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Excursions in ModernMathematics

Sixth Edition

Peter Tannenbaum

Edited by Ling Yeong Tyng

Faculty of Computer Science &

IT

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Chapter 15Chances, Probabilities, and Odds

Measuring

Uncertainty

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Outline/learning Objectives

To describe an appropriate sample space

of a random experiment. To apply the multiplication rule,

permutations, and combinations to

counting problems. To understand the concept of a

probability assignment.

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Outline/learning Objectives

To identify independent events and their 

properties. To use the language of odds in

describing probabilities of events.

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15.1 Random Experiments

and Sample Spaces

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Random experiment

Description of an activity or process whoseoutcome cannot be predicted ahead of time.

Sample space

 Associated with every random experiment is the

set of all of its possible outcomes. We will

consistently use the letter S to denote a sample

space and N to denote its size (the number of 

outcomes in S).

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Rolling the Dice: Part 1 cont’ 

When looking at the figure below you will notice that we

are treating the dice as distinguishable objects (as if one

were white and the other red), so that and

are considered different outcomes.

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Coin Tossing: Part 1

One simple random experiment is to toss a coin and

observe whether it lands heads or tails. The sample space

can be described by S = {H, T}, where H stands for Heads

and T stands for Tails. Here, N = 2.

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15.2 Counting Sample

Spaces

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Chances, Probabilities, and Odds

The Multiplication Rule

When something is done in stages, thenumber of ways it can be done is found

by multiplying the number of ways each

of the stages can be done.

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The Making of a Wardrobe: Part 2

Our strategy will be to think of an outfit as being put together instages and to draw a box for each of the stages. We then

separately count the number of choices at each stage and

enter that number in the corresponding box.

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The Making of a Wardrobe: Part 2

The last step is to multiply the numbers in each box. The finalcount for the number of different outfits is

 N = 3 x 7 x 27 x 3 = 1701

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15.3 Permutations

and Combinations

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Permutation

 A group of objects where the ordering of theobjects within the group makes a difference.

Combination

 A group of objects in which the ordering of theobjects is irrelevant .

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The Pleasures of Ice Cream: Part 1

Say you want a true doubleice cream cone… 

How many differentchoices do you have? 

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The Pleasures of Ice Cream: Part 1

Say you want a true double in a bowl –  how many differentchoices so you have?

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The Pleasures of Ice Cream: Part 1

The natural impulse is to count the number of choices using themultiplication rule (and a box model) as shown below. This

would give an answer of 930.

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The Pleasures of Ice Cream: Part 1

Unfortunately, this answer is double counting each of the truedoubles. Why?

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The Pleasures of Ice Cream: Part 1

When we use the multiplication rule, there is a well-definedorder to things, and a scoop of strawberry followed by ascoop of chocolate is counted separate from a scoop of chocolate followed by a scoop of strawberry.

The good news is that now we understand why the count of 930is wrong and we can fix it. All we have to do is divide theoriginal count by 2.

(31 x 30)/2 = 465

Ex

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5 * 7 * (13*12*11)/3! = 10010

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15.4 Probability

Spaces

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Event

 Any subset of the sample space.

Simple event

 An event that consists of just one outcome.

Impossible event

 A special case of the empty set { },

corresponding to an event with no outcomes.

Ex.

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Probability assignment

 A function that assigns to each event E a number between 0 and 1, which represents theprobability of the event E and which we denoteby Pr (E ).

Probability spaceOnce a specific probability assignment is madeon a sample space, the combination of thesample space and the probability assignment.

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Elements of a Probability Space

Sample space: S = {o1, o2,…., oN }

Probability assignment: Pr(o1),Pr(o2),… Pr(oN )

[Each of these is a number between 0 and 1 satisfying

Pr(o1) + Pr(o2) + … Pr(oN ) = 1]

Events: These are all the subsets of S, including { }and S  itself. The probability of an event is given by the

sum of the probabilities of the individual outcomes that

make up the event. [In particular, Pr({ }) = 0 and

Pr(S) =1]

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15.5 Equiprobable

Spaces

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Probabilities in Equiprobable Spaces

Pr(E ) = k /N  (where k denotes the size of theevent E and N denotes the size of the samplespace S).

 A probability space where each simple event hasan equal probability is called an equiprobable “equal opportunity” space. (i.e.: “honest” coin) 

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Chances, Probabilities, and OddsRolling the Dice: Part 2

The sample space has N = 36 individual outcomes (roll a pair of 

honest dice), each with probability 1/36. We will use thenotation T 2, T 3, …T 12 to describe the events “roll a total of 2,” “roll a total of 3,” …, “roll a total of 12,” respectively.We show you how to find Pr(T 7) and Pr(T 11),

T 11 = , Thus,Pr(T 11) = 2/36 0.056

T 7 = , Thus,

Pr(T 7

) = 6/36 = 1/6 0.167

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Tallying: E:”at least one of the dice comes up

an Ace”. 

We can just write down all the individual

outcomes in the event E and tally their number.

This approach gives

and Pr(E ) = 11/36.

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Complementary Event

Imagine that you are playing a game, and youwin if at least one of the two numbers comes

up an Ace (that’s event E ). Otherwise you lose

(call that event F ). The two events E and F are

called complementary events. The probabilities of complementary events add up

to 1. Thus,

Pr(E ) = 1 – Pr(F ).

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15.6 Odds

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Odds

Let E be an arbitrary event. If F denotes thenumber of ways that event E can occur (thefavorable outcomes or hits), and U denotes thenumber of ways that event E does not occur (the unfavorable outcomes, or misses), then

the odds of (also called the odds in favor of ),the event E are given by the ratio F to U , andthe odds against the event E are given by theratio U to F .

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Handicapping a Tennis Tournament.

Probability assignment for the tennistournament was Pr(Ana) = 0.08, Pr(Ivan) =

0.16, Pr(Luke) = 0.20, Pr(Roger) = 0.25,

Pr(Venus) = 0.16 and Pr(Serena) = 0.15.

Express each of these as odds. – Pr(Ana) = 0.08 = 8/100 = 2/25. The odds of Ana

winning the tournament are 2 to 23.

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Conclusion

Sample space

Random experimentEvents

Probability assignment

Equiprobable spaces