Copyright © Cengage Learning. All rights reserved.
3
Probability
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Copyright © Cengage Learning. All rights reserved.
3.4
Combinatorics and
Probability
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Objectives
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Combinatorics and Probability
You would probably guess that it’s pretty unlikely that two or more people in your math class share a birthday. It turns out, however, that it’s surprisingly likely. In this section, we’ll look at why that is.
Finding a probability involves finding the number of outcomes in an event and the number of outcomes in the sample space. So far, we have used probability rules as an alternative to excessive counting.
Another alternative is combinatorics—that is, permutations, combinations, and the Fundamental Principle of Counting.
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Combinatorics and Probability
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Example 1 – A Birthday Probability
A group of three people is selected at random. What is the probability that at least two of them will have the same birthday?
�Solution:
We will assume that all birthdays are equally likely, and for the sake of simplicity, we will ignore leap-year day (February 29).
• The experiment is to ask three people their birthdays. � One possible outcome is (May 1, May 3, August 23).�
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Example 1 – Solution
• The sample space S is the set of all possible lists of three� birthdays.
• Finding n (S) by counting the elements in S is impractical, � so we will use combinatorics as described in “Which � Combinatorics Method” above.
• The selected items are birthdays. They are selected � with replacement because people can share the same� birthday. This tells us to use the Fundamental Principle � of Counting:
cont’d
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Example 1 – Solution
• Each birthday can be any one of the 365 days in a year
• n (S) =
= 3653
• The event E is the set of all possible lists of three� birthdays in which some birthdays are the same. It is� difficult to compute n (E) directly; instead, we will compute� n (E ′) and use the Complement Rule. E ′ is the set of all� possible lists of three birthdays in which no birthdays are� the same.
cont’d
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Example 1 – Solution
• Finding n (E ′) using “Which Combinatorics Method”:
• The birthdays are selected without replacement,� because no birthdays are the same.
• There is only one category: birthdays.
• The order of selection does matter—(May 1, May 3,� August 23) is a different list from (May 3, August 23,� May1)—so use permutations
• n (E ′) = 365P3
cont’d
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Example 1 – Solution
• Finding p (E ′) and p (E):
p (E) = 1 – p(E ′)
= 0.008204 . . .
≈ 0.8%
using the previous results
the Complement Rule
using the above results
moving the decimal point
left two places
cont’d
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Example 1 – Solution
This result is not at all surprising. It means that two or more�people in a group of three share a birthday slightly less�than 1% of the time.
In other words, this situation is extremely unlikely. However, we will see in Exercise 1 that it is quite likely that two or more people in a group of thirty share a birthday.
cont’d
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Lotteries
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Lotteries
We’re going to look at one of the most common forms of�gambling in the United States: the lottery.
�Many people play the lottery, figuring that “someone’s got to win—why not me?” Is this is a reasonable approach to lottery games?
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Example 2 – Winning a Lottery
Arizona, Connecticut, Missouri, and Tennessee operate�6/44 lotteries. In this game, a gambler selects any six of the�numbers from 1 to 44.If his or her six selections match the six winning numbers, the player wins first prize. If his or her selections include five of the winning numbers, the player wins second prize. Find the probability of:
a. the event E, winning first prize
�b. the event F, winning second prize
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Example 2 – Solution
a. • The experiment is to select six numbers.
• The sample space S is the set of all possible lottery� tickets. That is, it is the set of all possible choices of six� numbers selected from the numbers 1 through 44.
• Finding n (S) by counting the elements in S is � impractical, so we will use combinatorics.
� • The selection is done without replacement, because� you can’t select the same lottery number twice.
• There is only one category—numbers.
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Example 2 – Solution
• Order does not matter, because the gambler can� choose the six numbers in any order.
• n (S) = 44C6
• Finding n (E) is easy, because there is only one winning � combination of numbers:�
n (E) = 1
• Finding p (E):
cont’d
using the above results
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Example 2 – Solution
This means that only one out of approximately seven� million combinations is the first–prize–winning � combinations. Let’s put this in perspective.
The probability of…� dying in a motor vehicle accident: 1/108� dying in an airplane accident: 1/7229� dying from a hornet, wasp, or bee bite: 1/71,107� dying from lightning strike: 1/126,158� being dealt a royal flush in poker: 1/649,740� dying from flesh-eating bacteria: 1/1,000,000� winning the 6/44 lottery: 1/7,000,000
Winning this lottery is an extremely unlikely event. It’s much less likely than dying from flesh-eating bacteria.
cont’d
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Example 2 – Solution
You probably don’t worry about dying this way, because �it essentially doesn’t happen. Sure, it happens to �somebody sometime, but not to anyone you’ve ever �known. Winning the lottery is similar.
b. To find the probability of event F, winning second prize, �we need to find n (F). In part(a), we didn’t have to do any work to find n(E)—we knew it was 1, because there is only one first-prize-winning ticket. We don’t have a shortcut like this to find n(F).
cont’d
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Example 2 – Solution
Instead, we’ll use the same procedure we use to find �n(S). With S, we were counting lottery tickets, without regard to winning numbers or losing numbers. With F, the win/lose distinction is important, and there are two categories—winning numbers and losing numbers. Use the Fundamental Principle of Counting.
• Finding n (F):
• To win second prize, we must select five winning� numbers and one losing number. This tells us to use� the Fundamental Principle of Counting:
cont’d
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Example 2 – Solution
• We will use combinations in each box, for the same� reasons that we used combinations to find n (S).
• The state selects six winning numbers, and the � second prize–winner must select five of them, so � there are 6C5 ways of selecting the five winning � numbers.
• There are 44 – 6 = 38 losing numbers, of which the� player selects 1, so there are 38C51 ways of selecting � one losing number.
cont’d
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Example 2 – Solution
• n (F) =
• Finding p (F):
•
How do we make sense of a decimal with so many � zeros in front of it?
cont’d
using the previous results
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Example 2 – Solution
One way is to round it off at the first nonzero digit and � convert the result to a fraction:
p(F) = 0.00003 . . .
= 3/100,000
This means that if you buy a lot of 6/44 lottery tickets, � you will win second prize approximately three times out � of every 100,000 times you play.
It also means that in any given game, there are about � three second–prize ticket for every 100,000 ticket � purchases.
cont’d
rounding to the first nonzero digit
converting to a fraction
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Example 2 – Solution
cont’d
• the number in front of the C’s add up correctly (6 winning � numbers + 38 losing numbers = 44 total numbers to � choose from)
• the numbers after the C’s add correctly (5 winning � numbers + 1 losing number = 6 total numbers to select)
• the numerator has two parts, because the event has two � categories—winning numbers and losing numbers
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Example 2 – Solution
cont’d
• the denominator has one part, because the sample space � has only one category—numbers
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Lotteries
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Keno
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Keno
The game of keno is a casino version of the lottery. In this game, the casino has a container filled with balls numbered from 1 to 80.
The player buys a keno ticket, with which he or she selects anywhere from 1 to 15 (usually 6, 8, 9, or 10) of those 80 numbers; the player’s selections are called “spots.”
The casino chooses 20 winning numbers, using a mechanical device to ensure a fair game. If a sufficient number of the player’s spots are winning numbers, the player receives an appropriate payoff.
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Example 4 – Winning at Keno
In the game of keno, if eight spots are marked, the player�wins if five or more of his or her spots are selected. Find�the probability of having five winning spots.
Solution:
The sample space S is the set of all ways in which a player can select eight numbers from the eighty numbers in the game.
• Finding n (S):
• Selection is done without replacement.
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Example 4 – Solution
• Order doesn’t matter, so use combinations.
• n (S) = 80C8
• The event E is the set of all ways in which an eight–spot� player can select five winning numbers and three losing� numbers.
• Finding n (E):
• There are two categories—winning numbers and � losing numbers—so use the Fundamental Principle � of Counting.
cont’d
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Example 4 – Solution
• Use combinations in each category, as with n (S).
• The casino selects twenty winning numbers, from which the gambler is to select five winning spots. There are 20C5 different ways of doing this.
• The casino selects 80 – 20 = 60 losing numbers, � from which the gambler is to select 8 – 5 = 3 losing � spots. There are 60C3 different ways of doing this.
• n(E) = 20C5 ● 60C3
cont’d
using the above results
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Example 4 – Solution
• Finding p (E):
≈ 1.8%
This means that if you play eight–spot keno a lot, you will�have five winning spots about 1.8% of the time. It also�means that in any given game, about 1.8% of the players�will have five winning spots.
one zero after the decimal point
writing as a percentage
cont’d
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Poker
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Poker
Poker is America’s card game. Its popularity surged during �the 2000s due to its exposure on TV, the Internet, and the movies. ESPN broadcasts the annual World Series of Poker, held in Las Vegas. Now, there’s also a World �Series of Poker in Europe, as well as one in Africa and one in Australia.
Poker’s popularity is due to the game’s reliance on skill, the �pleasures of strategic bluffing, and the fact that money is its means of keeping score. Poker is played in casinos, card rooms, and on the Internet.
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Poker
In 2012, a federal judge ruled that it’s predominantly a game of skill, not a game of chance. As a result, there is no federal law that prohibits U.S. players from playing poker online for money.
State laws vary quite a bit, but Washington is the only state that makes playing online a crime. Some states, most notably California, have legally sanctioned card rooms where one can play only poker.
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Example 5 – Getting Four Aces
Find the probability of being dealt four aces when playing five-card poker.
Solution:
The sample space consists of all possible five–card hands that can be dealt from a deck of fifty–two cards
• Finding n (S):
� • Selection is done without replacement
� • Order does not matter, so use combinations.
� • n (S) = 52C5
• The event E consists of all possible five–card hands that� include four aces and one non–ace.
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Example 5 – Solution
• There are two categories—aces and non-aces—so use� the Fundamental Principle of Counting.
• Use combinations as with n (S).
• The gambler is to be dealt four of four aces. This can� happen in 4C4 ways.
• The gambler is to be dealt one of 52 – 4 = 48 non-aces� This can be done in 48C1 ways.
• n (E) = 4C4 ● 48C1
cont’d
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Example 5 – Solution
• Finding p (E):
= 0.000018 . . .
≈ 0.00002
= 2/100,000
= 1/50,000
This means that if you play cards a lot, you will be dealt four aces about once every 50,000 deals.
more than two zeros after
the decimal point
rounding to the first nonzero digit
rewriting as a fraction
reducing
cont’d
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