arbitraryevent E in S is given by

P(E) =number of elements in Enumber of elements in S

=n(E)

n(S).





Example: Suppose an experiment consists of simultaneouslyrolling two fair six-sided dice (say, one red die and one greendie) and recording the values on each die.

Some possibilitiesare (Red=1, Green=5) or (Red=2, Green=2).

(1,5) (2,2)

(a) What is the probability that the sum on the two dice comesout to be 11?





Example: Suppose an experiment consists of simultaneouslyrolling two fair six-sided dice (say, one red die and one greendie) and recording the values on each die. Some possibilitiesare (Red=1, Green=5)

or (Red=2, Green=2).

(1,5)

(2,2)







(1,5) (2,2)







(1,5) (2,2)






Firstly, since the dice are fair, for each die, the numbers 1-6are all equally likely to turn up. So each possible pair ofnumbers (1,5), (3,2), etc, is just as likely as any other. So,we can make an equally likely assumption.

We know from earlier that n(S) = 36.E = {(5,6), (6,5)}, so n(E) = 2.Therefore

P(E) =n(E)

n(S)=

236

=1

18





Firstly, since the dice are fair, for each die, the numbers 1-6are all equally likely to turn up. So each possible pair ofnumbers (1,5), (3,2), etc, is just as likely as any other. So,we can make an equally likely assumption.We know from earlier that n(S) = 36.

E = {(5,6), (6,5)}, so n(E) = 2.Therefore

P(E) =n(E)

n(S)=

236

=1

18





Firstly, since the dice are fair, for each die, the numbers 1-6are all equally likely to turn up. So each possible pair ofnumbers (1,5), (3,2), etc, is just as likely as any other. So,we can make an equally likely assumption.We know from earlier that n(S) = 36.E = {(5,6), (6,5)}, so n(E) = 2.

ThereforeP(E) =

n(E)

n(S)=

236

=1

18





Firstly, since the dice are fair, for each die, the numbers 1-6are all equally likely to turn up. So each possible pair ofnumbers (1,5), (3,2), etc, is just as likely as any other. So,we can make an equally likely assumption.We know from earlier that n(S) = 36.E = {(5,6), (6,5)}, so n(E) = 2.Therefore

P(E) =n(E)

n(S)=

236

=1

18





(c) What is the probability that the numbers on the dice areequal?

The event here isF = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}So, n(F ) = 6Therefore,

P(F ) =n(F )

n(S)=

636

=16






The event here isF = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

So, n(F ) = 6Therefore,

P(F ) =n(F )

n(S)=

636

=16






The event here isF = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}So, n(F ) = 6

Therefore,

P(F ) =n(F )

n(S)=

636

=16






The event here isF = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}So, n(F ) = 6Therefore,

P(F ) =n(F )

n(S)=

636

=16





Example: 5 cards are drawn simultaneously from a standarddeck of 52 cards.

(a) Describe the sample space S. What is n(S)?

Each outcome is a set of 5 cards chosen from the 52 availablecards. So, the sample space S can be described as

S = {all possible 5 card hands}

How many 5-card hands can be drawn from a 52-card deck?From the previous chapter, we know this is

n(S) = C(52,5) = 2,598,960

When the cards are dealt, each card is just as likely as anyother, so any five card hand is just as likely as any other. Inother words, we can make an equally likely assumption.










n(S) = C(52,5) = 2,598,960










How many 5-card hands can be drawn from a 52-card deck?

From the previous chapter, we know this is

n(S) = C(52,5) = 2,598,960











n(S) = C(52,5) = 2,598,960











n(S) = C(52,5) = 2,598,960






(b)Find the probability that all of the cards are hearts.

The event “all of the cards are hearts” is the set

E = {all 5 card hands with only hearts}

Since there are 13 hearts in a standard deck of cards, we have

n(E) = C(13,5) = 1287

By the equally likely assumption

P(E) =n(E)

n(S)=

12872,598,960

≈ 0.000495

or about 0.05%.









n(E) = C(13,5) = 1287


P(E) =n(E)

n(S)=

12872,598,960

≈ 0.000495

or about 0.05%.









n(E) = C(13,5) = 1287


P(E) =n(E)

n(S)=

12872,598,960

≈ 0.000495

or about 0.05%.









n(E) = C(13,5) = 1287


P(E) =n(E)

n(S)=

12872,598,960

≈ 0.000495

or about 0.05%.





(c) Find the probability that all the cards are face cards (that is,jacks, queens or kings).

The event “all the cards are face cards” is the set

F = {all 5 card hands consisting only of face cards}

There are a total of 4× 3 = 12 face cards. So,

n(F ) = C(12,5) = 792


P(F ) =n(F )

n(S)=

7922,598,960

≈ 0.000305

or about 0.03%.









n(F ) = C(12,5) = 792


P(F ) =n(F )

n(S)=

7922,598,960

≈ 0.000305

or about 0.03%.









n(F ) = C(12,5) = 792


P(F ) =n(F )

n(S)=

7922,598,960

≈ 0.000305

or about 0.03%.









n(F ) = C(12,5) = 792


P(F ) =n(F )

n(S)=

7922,598,960

≈ 0.000305

or about 0.03%.





(d) Find the probability that all the cards are even. (Consideraces to be 1, jacks to be 11, queens to be 12 and kings to be13).

The event “all the cards are even is the set

G = {all 5 card hands consisting of only even cards}

There are 6 even cards per suit (2,4,6,8,10,Q); so there are atotal of 20 even cards in a deck. So,

n(G) = C(20,6) = 38,760.

By the qually likely assumption,

P(G) =n(G)

n(S)=

38,7602,598,960

≈ 0.0149

or about 14.9%.









n(G) = C(20,6) = 38,760.


P(G) =n(G)

n(S)=

38,7602,598,960

≈ 0.0149

or about 14.9%.









n(G) = C(20,6) = 38,760.


P(G) =n(G)

n(S)=

38,7602,598,960

≈ 0.0149

or about 14.9%.









n(G) = C(20,6) = 38,760.


P(G) =n(G)

n(S)=

38,7602,598,960

≈ 0.0149

or about 14.9%.Jason Aubrey Math 1300 Finite Mathematics

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Math 1300: Section 8-1 Sample Spaces, Events, and Probability

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