Chapter 4 Probability 4-1 Review and Preview

Chapter 4 Probability 4-1 Review and Preview
4-2 Basic Concepts of Probability 4-3 Addition Rule 4-4 Multiplication Rule: Basics 4-5 Multiplication Rule: Complements and Conditional Probability 4-6 Counting

Section 4-1 Review and Preview

Review Necessity of sound sampling methods.
Common measures of characteristics of data Mean Standard deviation

Rare Event Rule for Inferential Statistics:
Preview Rare Event Rule for Inferential Statistics: If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct. Statisticians use the rare event rule for inferential statistics.

Basic Concepts of Probability
Section 4-2 Basic Concepts of Probability

Part 1 Basics of Probability

Events and Sample Space
any collection of results or outcomes of a procedure Simple Event an outcome or an event that cannot be further broken down into simpler components Sample Space for a procedure consists of all possible simple events; that is, the sample space consists of all outcomes that cannot be broken down any further

Example A pair of dice are rolled. The sample space has 36 simple events: 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. Which elements of the sample space correspond to the event that the sum of each dice is 4?

Example Which elements of the sample space correspond to the event that the sum of each dice is 4? ANSWER: 3,1 2,2 1,3

P - denotes a probability. A, B, and C - denote specific events.
Notation for Probabilities P - denotes a probability. A, B, and C - denote specific events. P(A) - denotes the probability of event A occurring.

Computing Probability
Basic Rules for Computing Probability Rule 1: Relative Frequency Approximation of Probability Conduct (or observe) a procedure, and count the number of times event A actually occurs. Based on these actual results, P(A) is approximated as follows: P(A) = # of times A occurred # of times procedure was repeated

Example Problem 20 on page 149
F = event of a false negative on polygraph test Thus this is not considered unusual since it is more than (see page 146). The test is not highly accurate.

Rounding Off Probabilities
When expressing the value of a probability, either give the exact fraction or decimal or round off final decimal results to three significant digits. All digits are significant except for the zeros that are included for proper placement of the decimal point. Example: has four significant digits has two significant digits

F = event of a selecting a female senator NOTE: total number of senators=100 Thus this does not agree with the claim that men and women have an equal (50%) chance of being selected as a senator.

A = event that Delta airlines passenger is involuntarily bumped from a flight Thus this is considered unusual since it is less than 0.05 (see directions on page 149). Since probability is very low, getting bumped from a flight on Delta is not a serious problem.

P(A) = Basic Rules for Computing Probability
Rule 2: Classical Approach to Probability (Requires Equally Likely Outcomes) Assume that for a given procedure each simple event has an equal chance of occurring. P(A) = number of ways A can occur number of different simple events in the sample space

Example What is the probability of rolling two die and getting a sum of 4? A = event that sum of the dice is 4 Assume each number is equally likely to be rolled on the die. Rolling a sum of 4 can happen in one of three ways (see previous slide) with 36 simple events so:

Example What is the probability of getting no heads when a “fair” coin is tossed three times? (A fair coin has an equal probability of showing heads or tails when tossed.) A = event that no heads occurs in three tosses Sample Space (in order of toss):

Example Sample space has 8 simple events.
Event A corresponds to TTT only so that:

Example Problem 36 on page 151 Let:
S = event that son inherits disease (xY or Yx) D = event that daughter inherits disease (xx)

Example Problem 36 on page 151 (a) Father: xY Mother: XX
Sample space for a son: YX YX Sample space has no simple events that represent a son that has the disease so:

Example Problem 36 on page 151 (b) Father: xY Mother: XX
Sample space for a daughter: xX xX Sample space has no simple events that represent a daughter that has the disease so:

Example Problem 36 on page 151 (c) Father: XY Mother: xX
Sample space for a son: Yx YX Sample space has one simple event that represents a son that has the disease so:

Example Problem 36 on page 151 (d) Father: XY Mother: xX
Sample space for a daughter: Xx XX Sample space has no simple event that represents a daughter that has the disease so:

Example Problem 18 on page 149 Table 4-1 on page 137 (polygraph data)
Did Not Lie Did Lie Positive Test Result 15 (false positive) 42 (true positive) NegativeTest Result 32 (true negative) 9 (false negative)

Example It is helpful to first total the data in the table:
(a) How many responses were lies: ANSWER: 51 Did Not Lie Did Lie TOTALS Positive Test Result 15 (false positive) 42 (true positive) 57 NegativeTest Result 32 (true negative) 9 (false negative) 41 47 51 98

Example If one response is randomly selected, what is the probability it is a lie? L = event of selecting one of the lie responses (c)

Computing Probability - continued
Basic Rules for Computing Probability - continued Rule 3: Subjective Probabilities P(A), the probability of event A, is estimated by using knowledge of the relevant circumstances.

Probability should be high based on experience (it is rare to be delayed because of an accident). Guess: P=99/100 (99 out of 100 times you will not be delayed because of an accident) ANSWERS WILL VARY

Example: Classical probability predicts the probability of flipping a (non-biased) coin and it coming up heads is ½=0.5 Ten coin flips will sometimes result in exactly 5 heads and a frequency probability of heads 5/10=0.5; but often you will not get exactly 5 heads in ten flips.

Law of Large Numbers As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability. Example: If we flip a coin 1 million times the frequency probability should be approximately 0.5

Probability Limits Always express a probability as a fraction or decimal number between 0 and 1. The probability of an impossible event is 0. The probability of an event that is certain to occur is 1. For any event A, the probability of A is between 0 and 1 inclusive. That is: 0  P(A)  1

Possible Values for Probabilities

Complementary Events The complement of event A, denoted by A, consists of all outcomes in which the event A does not occur.

Example If a fair coin is tossed three times and
A = event that exactly one heads occurs Find the complement of A.

Example Sample space: Event A corresponds to HTT, THT, TTH
Therefore, the complement of A are the simple events: HHH, HHT, HTH, THH, TTT

Beyond the Basics of Probability: Odds
Part 2 Beyond the Basics of Probability: Odds

payoff odds against event A = (net profit) : (amount bet)
The actual odds in favor of event A occurring are the ratio P(A)/ P(A), usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors. The actual odds against event A occurring are the ratio P(A)/P(A), which is the reciprocal of the actual odds in favor of the event. If the odds in favor of A are a:b, then the odds against A are b:a. The payoff odds against event A occurring are the ratio of the net profit (if you win) to the amount bet. payoff odds against event A = (net profit) : (amount bet)

Example W = simple event that you win due to an odd number
Problem 38, page 149 W = simple event that you win due to an odd number Sample Space 00, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36 (a) There are 18 odd numbers so that P(W) = 18/38

Example Problem 38, page 149 There are 20 events that correspond to winning from a number that is not odd (i.e. you do not win due to an odd number) so: (b) Odds against winning are

Example (c) Payoff odds against winning are 1:1
Problem 38, page 149 (c) Payoff odds against winning are 1:1 That is, $1 net profit for every $1 bet Thus, if you bet $18 and win, your net profit is $18 which can be found by solving the proportion: The casino returns $18+$18=$36 to you.

Example (d) Actual odds against winning are 10:9
Problem 38, page 149 (d) Actual odds against winning are 10:9 That is, $10 net profit for every $9 bet Thus, if you bet $18 and win, your net profit is $20 which can be found by solving the proportion: The casino returns $18+$20=$38 to you.

Recap In this section we have discussed:
Rare event rule for inferential statistics. Probability rules. Law of large numbers. Complementary events. Rounding off probabilities. Odds.

Section 4-3 Addition Rule

Key Concept This section presents the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of the procedure. The key word in this section is “or.” It is the inclusive or, which means either one or the other or both.

any event combining 2 or more simple events
Compound Event Compound Event any event combining 2 or more simple events Notation P(A or B) = P (in a single trial, event A occurs or event B occurs or they both occur)

General Rule for a Compound Event
When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find that total in such a way that no outcome is counted more than once.

Example A random survey of members of the class of 2005 finds the following: What is the probability the student did not graduate or was a man? Number of students who graduated Number of students who did not graduate TOTALS Women 672 22 694 Men 582 19 601 1254 41 1295

Example Method 1: directly add up those who did not graduate and those who are men (without counting men twice): = 623 Method 2: add up the total number who did not graduate and the total number of men, then subtract the double count of men: = 623

Example The probability the student did not graduate or was a man :
623/1295 = 0.481

P(A or B) = P(A) + P(B) – P(A and B)
Compound Event Formal Addition Rule P(A or B) = P(A) + P(B) – P(A and B) where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure.

Previous Example N = event that student did not graduate
M = event that student was a man P(N) = 41/1295 = P(M) = 601/1295 = 0.464 P(N and M) = 19/1295 = P(N or M) = P(N) + P(M) - P(N and M) = = 0.481

What is the probability the subject had a negative test result or lied? Did Not Lie Did Lie TOTALS Positive Test Result 15 (false positive) 42 (true positive) 57 NegativeTest Result 32 (true negative) 9 (false negative) 41 47 51 98

Example (cont.) N = event that there is a negative test result
L = event that the subject lied P(N) = 41/98 = 0.418 P(L) = 51/98 = 0.520 P(N and L) = 9/98 = P(N or L) = P(N) + P(L) - P(N and L) = = 0.846

Example Problem 30 on page 158 N=event that person refuses to respond
F=event that person’s age is 60 or older Note: total number of people in the study is 1205 total number who refused to respond is 156 so: P(N)=156/1205 total number who are 60 or older is 251 so: P(F)=251/1205 total number who refuse to respond and 60 or older is 49 so: P(N and F)=49/1205 P(N OR F) = 156/ /1205 – 49/1205 = 358/1205 = 0.297

Disjoint or Mutually Exclusive
Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.) Previous Example: G = students who graduated N = students who did not graduate N and G are disjoint because no student who did not graduate also graduated

Disjoint or Mutually Exclusive
Events A and B are not disjoint if they overlap. Previous Example: M = male students G = students who graduated M and G are not disjoint because some students who graduated are also male

Example Problem 10 and 12 on page 157
These are disjoint since a subject treated with Lipitor cannot be a subject given no medication. 12. These are not disjoint since it is possible for a homeless person to be a college graduate.

Venn Diagram A Venn diagram is a way to picture how sets overlap.
Venn Diagram for Events That Are Not Disjoint and overlap. Venn Diagram for Disjoint Events which do not overlap.

Complementary Events It is impossible for an event and its complement to occur at the same time. That is, for any event A, are disjoint

Probability Rule of Complementary Events
P(A) + P(A) = 1 = 1 – P(A) P(A) = 1 – P(A) P(A)

Venn Diagram for the Complement of Event A

is the probability that a screened driver is not intoxicated

Recap In this section we have discussed: Compound events.
Formal addition rule. Intuitive addition rule. Disjoint events. Complementary events.

Multiplication Rule: Basics
Section 4-4 Multiplication Rule: Basics

Tree Diagrams A tree diagram is a picture of the possible outcomes of a procedure, shown as line segments emanating from one starting point. These diagrams are sometimes helpful in determining the number of possible outcomes in a sample space, if the number of possibilities is not too large.

Tree Diagrams This figure summarizes the possible outcomes
for a true/false question followed by a multiple choice question. Note that there are 10 possible combinations.

Example: Tree Diagrams
A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag and after the first marble is drawn, it is put back into the bag before the second marble is drawn. Construct a tree diagram that depicts all possible outcomes.

Example: Computing Probability withTree Diagrams
Use the previous example of drawing two marbles with replacement to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.

P(Y) = Example: Computing Probability withTree Diagrams
Y = event of drawing red first then blue P(Y) = number of ways Y can occur number of different simple events in the sample space

Consider each event separately: R = event of drawing red on first draw B = event of drawing blue on second draw

Notation P(A and B) = P(event A occurs in a first trial and
event B occurs in a second trial) To differentiate between the addition rule and the multiplication rule, the addition rule will use the word ‘or’ - P(A or B) - , and the multiplication rule will use the word ‘and’ - P(A and B). Page 159 of Elementary Statistics, 10th Edition

Key Concept The basic multiplication rule is used for finding P(A and B), the probability that event A occurs in a first trial and event B occurs in a second trial.

The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:

Key Concept NEXT EXAMPLE SHOWS:
If the outcome of the first event A somehow affects the probability of the second event B, it is important to adjust the probability of B to reflect the occurrence of event A. This is Conditional Probability

A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag without replacing the first marble after it is drawn. Construct a tree diagram that depicts all possible outcomes.

Use the previous example to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.

P(Y) = Example: Computing Probability withTree Diagrams
Y = event of drawing red first then blue P(Y) = number of ways Y can occur number of different simple events in the sample space

Multiplication Method R = event of drawing red on first draw B = event of drawing blue on second draw (given that there are now only two marbles in the bag)

The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:

Conditional Probability
Key Point Without replacement, we must adjust the probability of the second event to reflect the outcome of the first event.

Important Principle The probability for the second event B should take into account the fact that the first event A has already occurred.

Notation for Conditional Probability
P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B|A as “B given A.”)

Multiplication Rule and Conditional Probability
P(A and B) = P(A) • P(B|A)

Intuitive Multiplication Rule
When finding the probability that event A occurs in one trial and event B occurs in the next trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B takes into account the previous occurrence of event A. Many students find that they understand the multiplication rule more intuitively than by using the formal rule and formula. Example on page 162 of Elementary Statistics, 10th Edition

If three are selected without replacement, what is probability they all had false positive test results? Did Not Lie Did Lie TOTALS Positive Test Result 15 (false positive) 42 (true positive) 57 NegativeTest Result 32 (true negative) 9 (false negative) 41 47 51 98

On first selection there are 98 subjects, 15 of which are false positive:

On second selection there are 97 subjects (without replacement) If the first selection was false positive, there are 14 false positives left:

On third selection there are 96 subjects (without replacement) If the first and second selections were false positive, there are 13 false positives left:

Example Problem 14 on page 168 ANSWER:
This is considered unusual since probability is less than 0.05

Dependent and Independent
Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are similarly independent if the occurrence of any does not affect the probabilities of the occurrence of the others.) If A and B are not independent, they are said to be dependent. Page 162 of Elementary Statistics, 10th Edition

Examples Problem 10 on page 168 Finding that your calculator works
Finding that your computer works Assuming your calculator and your computer are not both running off the same source of power, these are independent events.

Examples Previous example of drawing two marbles with replacement
Since the probability of drawing the second marble is not affected by drawing the first marble, these are independent events.

Dependent Events Two events are dependent if the occurrence of one of them affects the probability of the occurrence of the other, but this does not necessarily mean that one of the events is a cause of the other. Page 162 of Elementary Statistics, 10th Edition

Examples Previous example of drawing two marbles without replacement
Since the probability of drawing the second marble is affected by drawing the first marble, these are dependent events.

Multiplication Rule for Independent Events
Note that if A and B are independent events, then P(B|A)=P(B) and the multiplication rule is then: P(A and B) = P(A) • P(B)

Applying the Multiplication Rule

Caution When applying the multiplication rule, always consider whether the events are independent or dependent, and adjust the calculations accordingly.

Multiplication Rule for Several Events
In general, the probability of any sequence of independent events is simply the product of their corresponding probabilities.

Example Page 169, problem 26 These events are independent since the probability of getting a girl on any try is not affected by the occurence of getting a girl on a previous try.

Example Page 169, problem 26 ten factors of 1/2

Example Page 169, problem 26 Calculator: use “hat key” to evaluate powers:

Example Page 169, problem 26 Since: We see that getting 10 girls by chance alone is unusual and conclude that the gender selection method is effective. Calculator: use “hat key” to evaluate powers:

Treating Dependent Events as Independent
Some calculations are cumbersome, but they can be made manageable by using the common practice of treating events as independent when small samples are drawn from large populations. In such cases, it is rare to select the same item twice (sample with replacement). Since independent events where the probability is the same for each event is much easier to compute (using exponential notation) than dependent events where each probability fraction will be different, the rule for small samples and large population is often used by pollsters. Page 163 of Elementary Statistics, 10th Edition

The 5% Guideline for Cumbersome Calculations
If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent). Since independent events where the probability is the same for each event is much easier to compute (using exponential notation) than dependent events where each probability fraction will be different, the rule for small samples and large population is often used by pollsters. Page 163 of Elementary Statistics, 10th Edition

Example Page 169, problem 30 a) If we select without replacement, then randomly selecting an ignition system are not independent. But since 3/200 = =1.5%, we could use the 5% guideline and regard these events as independent.

Example Page 169, problem 30 b) If these events are not independent (dependent) then:

Example Page 169, problem 30 c) If these events are independent then:

Example Page 169, problem 30 d) The answer from part (b) is exact so it is better.

Summary of Fundamentals
In the addition rule, the word “or” in P(A or B) suggests addition. Add P(A) and P(B), being careful to add in such a way that every outcome is counted only once. In the multiplication rule, the word “and” in P(A and B) suggests multiplication. Multiply P(A) and P(B), but be sure that the probability of event B takes into account the previous occurrence of event A.

Recap In this section we have discussed: Notation for P(A and B).
Tree diagrams. Notation for conditional probability. Independent events. Formal and intuitive multiplication rules.

Complements and Conditional Probability
Section 4-5 Multiplication Rule: Complements and Conditional Probability

Key Concepts Probability of “at least one”: Find the probability that among several trials, we get at least one of some specified event. Conditional probability: Find the probability of an event when we have additional information that some other event has already occurred.

Complements: The Probability of “At Least One”
“At least one” is equivalent to “one or more.” The complement of getting at least one item of a particular type is that you get no items of that type.

Example Page 175, problems 6 If not all 6 are free from defects, that means that at least one of them is defective.

Example Page 175, problems 8 If it is not true that at least one of the five accepts an invitation, then all five did not accept the invitation.

Finding the Probability of “At Least One”
To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is, P(at least one) = 1 – P(none).

Example Page 175, problem 10 P(at least one girl) = 1 – P(all boys)

Example Page 175, problem 10 The probability of having 8 children and none of them are girls (all boys) is which means this is a rare event.

Example Page 175, problem 12 P(at least one working calculator)
= 1 – P(all calculators fail) Note: these are independent events and P(a calculator fails) = 1 – P(a calculator does not fail)

Example Page 175, problem 12

Example Page 175, problem 12 With one calculator,
P(working calculator) = 0.96 With two calculators, P(at least one working calculator) = 0.998 The increase in chance of a working calculator might be worth the effort.

A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred. P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred.

Intuitive Approach to Conditional Probability
The conditional probability of B given A can be found by assuming that event A has occurred, and then calculating the probability that event B will occur.

Example Problem 22 on page 175 Table 4-1 on page 137 (polygraph data)
Did Not Lie Did Lie Positive Test Result 15 (false positive) 42 (true positive) NegativeTest Result 32 (true negative) 9 (false negative)

Example Page 175, problem 22 There are 47 subjects who did not lie, 32 of which had a negative test result.

Conditional Probability Formula
Conditional probability of event B occurring, given that event A has already occurred P(B A) = P(A and B) P(A)

Example Page 175, problem 22 Using the formula

Confusion of the Inverse
To incorrectly believe that P(A|B) and P(B|A) are the same, or to incorrectly use one value for the other, is often called confusion of the inverse.

Example Page 175, problem 22 (b) Using the formula

Example Page 175, problem 22 (c) The results from parts (a) and (b) are not equal.

Recap In this section we have discussed: Concept of “at least one.”
Conditional probability. Intuitive approach to conditional probability.

Section 4-6 Counting

Key Concept In many probability problems, the big obstacle is finding the total number of outcomes, and this section presents several methods for finding such numbers without directly listing and counting the possibilities.

Fundamental Counting Rule
For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways.

Example A coin is flipped and then a die is rolled. What are the total number of outcomes?

Example A coin is flipped and then a die is rolled. What are the total number of outcomes? ANSWER: There are 2 outcomes for the coin flip and 6 outcomes for the die roll. Total number of outcomes:

Example Page 185, Problem 32 (a)
If 20 newborn babies are randomly selected, how many different gender sequences are possible?

Example Page 185, Problem 32 (a) ANSWER:
This is a sequence of 20 events each of which has two possible outcomes (boy or girl). By the fundamental counting rule, total number of gender sequences will be:

Example Page 185, Problem 28 A safe combination consists of four numbers between 0 and 99. If four numbers are randomly selected, what is the probability of getting the correct combination on the first attempt?

Example Page 185, Problem 28 ANSWER:
Total number of possible combinations is Since there is only one correct combination: It is not feasible to try opening the safe this way.

By special definition, 0! = 1.
Notation The factorial symbol ! denotes the product of decreasing positive whole numbers. For example, By special definition, 0! = 1.

Factorial Rule A collection of n different items can be arranged in order n! different ways. (This factorial rule reflects the fact that the first item may be selected in n different ways, the second item may be selected in n – 1 ways, and so on.)

Example Page 183, Problem 6 Find the number of different ways that the nine players on a baseball team can line up for the National Anthem by evaluating 9 factorial.

Factorial on Calculator
9 MATH PRB 4:! To get: Then Enter gives the result

(when items are all different)
Permutations Rule (when items are all different) Requirements: There are n different items available. (This rule does not apply if some of the items are identical to others.) We select r of the n items (without replacement). We consider rearrangements of the same items to be different sequences. (The permutation of ABC is different from CBA and is counted separately.) If the preceding requirements are satisfied, the number of permutations (or sequences) of r items selected from n available items (without replacement) is (n - r)! n r P = n!

Example A bag contains 4 colored marbles: red, blue, green, yellow. If we select 3 of the four marbles from the bag without replacement, how many different color orders are there?

Example ANSWER:

Example Page 183, Problem 12 In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Find the number of different possible trifecta bets in a race with ten horses by evaluating

Example Page 183, Problem 12 ANSWER:

Permutations on Calculator
10 MATH PRB 2:nPr 3 To get: Then Enter gives the result

(when some items are identical to others)
Permutations Rule (when some items are identical to others) Requirements: There are n items available, and some items are identical to others. We select all of the n items (without replacement). We consider rearrangements of distinct items to be different sequences. If the preceding requirements are satisfied, and if there are n1 alike, n2 alike, nk alike, the number of permutations (or sequences) of all items selected without replacement is n1! . n2! nk! n!

Example Page 185, Problem 26 Find the number of different ways the letters AGGYB can be arranged.

Example Page 185, Problem 26 ANSWER: there are two letters of the five that are alike. Then the total number of arrangements will be

Example Page 185, Problem 32 (b)
How many different ways can 10 girls and 10 boys be arranged in a sequence?

ANSWER: there are twenty total children in the arrangement. There are two groups of 10 that are alike. This gives a total number of arrangements:

Example Page 185, Problem 32 (c)
What is the probability of getting 10 girls and 10 boys when twenty babies are born?

Example Page 185, Problem 32 (c) ANSWER: as found on previous slides
The total number of boy/girl arrangements of 20 newborns is: The total number of ways 10 girls and 10 boys can be arranged in a sequence is

Example Page 185, Problem 32 (c) ANSWER:

Combinations Rule n! nCr = (n - r )! r! Requirements:
There are n different items available. We select r of the n items (without replacement). We consider rearrangements of the same items to be the same. (The combination of ABC is the same as CBA.) If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is (n - r )! r! n! nCr =

Example Page 183, Problem 8 Find the number of different possible 5 card poker hands by evaluating

Example Page 183, Problem 8 ANSWER:

Combinations on Calculator
52 MATH PRB 3:nCr 5 To get: Then Enter gives the result

Example Page 184, Problem 16 (a)
What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31. Note: numbers selected are different and order does not matter.

Example Page 184, Problem 16 (a) ANSWER:
Total number of possible combinations is: Since only one combination wins:

What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31. Note: assume now that you must select the numbers in the same order they were drawn so that order does matter.

Example Page 184, Problem 16 (b) ANSWER:
Total number of possible combinations is: Since only one permutation wins:

Permutations versus Combinations
When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.

Recap In this section we have discussed:
The fundamental counting rule. The factorial rule. The permutations rule (when items are all different). The permutations rule (when some items are identical to others). The combinations rule.

Chapter 4 Probability 4-1 Review and Preview

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