15.1 Random Experiments and Sample Spacesjga001/chapter 15.1 to 15.3.pdfCopyright © 2010 Pearson...

Excursions in Modern Mathematics, 7e: 15.1 - 1 Copyright © 2010 Pearson Education, Inc.

15 Chances, Probabilities, and Odds

15.1 Random Experiments and Sample

Spaces

15.2 Counting Outcomes in Sample

Spaces

15.3 Permutations and Combinations

15.4 Probability Spaces

15.5 Equiprobable Spaces

15.6 Odds


• Probability is the quantification of uncertainty.

• We will use the term random experiment to describe an activity or a process whose outcome cannot be predicted ahead of time.

• Examples of random experiments: tossing a coin, rolling a pair of dice, drawing cards out of a deck of cards, predicting the result of a football game, and forecasting the path of a hurricane.

Random Experiment


• Associated with every random

experiment is the set of all of its possible

outcomes, called the sample space of

the experiment.

• For the sake of simplicity, we will

concentrate on experiments for which

there is only a finite set of outcomes,

although experiments with infinitely many

outcomes are both possible and

important.

Sample Space


• We use the letter S to denote a sample

space and the letter N to denote the size

of the sample space S (i.e., the number

of outcomes in S).

Sample Space - Set Notation


One simple random experiment is to toss a

quarter 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

for Tails. Here N = 2.

Example 15.1 Tossing a Coin


Suppose we toss a coin twice and record the

outcome of each toss (H or T) in the order it

happens. What is the sample space?

Example 15.2 More Coin Tossing


The sample space now is

S = {HH, HT, TH, TT}, where HT means that

the first toss came up H and the second toss

came up T, which is a different outcome from

TH (first toss T and second toss H). In this

sample space N = 4.



Suppose now we toss two distinguishable

coins (say, a nickel and a quarter) at the

same time. What is the sample space?



The sample space is still

S = {HH, HT, TH, TT}. (Here we must agree

what the order of the symbols is–for example,

the first symbol describes the quarter and the

second the nickel.)



Since they have the same sample space, we

will consider the two previous random

experiments as the same random experiment.



Suppose we toss a coin twice, but we only

care now about the number of heads that

come up. What is the sample space?



Here there are only three possible outcomes

(no heads, one head, or both heads), and

symbolically we might describe this sample

space as

S = {0, 1, 2}.



The experiment is to roll a pair of dice. What is the sample space?

Example 15.5 Dice Rolling


Example 15.5 More Dice Rolling


Here we have a sample space with 36 different outcomes. Notice that the dice are colored white and red, a symbolic way to emphasize the fact that we are treating the dice as distinguishable objects. That is why the following rolls are distinguishable.

Example 15.5 More Dice Rolling


Example 15.5 More Dice Rolling The sample space has 36 possible outcomes:

{(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)}

where the pairs represent the numbers rolled on

each dice (white, red).


Roll a pair of dice and consider the total of the

two numbers rolled. What is the sample

space?

Example 15.4 Rolling a Pair of Dice


The possible outcomes in this scenario range

from “rolling a two” to “rolling a twelve,” and

the sample space can be described by

S = {2, 3, 4, 5, 6, 7, 8, 9, 10,11,12}.

Example 15.4 Rolling a Pair of Dice


• Page 577, problem 3

Examples



Solution:

• {ABCD, ABDC, ACBD, ACDB, ADBC,

ADCB, BACD, BADC, BCAD, BCDA,

BDAC, BDCA, CABD, CADB, CBAD,

CBDA, CDAB, CDBA, DABC, DACB,

DBAC, DBCA, DCAB, DCBA}

• There are 24 outcomes.

Examples


We would like to understand what the

sample space looks like without necessarily

writing all the outcomes down. Our real goal

is to find N, the size of the sample space. If

we can do it without having to list all the

outcomes, then so much the better.

Not Listing All of the Outcomes




Spaces


Spaces




15.6 Odds


• If we toss a coin three times and separately record the outcome of each toss, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

• Here we can just count the outcomes and get N = 8.

Example 15.7 Tossing More Coins


• Toss a coin 10 times

• In this case the sample space S is too big to write down

• We can “count” the number of outcomes in S without having to tally them one by one using the multiplication rule.



• Suppose an activity consists of a series of events in which there are a possible outcomes for the first event, b possible outcomes for the second event, c possible outcomes for the third event, and so on.

• Then the total number of different possible outcomes for the series of events is: a · b · c · …

Multiplication Rule


• Toss a coin ten times. How many

outcomes are in the sample space?

• There are two outcomes on the first toss

• There are two outcomes on the second

toss, etc.

• The total number of possible outcomes is

found by multiplying ten two’s together.



• Total number of outcomes if a coin is

tossed ten times:

• Thus N = 210 = 1024.


factors 10

2222N


Dolores is a young saleswoman planning her

next business trip. She is thinking about

packing three different pairs of shoes, four

skirts, six blouses, and two jackets. How

many different outfits will she be able to

create by combining these items? (Assume

that an outfit consists of one pair of shoes,

one skirt, one blouse, and one jacket.)

Example 15.8 The Making of a

Wardrobe


Let’s assume that an outfit consists of one

pair of shoes, one skirt, one blouse, and one

jacket. Then to make an outfit Dolores must

choose a pair of shoes (three choices), a skirt

(four choices), a blouse (six choices), and a

jacket (two choices). By the multiplication rule

the total number of possible outfits is

3 4 6 2 = 144.

Example 15.8 The Making of a

Wardrobe


Five candidates are running in an election,

with the top three vote getters elected (in

order) as President, Vice President, and

Secretary. How many different ways are there

to choose the five candidates to fill these

three positions?

Example 15.10 Ranking the Candidate

in an Election: Part 2


Example 15.10 Ranking the Candidate

in an Election: Part 2



Examples


• Solution to part (a)

Examples

320,4012345678


• Solution to part (a)

Examples

160,2012345674


• Solution to part (c)

Examples

57611223344




Spaces


Spaces




15.6 Odds


Many counting problems can be reduced to

a question of counting the number of ways

in which we can choose groups of objects

from a larger group of objects. Often these

problems require somewhat more

sophisticated counting methods than the

multiplication rule.

Counting Problems


In this section we will discuss the dual

concepts of permutation (a group of objects

in which the ordering of the objects within

the group makes a difference) and

combination (a group of objects in which the

ordering of the objects is irrelevant).

Counting Problems


Suppose that we have a set of n distinct

objects and we want to select r different

objects from this set. The number of ways

that this can be done depends on whether

the selections are ordered or unordered.

Permutation versus Combination


To distinguish between these two scenarios,

we use the terms permutation to describe

an ordered selection and combination to

describe an unordered selection.



For a given number of objects n and a given

selection size r (where 0 ≤ r ≤ n) we can talk

about the “number of permutations of n

objects taken r at a time” and the “number of

combinations of n objects taken r at a time,”

and these two extremely important families

of numbers are denoted nPr and nCr,

respectively. (Some calculators use

variations of this notation, such as Pn,r and

Cn,r respectively.)



Factorial symbol

• For any integer n ≥ 0, the factorial symbol n! is defined as follows:

• 0! = 1

• 1! = 1

• n! = n(n - 1)(n - 2) · · · 3 · 2 · 1


Find each of the following

1. 4!

2. 7!

Example


ANSWER

Example

50401234567!7 .2

241234!4 1.


Summary of essential facts about the

numbers nPr and nCr,.



Evaluate each:

4738 and CP

Example


ANSWER:

Example

336

678

12345

12345678

5!

8!

)!38(

!838 P


Permutations on the Calculator

8 MATH PRB 2:nPr 3

To get:

Then Enter gives 336

38 P


ANSWER:

Example

35

57

123

567

)123()1234(

1234567

3!4!

7!

)!47(!4

!747 C


Combinations on the Calculator

7 MATH PRB 3:nCr 4

To get:

Then Enter gives 35

47 C


We will compare two types of games: five-

card stud poker and five-card draw poker. In

both of these games a player ends up with

five cards, but there is an important difference

when analyzing the mathematics behind the

games: In five-card draw the order in which

the cards come up is irrelevant; in five-card

stud the order in which the cards come up is

extremely relevant.

Example 15.13 Five-Card Poker Hands


The reason for this is that in five-card draw all

cards are dealt down, but in five-card stud

only the first card is dealt down–the remaining

four cards are dealt up, one at a time. This

means that players can assess the relative

strengths of the other players’ hands as the

game progresses and play their hands

accordingly.



Counting the number of five-card stud poker

hands is a direct application of the

multiplication rule: 52 possibilities for the first

card, 51 for the second card, 50 for the third

card, 49 for the fourth card, and 48 for the fifth

card, for an awesome total of

52 51 50 49 48 = 311,875,200

possible hands.



Counting the number of five-card draw poker

hands requires a little more finesse. Here a

player gets five down cards and the hand is

the same regardless of the order in which the

cards are dealt. There are 5! = 120 different

ways in which the same set of five cards can

be ordered, so that one draw hand

corresponds to 120 different stud hands.



Thus, the stud hands count is exactly 120

times bigger than the draw hands count.

Great! All we have to do then is divide the

311,875,200 (number of stud hands) by 120

and get our answer: There are 2,598,960

possible five-card draw hands.

As before, it’s more telling to look at this

answer in the uncalculated form

(52 51 50 49 48)/5! = 2,598,960.



Like many other state lotteries, the Florida

Lotto is a game in which for a small

investment of just one dollar a player has a

chance of winning tens of millions of dollars.

Enormous upside, hardly any downside–

that’s why people love playing the lottery and,

like they say, “everybody has to have a

dream.” But, in general, lotteries are a very

bad investment, even if it’s only a dollar, and

the dreams can turn to nightmares.

Why so?

Example 15.14 The Florida Lotto


In a Florida Lotto ticket, one gets to select six

numbers from 1 through 53. To win the

jackpot (there are other lesser prizes we

won’t discuss here), those six numbers have

to match the winning numbers drawn by the

lottery in any order. Since a lottery draw is

just an unordered selection of six objects (the

winning numbers) out of 53 objects (the

numbers 1 through 53), the number of

possible draws is 53C6 .



Doesn’t sound too bad until we do the

computation (or use a calculator) and realize

that


53C

6

53 52 51 50 49 48

6!

22,957,480


A safe combination consists of four numbers

between 0 and 99. If four numbers are

randomly selected, determine the number of

possible combinations. Assume you are not

allowed to repeat numbers so that a

combination such as:

1-2-1-0

is not allowed. Verify that this is a permutation.

Identify both r and n.

Example


An arrangement of numbers, such that

• 4 numbers are chosen at a time from 100 distinct numbers.

• repetition of numbers is not allowed (each number is distinct)

• the order of the numbers is important.

Example


Solution:

Examples

4100 P

400,109,94979899100



Examples


34(a)

Examples

720310 P


34(b)

Examples

120710C

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