Chances Probabilities and Odds
Transcript of Chances Probabilities and Odds
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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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Chances, Probabilities, and Odds
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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Chances, Probabilities, and Odds
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