copyright © 2013, 2010 and 2007 pearson education, inc. chapter probability 5

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

Probability

5


Section

Putting It Together: Which Method Do I Use?

5.6


Objectives

1. Determine the appropriate probability rule to use

2. Determine the appropriate counting technique to use

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Objective 1

• Determine the Appropriate Probability Rule to Use

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In the game show Deal or No Deal?, a contestant is presented with 26 suitcases that contain amounts ranging from $0.01 to $1,000,000. The contestant must pick an initial case that is set aside as the game progresses. The amounts are randomly distributed among the suitcases prior to the game. Consider the following breakdown:

EXAMPLE Probability: Which Rule Do I Use?EXAMPLE Probability: Which Rule Do I Use?

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The probability of this event is not compound. Decide among the empirical, classical, or subjective approaches. Each prize amount is randomly assigned to one of the 26 suitcases, so the outcomes are equally likely. From the table we see that 7 of the cases contain at least $100,000. Letting E = “worth at least $100,000,” we compute P(E) using the classical approach.


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P E N E N S

7

260.269


The chance the contestant selects a suitcase worth at least $100,000 is 26.9%. In 100 different games, we would expect about 27 games to result in a contestant choosing a suitcase worth at least $100,000.


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P E N E N S

7

260.269


According to a Harris poll in January 2008, 14% of adult Americans have one or more tattoos, 50% have pierced ears, and 65% of those with one or more tattoos also have pierced ears. What is the probability that a randomly selected adult American has one or more tattoos and pierced ears?


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The probability of a compound event involving ‘AND’. Letting E = “one or more tattoos” and F = “ears pierced,” we are asked to find P(E and F). The problem statement tells us that P(F) = 0.50 and P(F|E) = 0.65. Because P(F) ≠ P(F|E), the two events are not independent. We can find P(E and F) using the General Multiplication Rule.

P E and F P E P F | E 0.14 0.65 0.091

So, the chance of selecting an adult American at random who has one or more tattoos and pierced ears is 9.1%.


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Objective 2

• Determine the Appropriate Counting Technique to Use

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The Hazelwood city council consists of 5 men and 4 women. How many different subcommittees can be formed that consist of 3 men and 2 women?

Sequence of events to consider: select the men, then select the women. Since the number of choices at each stage is independent of previous choices, we use the Multiplication Rule of Counting to obtain

N(subcommittees) = N(ways to pick 3 men) • N(ways to pick 2 women)

EXAMPLE Counting: Which Technique Do I Use?EXAMPLE Counting: Which Technique Do I Use?

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To select the men, we must consider the number of arrangements of 5 men taken 3 at a time. Since the order of selection does not matter, we use the combination formula.

N(subcommittees) = 10 • 3 = 30. There are 30 possible subcommittees that contain 3 men and 2 women.

To select the women, we must consider the number of arrangements of 3 women taken 2 at a time. Since the order of selection does not matter, we use the combination formula again.

N ways to pick 3 women 3C2 3!

2!1!3


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N ways to pick 3 men 5 C3 5!

3!2!10


On February 17, 2008, the Daytona International Speedway hosted the 50th running of the Daytona 500. Touted by many to be the most anticipated event in racing history, the race carried a record purse of almost $18.7 million. With 43 drivers in the race, in how many different ways could the top four finishers (1st, 2nd, 3rd, and 4th place) occur?


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The number of choices at each stage is independent of previous choices, so we can use the Multiplication Rule of Counting. The number of ways the top four finishers can occur is

N(top four) = 43 • 42 • 41 • 40 = 2,961,840We could also approach this problem as an arrangement of units. Since each race position is distinguishable, order matters in the arrangements. We are arranging the 43 drivers taken 4 at a time, so we are only considering a subset of r = 4 distinct drivers in each arrangement. Using our permutation formula, we get

N top four 43 P4 43!

43 4 ! 43424140 2,961,840


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copyright © 2013, 2010 and 2007 pearson education, inc. chapter probability 5

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