- 161 Views
- Uploaded on
- Presentation posted in: General

Chapter 10

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 10

Counting Methods

2012 Pearson Education, Inc.

- 10.1 Counting by Systematic Listing
- 10.2 Using the Fundamental Counting Principle
- 10.3 Using Permutations and Combinations
- 10.4 Using Pascal’s Triangle
- 10.5Counting Problems Involving “Not” and “Or”

2012 Pearson Education, Inc.

- Counting by Systematic Listing

2012 Pearson Education, Inc.

- One-Part Tasks
- Product Tables for Two-Part Tasks
- Tree Diagrams for Multiple-Part Tasks
- Other Systematic Listing Methods

2012 Pearson Education, Inc.

The results for simple, one-part tasks can often be listed easily. For the task of tossing a fair coin, the list is heads, tails, with two possible results. For the task of rolling a single fair die the list is 1, 2, 3, 4, 5, 6, with six possibilities.

2012 Pearson Education, Inc.

Consider a club N with four members:

N = {Mike, Adam, Ted, Helen} or in abbreviated formN = {M, A, T, H}

In how many ways can this group select a president?

Solution

The task is to select one of the four members as president. There are four possible results: M, A, T, and H.

2012 Pearson Education, Inc.

Determine the number of two-digit numbers that can be written using the digits from the set {2, 4, 6}.

Solution

The task consists of two parts:

1. Choose a first digit

2. Choose a second digit

The results for a two-part task can be pictured in a product table, as shown on the next slide.

2012 Pearson Education, Inc.

Solution(continued)

From the table we obtain the list of possible results: 22, 24, 26, 42, 44, 46, 62, 64, 66.

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

Find the number of ways club N (previous slide) can elect a president and secretary.

Solution

The task consists of two parts:

1. Choose a president

2. Choose a secretary

The product table is pictured on the next slide.

2012 Pearson Education, Inc.

Solution(continued)

From the table we see that there are 12 possibilities.

2012 Pearson Education, Inc.

Find the number of ways club N (previous slide) can appoint a committee of two members.

Solution

Using the table on the previous slide, this time the order of the letters (people) in a pair makes no difference. So there are 6 possibilities: MA, MT, MH, AT, AH, TH.

2012 Pearson Education, Inc.

A task that has more than two parts is not easy to analyze with a product table. Another helpful device is a tree diagram, as seen in the next example.

2012 Pearson Education, Inc.

Find the number of three digit numbers that can be written using the digits from the set {2, 4, 6} assuming repeated digits are not allowed.

Solution

Second

Third

First

4

6

6

4

246

264

426

462

624

642

2

4

6

2

6

6

2

6 possibilities

2

4

4

2

2012 Pearson Education, Inc.

There are additional systematic ways to produce complete listings of possible results besides product tables and tree diagrams. One of these ways is shown in the next example.

2012 Pearson Education, Inc.

How many triangles (of any size) are in the figure below?

D

Solution

E

C

One systematic approach is to label the points as shown, begin with A, and proceed in alphabetical

F

A

B

order to write all 3-letter combinations (like ABC, ABD, …), then cross out ones that are not triangles. There are 12 different triangles.

2012 Pearson Education, Inc.

- Using the Fundamental Counting Principle

2012 Pearson Education, Inc.

- Uniformity and the Fundamental Counting Principle
- Factorials
- Arrangements of Objects

2012 Pearson Education, Inc.

A multiple-part task is said to satisfy the uniformity criterion if the number of choices for any particular part is the same no matter which choices were selected for the previous parts.

2012 Pearson Education, Inc.

When a task consists of k separate parts and satisfies the uniformity criterion, if the first part can be done in n1 ways, the second part can be done in n2 ways, and so on through the k th part, which can be done in nk ways, then the total number of ways to complete the task is given by the product

2012 Pearson Education, Inc.

How many two-digit numbers can be made from the set {0, 1, 2, 3, 4, 5}? (numbers can’t start with 0.)

Solution

There are 5(6) = 30 two-digit numbers.

2012 Pearson Education, Inc.

How many two-digit numbers that do not contain repeated digits can be made from the set {0, 1, 2, 3, 4, 5} ?

Solution

There are 5(5) = 25 two-digit numbers.

2012 Pearson Education, Inc.

How many ways can you select two letters followed by three digits for an ID?

Solution

There are 26(26)(10)(10)(10) = 676,000 IDs possible.

2012 Pearson Education, Inc.

For any counting number n, the product of all counting numbers from n down through 1 is called n factorial, and is denoted n!.

2012 Pearson Education, Inc.

For any counting number n, the quantity n factorial is given by

2012 Pearson Education, Inc.

Evaluate each expression.

a) 4! b) (4 – 1)!c)

Solution

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

When finding the total number of ways to arrange a given number of distinct objects, we can use a factorial.

2012 Pearson Education, Inc.

The total number of different ways to arrange n distinct objects is n!.

2012 Pearson Education, Inc.

How many ways can you line up 6 different books on a shelf?

Solution

The number of ways to arrange 6 distinct objects is 6! = 720.

2012 Pearson Education, Inc.

The number of distinguishable arrangements of n objects, where one or more subsets consist of look-alikes (say n1 are of one kind, n2 are of another kind, …, and nk are of yet another kind), is given by

2012 Pearson Education, Inc.

Determine the number of distinguishable arrangements of the letters of the word INITIALLY.

Solution

9 letters total

3 I’s and 2 L’s

2012 Pearson Education, Inc.

- Using Permutations and Combinations

2012 Pearson Education, Inc.

- Permutations
- Combinations
- Guidelines on Which Method to Use

2012 Pearson Education, Inc.

In the context of counting problems, arrangements are often called permutations; the number of permutations of n things taken r at a time is denoted nPr. Applying the fundamental counting principle to arrangements of this type gives

nPr= n(n – 1)(n – 2)…[n – (r – 1)].

2012 Pearson Education, Inc.

The number of permutations, or arrangements, of n distinct things taken r at a time, where rn, can be calculated as

2012 Pearson Education, Inc.

Evaluate each permutation.

a) 5P3b) 6P6

Solution

2012 Pearson Education, Inc.

How many ways can you select two letters followed by three digits for an ID if repeats are not allowed?

Solution

There are two parts:

1. Determine the set of two letters.

2. Determine the set of three digits.

Part 1

Part 2

2012 Pearson Education, Inc.

How many four-digit numbers can be written using the numbers from the set {1, 3, 5, 7, 9} if repetitions are not allowed?

Solution

Repetitions are not allowed and order is important, so we use permutations:

2012 Pearson Education, Inc.

In the context of counting problems, subsets, where order of elements makes no difference, are often called combinations; the number of combinations of n things taken r at a time is denoted nCr.

2012 Pearson Education, Inc.

The number of combinations, or subsets, of n distinct things taken r at a time, where rn, can be calculated as

Note:

2012 Pearson Education, Inc.

Evaluate each combination.

a) 5C3b) 6C6

Solution

2012 Pearson Education, Inc.

Find the number of different subsets of size 3 in the set {m, a, t, h, r, o, c, k, s}.

Solution

A subset of size 3 must have 3 distinct elements, so repetitions are not allowed. Order is not important.

2012 Pearson Education, Inc.

A common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible?

Solution

Repetitions are not allowed and order is not important.

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

- Using Pascal’s Triangle

2012 Pearson Education, Inc.

- Pascal’s Triangle
- Applications

2012 Pearson Education, Inc.

The triangular array on the next slide represents “random walks” that begin at START and proceed downward according to the following rule. At each circle (branch point), a coin is tossed. If it lands heads, we go downward to the left. If it lands tails, we go downward to the right. At each point, left an right are equally likely. In each circle the number of different routes that could bring us to that point are recorded.

2012 Pearson Education, Inc.

START

1

1

1

1

2

1

1

1

3

3

1

4

1

4

6

2012 Pearson Education, Inc.

Another way to generate the same pattern of numbers is to begin with 1s down both diagonals and then fill in the interior entries by adding the two numbers just above the given position. The pattern is shown on the next slide. This unending “triangular array of numbers is called Pascal’s triangle.

2012 Pearson Education, Inc.

row

and so on

2012 Pearson Education, Inc.

The “triangle” possesses may properties. In counting applications, entry number r in row number n is equal to nCr – the number of combinations of n things taken r at a time. The next slide shows part of this correspondence.

2012 Pearson Education, Inc.

row

and so on

2012 Pearson Education, Inc.

A group of seven people includes 3 women and 4 men. If five of these people are chosen at random, how many different samples of five people are possible?

Solution

Since this is really selecting 5 from a set of 7. We can read 7C5 from row 7 of Pascal’s triangle. The answer is 21

2012 Pearson Education, Inc.

Among the 21 possible samples of five people in the last example, how many of them would consist of exactly 2 women?

Solution

To select the women (2), we have 3C2 ways. To select the men (3), we have 4C3 ways. This gives a total of

2012 Pearson Education, Inc.

If six fair coins are tossed, in how many different ways could exactly four heads be obtained?

Solution

There are various “ways” of obtaining exactly four heads because the four heads can occur on different subsets of coins. The answer is the number of size-four subsets of a size-six subset. This answer is from row 6 of Pascal’s triangle:

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

- Counting Problems Involving “Not” and “Or”

2012 Pearson Education, Inc.

- Problems Involving “Not”
- Problems Involving “Or”

2012 Pearson Education, Inc.

The counting techniques in this section, which can be thought of as indirect techniques, are based on some correspondences between set theory, logic, and arithmetic as shown on the next slide.

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

Suppose U is the set of all possible results of some type. Let A be the set of all those results that satisfy a given condition. The figure below suggests that

U

A

2012 Pearson Education, Inc.

The number of ways a certain condition can be satisfied is the total number of possible results minus the number of ways the condition would not be satisfied. Symbolically, if A is any set within the universal set U, then

2012 Pearson Education, Inc.

For the set S = {c, a, l, u, t, o, r}, find the number of proper subsets.

Solution

A proper subset has less than seven elements. Subsets of many sizes would satisfy this condition. It is easier to consider the one subset that is not proper, namely S itself. S has a total of 27 = 128 subsets. From the complements principle, the number of proper subsets is 128 – 1 = 127.

2012 Pearson Education, Inc.

If five fair coins are tossed, in how many ways can at least one tail be obtained?

Solution

By the fundamental counting principle, there are 25 = 32 different results possible. Exactly one of these fails to satisfy “at least one tail.” So from the complement principle we have the answer: 32 – 1 = 31.

2012 Pearson Education, Inc.

Another technique to count indirectly is to count the elements of a set by breaking that set into simpler component parts. If the cardinal number formula (Section 2.4) says to find the number of elements in S by adding the number in A to the number in B. We must then subtract the number in the intersection if A and B are not disjoint. If A and B are disjoint, the subtraction is not necessary.

2012 Pearson Education, Inc.

The number of ways that one or the other of two conditions could be satisfied is the number of ways one of them could be satisfied plus the number of ways the other could be satisfied minus the number of ways they could both be satisfied together.

If A and B are any two sets, then

If A and B are disjoint, then

2012 Pearson Education, Inc.

How many five-card hands consist of all hearts or all black cards?

Solution

The sets all hearts and all black cards are disjoint. n(all hearts or all black cards)

= n(all hearts) + n(all black cards)

= 13C5 + 26C5

= 1,287 + 65,780 = 67,067

2012 Pearson Education, Inc.

A diplomatic delegation of 20 congressional members are categorized as to political party and gender. If one of the members is chosen randomly to be spokesperson for the group, in how many ways could that person be a Democrat or a man?

2012 Pearson Education, Inc.

Solution

2012 Pearson Education, Inc.

How many three-digit counting numbers are multiples of 2 or multiples of 5?

Solution

There are 9(10)(5) = 450 three-digit multiples of 2. A multiple of 5 ends in a 0 or 5, so there are 9(10)(2) = 180 of those. A multiple of 2 and 5 must end in a 0. There are 9(10)(1) = 90 of those.

So we have 450 + 180 – 90 = 540 three-digit counting numbers that are multiples of 2 or 5.

2012 Pearson Education, Inc.

A single card is drawn from a standard 52-card deck. a) In how many ways could it be a club or a queen?

b) In how many ways could it be a red card or a face

card?

Solution

a) club + queen – queen of clubs = 13 + 4 – 1 = 16?

b) red card + face card – red face cards

= 26 + 12 – 6 = 32

2012 Pearson Education, Inc.

- 11.1 Basic Concepts
- 11.2 Events Involving “Not” and “Or”
- 11.3 Conditional Probability; Events Involving “And”
- 11.4 Binomial Probability
- 11.5Expected Value

2012 Pearson Education, Inc.

- Basic Concepts

2012 Pearson Education, Inc.

- Historical Background
- Probability
- The Law of Large Numbers
- Probability in Genetics
- Odds

2012 Pearson Education, Inc.

Much of the early work in probability concerned games and gambling. One of the first to apply probability to matters other than gambling was Pierre Simon de Laplace, who is often credited with being the “father” of probability theory. In the twentieth century a coherent mathematical theory of probability was developed through people such as Chebyshev, Markov, and Kolmogorov.

2012 Pearson Education, Inc.

The study of probability is concerned with random phenomena. Even though we cannot be certain whether a given result will occur, we often can obtain a good measure of its likelihood, or probability.

2012 Pearson Education, Inc.

In the study of probability, any observation, or measurement, of a random phenomenon is an experiment. The possible results of the experiment are called outcomes, and the set of all possible outcomes is called the sample space.

Usually we are interested in some particular collection of the possible outcomes. Any such subset of the sample space is called an event.

2012 Pearson Education, Inc.

If a single fair coin is tossed, find the probability that it will land heads up.

Solution

The sample space S = {h, t}, and the event whose probability we seek is E = {h}.

P(heads) = P(E) = 1/2.

Since no coin flipping was actually involved, the desired probability was obtained theoretically.

2012 Pearson Education, Inc.

If all outcomes in a sample space S are equally likely, and E is an event within that sample space, then the theoretical probability of the event E is given by

2012 Pearson Education, Inc.

A cup is flipped 100 times. It lands on its side 84 times, on its bottom 6 times, and on its top 10 times. Find the probability that it will land on its top.

Solution

From the experiment it appears that

P(top) = 10/100 = 1/10.

This is an example of experimental, or empirical probability.

2012 Pearson Education, Inc.

If E is an event that may happen when an experiment is performed, then the empirical probability of event E is given by

2012 Pearson Education, Inc.

There are 2,598,960 possible hands in poker. If there are 36 possible ways to have a straight flush, find the probability of being dealt a straight flush.

Solution

2012 Pearson Education, Inc.

A school has 820 male students and 835 female students. If a student from the school is selected at random, what is the probability that the student would be a female?

Solution

2012 Pearson Education, Inc.

As an experiment is repeated more and more times, the proportion of outcomes favorable to any particular event will tend to come closer and closer to the theoretical probability of that event.

2012 Pearson Education, Inc.

Gregor Mendel, an Austrian monk used the idea of randomness to establish the study of genetics. To study the flower color of certain pea plants he found that: Pure red crossed with pure white produces red.

Mendel theorized that red is “dominant” (symbolized by R), while white is recessive (symbolized by r). The pure red parent carried only genes for red (R), and the pure white parent carried only genes for white (r).

2012 Pearson Education, Inc.

Every offspring receives one gene from each parent which leads to the tables below. Every second generation is red because R is dominant.

1st to 2nd Generation

2nd to 3rd Generation

offspring

offspring

2012 Pearson Education, Inc.

Referring to the 2nd to 3rd generation table (previous slide), determine the probability that a third generation will be

a) redb) white

Base the probability on the sample space of equally likely outcomes: S = {RR, Rr, rR, rr}.

2012 Pearson Education, Inc.

Solution

a)Since red dominates white, any combination with R will be red. Three out of four have an R, so P(red) = 3/4.

b)Only one combination rr has no gene for red, so P(white) = 1/4.

2012 Pearson Education, Inc.

Odds compare the number of favorable outcomes to the number of unfavorable outcomes. Odds are commonly quoted in horse racing, lotteries, and most other gambling situations.

2012 Pearson Education, Inc.

If all outcomes in a sample space are equally likely, a of them are favorable to the event E, and the remaining b outcomes are unfavorable to E, then the oddsin favor of E are a to b, and the odds against E are b to a.

2012 Pearson Education, Inc.

200 tickets were sold for a drawing to win a new television. If Matt purchased 10 of the tickets, what are the odds in favor of Matt’s winning the television?

Solution

Matt has 10 chances to win and 190 chances to lose. The odds in favor of winning are 10 to 190, or 1 to 19.

2012 Pearson Education, Inc.

Suppose the probability of rain today is .43. Give this information in terms of odds.

Solution

We can say that

43 out of 100 outcomes are favorable, so 100 – 43 = 57 are unfavorable. The odds in favor of rain are 43 to 57 and the odds against rain are 57 to 43.

2012 Pearson Education, Inc.

Your odds of completing College Algebra class are 16 to 9. What is the probability that you will complete the class?

Solution

There are 16 favorable outcomes and 9 unfavorable. This gives 25 possible outcomes. So

2012 Pearson Education, Inc.

- Events Involving “Not” and “Or”

2012 Pearson Education, Inc.

- Properties of Probability
- Events Involving “Not”
- Events Involving “Or”

2012 Pearson Education, Inc.

Let E be an event from the sample space S. That is, E is a subset of S. Then the following properties hold.

(The probability of an event is between 0 and 1, inclusive.)

(The probability of an impossible event is 0.)

(The probability of a certain event is 1.)

2012 Pearson Education, Inc.

When a single fair die is rolled, find the probability of each event.

a) the number 3 is rolled

b) a number other than 3 is rolled

c) the number 7 is rolled

d) a number less than 7 is rolled

2012 Pearson Education, Inc.

Solution

The outcome for the die has six possibilities: {1, 2, 3, 4, 5, 6}.

2012 Pearson Education, Inc.

The table on the next slide shows the correspondences that are the basis for the probability rules developed in this section. For example, the probability of an event not happening involves the complement and subtraction.

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

The probability that an event E will not occur is equal to one minus the probability that it will occur.

E

S

So we have

and

2012 Pearson Education, Inc.

When a single card is drawn from a standard 52-card deck, what is the probability that is will not be an ace?

Solution

2012 Pearson Education, Inc.

Probability of one event or another should involve the union and addition.

2012 Pearson Education, Inc.

Two events A and B are mutually exclusive events if they have no outcomes in common. (Mutually exclusive events cannot occur simultaneously.)

2012 Pearson Education, Inc.

If A and B are any two events, then

If A and B are mutually exclusive, then

2012 Pearson Education, Inc.

When a single card is drawn from a standard 52-card deck, what is the probability that it will be a king or a diamond?

Solution

2012 Pearson Education, Inc.

If a single die is rolled, what is the probability of a 2 or odd?

Solution

These are mutually exclusive events.

2012 Pearson Education, Inc.

- 11.1 Basic Concepts
- 11.2 Events Involving “Not” and “Or”
- 11.3 Conditional Probability; Events Involving “And”
- 11.4 Binomial Probability
- 11.5Expected Value

2012 Pearson Education, Inc.

- Conditional Probability; Events Involving “And”

2012 Pearson Education, Inc.

- Conditional Probability
- Events Involving “And”

2012 Pearson Education, Inc.

Sometimes the probability of an event must be computed using the knowledge that some other event has happened (or is happening, or will happen – the timing is not important). This type of probability is called conditional probability.

2012 Pearson Education, Inc.

The probability of event B, computed on the assumption that event A has happened, is called the conditional probability of B, given A, and is denoted P(B | A).

2012 Pearson Education, Inc.

From the sample space S = {2, 3, 4, 5, 6, 7, 8, 9}, a single number is to be selected randomly. Given the events

A: selected number is odd, and

B selected number is a multiple of 3.

find each probability.

a) P(B)

b) P(A and B)

c) P(B | A)

2012 Pearson Education, Inc.

Solution

a) B = {3, 6, 9}, so P(B) = 3/8

b) P(A and B) = {3, 5, 7, 9} {3, 6, 9} = {3, 9}, so

P(A and B) = 2/8 = 1/4

c) The given condition A reduces the sample space

to {3, 5, 7, 9}, so P(B | A) = 2/4 = 1/2

2012 Pearson Education, Inc.

The conditional probability of B, given A, and is given by

2012 Pearson Education, Inc.

Given a family with two children, find the probability that both are boys, given that at least one is a boy.

Solution

Define S = {gg, gb, bg, bb}, A = {gb, bg, bb}, and B = {bb}.

2012 Pearson Education, Inc.

Two events A and B are called independent events if knowledge about the occurrence of one of them has no effect on the probability of the other one, that is, if

P(B | A) = P(B), or equivalently

P(A | B) = P(A).

2012 Pearson Education, Inc.

A single card is to be drawn from a standard 52-card deck. Given the events

A: the selected card is an ace

B: the selected card is red

a) Find P(B).

b) Find P(B | A).

c) Determine whether events A and B are independent.

2012 Pearson Education, Inc.

Solution

c. Because P(B | A) = P(B), events A and B are independent.

2012 Pearson Education, Inc.

If we multiply both sides of the conditional probability formula by P(A), we obtain an expression for P(A and B). The calculation of P(A and B) is simpler when A and B are independent.

2012 Pearson Education, Inc.

If A and B are any two events, then

If A and B are independent, then

2012 Pearson Education, Inc.

Jeff draws balls from the jar below. He draws two balls without replacement. Find the probability that he draws a red ball and then a blue ball, in that order.

4 red

3 blue

2 yellow

2012 Pearson Education, Inc.

Solution

2012 Pearson Education, Inc.

Jeff draws balls from the jar below. He draws two balls, this time with replacement. Find the probability that he gets a red and then a blue ball, in that order.

4 red

3 blue

2 yellow

2012 Pearson Education, Inc.

Solution

Because the ball is replaced, repetitions are allowed. In this case, event B2 is independent of R1.

2012 Pearson Education, Inc.

- Binomial Probability

2012 Pearson Education, Inc.

- Binomial Probability Distribution
- Binomial Probability Formula

2012 Pearson Education, Inc.

The spinner below is spun twice and we are interested in the number of times a 2 is obtained (assume each sector is equally likely).

Think of outcome 2 as a “success” and outcomes 1 and 3 as “failures.” The sample space is

2

1

3

S = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2),

(2, 3), (3, 1), (3, 2), (3, 3)}.

2012 Pearson Education, Inc.

When the outcomes of an experiment are divided into just two categories, success and failure, the associated probabilities are called “binomial.” Repeated trials of the experiment, where the probability of success remains constant throughout all repetitions, are also known as Bernoulli trials.

2012 Pearson Education, Inc.

If x denotes the number of 2s occurring on each pair of spins, then x is an example of a random variable. In S, the number of 2s is 0 in four cases, 1 in four cases, and 2 in one case. Because the table on the next slide includes all the possible values of x and their probabilities it is an example of a probability distribution. In this case, it is a binomial probability distribution.

2012 Pearson Education, Inc.

2012 Pearson Education, Inc.

In general, let

n = the number of repeated trials,

p = the probability of success on any given trial,

q = 1 – p = the probability of failure on any given trial,

andx = the number of successes that occur.

Note that p remains fixed throughout all n trials. This means that all trials are independent. In general, x, successes can be assigned among n repeated trials in nCxdifferent ways.

2012 Pearson Education, Inc.

When n independent repeated trials occur, where

p = probability of success and

q = probability of failure

with p and q (where q = 1 – p) remaining constant throughout all n trials, the probability of exactly x successes is given by

2012 Pearson Education, Inc.

Find the probability of obtaining exactly three heads in five tosses of a fair coin.

Solution

This is a binomial experiment with n = 5, p = 1/2, q = 1/2, and x = 3.

2012 Pearson Education, Inc.

Find the probability of obtaining exactly two 3’s in six rolls of a fair die.

Solution

This is a binomial experiment with n = 6, p = 1/6, q = 5/6, and x = 2.

2012 Pearson Education, Inc.

Find the probability of obtaining less than two 3’s in six rolls of a fair die.

Solution

We have n = 6, p = 1/6, q = 5/6, and x < 2.

2012 Pearson Education, Inc.

A baseball player has a well-established batting average of .250. In the next series he will bat 10 times. Find the probability that he will get more than two hits.

Solution

In this case n = 10, p = .250, q = .750, and x > 2.

2012 Pearson Education, Inc.

Solution(continued)

2012 Pearson Education, Inc.

- Expected Value

2012 Pearson Education, Inc.

- Expected Value
- Games and Gambling
- Investments
- Business and Insurance

2012 Pearson Education, Inc.

Children in third grade were surveyed and told to pick the number of hours that they play electronic games each day. The probability distribution is given below.

2012 Pearson Education, Inc.

Compute a “weighted average” by multiplying each possible time value by its probability and then adding the products.

1.1 hours is the expected value (or the mathematical expectation) of the quantity of time spent playing electronic games.

2012 Pearson Education, Inc.

If a random variable x can have any of the values x1, x2 , x3 ,…, xn, and the corresponding probabilities of these values occurring are

P(x1), P(x2), P(x3), …, P(xn), then the expected value of xis given by

2012 Pearson Education, Inc.

Find the expected number of boys for a three-child family. Assume girls and boys are equally likely.

Solution

S = {ggg, ggb, gbg, bgg, gbb, bgb, bbg, bbb}

The probability distribution is on the right.

2012 Pearson Education, Inc.

Solution(continued)

The expected value is the sum of the third column:

So the expected number of boys is 1.5.

2012 Pearson Education, Inc.

A player pays $3 to play the following game: He rolls a die and receives $7 if he tosses a 6 and $1 for anything else. Find the player’s expected net winnings for the game.

2012 Pearson Education, Inc.

Solution

The information for the game is displayed below.

Expected value: E(x) = –$6/6 = –$1.00

2012 Pearson Education, Inc.

A game in which the expected net winnings are zero is called a fair game. A game with negative expected winnings is unfair against the player. A game with positive expected net winnings is unfair in favor of the player.

2012 Pearson Education, Inc.

What should the game in the previous example cost so that it is a fair game?

Solution

Because the cost of $3 resulted in a net loss of $1, we can conclude that the $3 cost was $1 too high. A fair cost to play the game would be $3 – $1 = $2.

2012 Pearson Education, Inc.

Expected value can be a useful tool for evaluating investment opportunities.

2012 Pearson Education, Inc.

Mark is going to invest in the stock of one of the two companies below. Based on his research, a $6000 investment could give the following returns.

2012 Pearson Education, Inc.

Find the expected profit (or loss) for each of the two stocks.

Solution

ABC: –$400(.2) + $800(.5) + $1300(.3) = $710

PDQ: $600(.8) + $1000(.2) = $680

2012 Pearson Education, Inc.

Expected value can be used to help make decisions in various areas of business, including insurance.

2012 Pearson Education, Inc.

A lumber wholesaler is planning on purchasing a load of lumber. He calculates that the probabilities of reselling the load for $9500, $9000, or $8500 are .25, .60, and .15, respectfully. In order to ensure an expected profit of at least $2500, how much can he afford to pay for the load?

2012 Pearson Education, Inc.

Solution

The expected revenue from sales can be found below.

Expected revenue: E(x) = $9050

2012 Pearson Education, Inc.

Solution(continued)

profit = revenue – cost or cost = profit – revenue

To have an expected profit of $2500, he can pay up to $9050 – $2500 = $6550.

2012 Pearson Education, Inc.