Chapter 1: IntroductionSection 1.2: Combinatorial MethodsTheorem 1.1 p.2If an operation consists of two steps, of which the first can done in n1 ways and the second can be done in n2 ways, then the whole operation can be done in n1 n2 ways.

Theorem 1.2p.4If an operation consists of k steps, of which the first can be done in n1 ways, for each of these the second step can be done in n2 ways, for each of the first two, the third step can be done in n3 ways, and so forth, the whole operation can be done in n1 n2 nk ways.

Example: At a food truck you have a choice of 2 types of bread, 4 kinds of meat and three varieties of cheese.

How many kinds of sandwiches can be made? Make the tree diagram.

Example: In how many ways can five multiple-choice questions be answered if each question has four choices?

Combination for n objects, a combination is the set or collection of those objects. The collection of objects is in no particular order. ABC and CAB would be the same combination of letters.Permutation for n objects, a permutation is an arrangement of the n objects. ABC, ACB, BAC, BCA, CAB, and CBA are the six permutations for the letters A, B, and C. The combination is ABC in no particular order.Theorem 1.3p. 5The number of permutations of n distinct objects is n!.

1. In how many ways can five tasks be assigned to five people, if each person is assigned exactly one task?

2. In how many ways can two members of a committee of six be chosen to fill the positions of chair and secretary?

3. In how many ways can three people be chosen from a group of 12 to be on a committee?

4. In how many ways can we get 6 people to stand in a straight line? In how many ways can they be seated around a table?

Theorem 1.4p. 5The number of permutations of n distinct objects taken r at a time is

for r = 0, 1, 2, . . . n.Theorem 1.5p. 7The number of permutations of n distinct objects arranged in a circle is (n 1)!Theorem 1.6p. 7The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . nk are of a kth kind and n1 + n2 + . . . + nk = n is

Theorem 1.7p. 8The number of combinations of n distinct objects taken r at a time is

for r = 0, 1, 2, . . . n.

Theorem 1.8p. 10The number of ways in which a set of n distinct objects can be partitioned into k subsets with n1 objects in the first subset, n2 objects in the second subset, . . . , nk objects in the kth subset is

1. How many words can be made using all and only the letters a, a, b, b, d, d, d, d, and r?

2. In how many ways can 8 people be assigned to three committees needing 1, 1, and 6 new members?

3. Find the expression for how many ways that r indistinguishable objects can be distributed among n cells. First, look at how 7 loaves of bread be can distributed among 4 customers?

4. In how many ways can 12 loaves of bread be distributed among 5 customers, if each customer gets at least one loaf of bread?

Chapter 1: IntroductionSection 1.3: Binomial CoefficientsTheorem 1.9p. 11

for any positive integer n.Find (x + y)5.

Theorem1.10p. 11For any positive integers n and r = 0, 1, 2, . . . , n,

Theorem 1.11p.12For any positive integers n and r = 0, 1, 2, . . . , n 1,

Proof: Algebraic Proof

Theorem 1.11p.12For any positive integers n and r = 0, 1, 2, . . . , n 1,

Proof: Using Theorem 1.9

Pascals TriangleThis is an important application of Theorem 1.1111112113311464115101051

You can also use Pascals triangle to get the coefficients of (x + y)n, where n is a positive integer.How?How is this an application of Theorem 1.11?

Theorem 1.12p.14

Multinomial Coefficient of (x1 + x2 + +xk)n

Find the coefficient of x2y1z8 in the expansion of (x + y + z)11.

Chapter 2: ProbabilitySection 2.1: Introduction Classical ProbabilityAll outcomes are equally likely. Number of SuccessesP(Success) = ------------------------------------ Total Number of Outcomes

nP(Success) = ---- N

Example: An urn contains balls that are numbered from 1 to 40. Find the probability of a multiple of 7 being drawn.

Frequency InterpretationProbability is defined as the proportion of time that some outcome occurs.Example. We might say that a person is at her home 40% of the time or there is a 0.40 probability that the person will be home.

Axiomatic Approach . . . the axiomatic approach in which probabilities are defined as mathematical objects that behave according to certain well defined rules. P. 24This is the approach used in this chapter.

Chapter 2: ProbabilitySection 2.2: Sample SpacesExperiment: This refers to the process of observation or measurement.This should not be confused with the term experiment as used in research.Flipping a coin, rolling a die, and picking a card from a deck are all experiments.

Outcome: This is the most basic measurement. In rolling two pair of dice, an outcome could be defined in more than one way.Outcome: The sum of the dots on the sides facing up.Sample Space: This is the set of all possible outcomes.For the previous example:Outcome: The sum of the dots on the sides facing up.Sample Space: S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Outcome: The pairs of the sums of the dots on the red die and green die.Sample Space: S = {(x, y)| x = 1, 2, . . . , 6, y = 1, 2, . . . , 6}

Both of these sample spaces have a finite number of elements.

For rolling the experiment of rolling dice, which type of outcome do we usually use?

Do we have sample spaces that have an infinite number of elements or outcomes?

Suppose an outcome is how far the length of a towel made at a factory? In this case, the sample space is continuous.

A sample space is countable if can be matched up one-to-one with some subset of the set of whole numbers. Example: The experiment of choosing a card from standard deck of 52 cards.

It can even be a set with an infinite number of elements. Example: A good example is given in the book. The experiment is flipping a coin until you get heads.{H, TH, TTH, TTTH, TTTTH, . . . . }

A sample space that contains a countable number of elements is said to be discrete.

Chapter 2: ProbabilitySection 2.3: EventsEvent: A subset of the sample space. Consider rolling a red die and green die.See figure 2.1 on p. 27The sample space could be considered as a set of the sample points. These could be represented by ordered pairs.Some possible events:A: both die show the same number of dots.B: The sum of both die is even.

Find the sample space for the experiment of tossing a coin three times or tossing three coins.

A: two headsB: at least two headsC: no headsD: 3 tails.

P(A) = __________

P(B) = ___________

P(C) = ___________

P (D) = ____________

Suppose A is a subset of sample space S.The complement of A is A, the set of elements not included in but in S.A A = ________A A = ________P (A) + P (A) = _________P (A) = _____________P (A) = ____________Two events are said to be mutually exclusive if they no outcomes in common.For rolling two dice when events are mutually exclusive.A: Rolling an 8B: Rolling an odd numberC: Rolling the same number on each dieD: Rolling a sum of at most 5E: Rolling a 4

A and B are mutually exclusive events. P(A) = 0.3 and P(B) = 0.5.1. P(A B) = ____________________

2. P(A) = _________________

3. P[(A B)] = ________________

4.

Venn Diagrams

Mr. Valk is responsible for advising 94 freshmen. Sixty-two of the students are taking English and 70 of the students are taking mathematics. Forty seven of the students are taking both English and mathematics courses.

a. Draw the Venn diagram.

b. What is the probability of randomly choosing a student who is taking English but not mathematics?

c. What is the probability of randomly choosing a student who is taking neither English nor mathematics courses?

2.72. Ride ProbabilityJungle Cruise (J)0.74Monorail (M)0.70Matterhorn (H)0.62J and M0.52J and H0.46M and H0.44J, M, and H0.34

Draw the Venn diagram.

1. Find the probability that s/he will go on at least one ride.

2. Find the probability that s/he will not go on any of these rides

3. P (J M) = ______________

4. P[(J M)(MH)] = ______________

5. P(J MH) = _________________

Know the results from exercises 2.1 through 2.4.

Show: P[ (A B) (A B)] < P( A B)

Chapter 2: ProbabilitySection 2.4: The Probability of an Event

Postulate 1: The probability of an event is a nonnegative real number; that is, P (A) > 0 for any subset A of S.Postulate 2: P(S) = 1Postulate 3: If A1, A2, A3, . . . . Is a finite or infinite sequence of mutually exclusive events of S, then P(A1 A2 A3 ) = P(A1) + P(A2) + P(A3) + Recall, postulates are statements that require no proof. These will be used to prove theorems about probability.

Theorem 2.1. If A is an event in a discrete sample space S, then P(A) equals the sum of the probabilities of the individual outcomes comprising A.

Example: Find the probability of rolling dice and getting a sum of 6.

1Is P(x) = ------- a probability function where x = 1, 2, 3, 4, . . . ? x + 1

Theorem 2.2. If an experiment can result in any of N different equally likely outcomes, and if n of these outcomes together constitute event A, then the probability of event A is nP (A) = ----- N Find the probability of rolling two dice and getting a sum of at least 10.

Find the probability of being dealt a three of a kind in a six card hand.

Find the probability of being dealt a three of a kind and a pair (but not three of a kind) six card hand.

Find the probability of being dealt exactly two pairs (but not as part of three of a kind or four of a kinds) in a six card hand.

Suppose you roll seven dice. What is the probability you roll exactly 2 pairs but not including a three of a kind higher?

Chapter 2: ProbabilitySection 2.5: Some Rules for Probability

Theorem 2.3: If A and A are complementary events in a sample space S, then P (A) = 1 P (A)

Theorem 2.4 P () = 0 for any sample space S.

Theorem 2.5. If A and B are events in a sample space S and A B, then P (A) < P (B).

Theorem 2.6. 0 < P(A) < 1 for any event A.

Theorem 2.7. If A and B are any two events in a sample space S, P (A B) = P (A) + P (B) P (A B)See proof on p. 36.

Theorem 2.8. If A, B, and C are three events in a sample space S, then P (A B C) = P(A) + P(B) + P(C) P(A B) P(A C) P(B C) + P(A B C )

In an urn, 80 of the balls are red, 90 of the balls are blue, and 30 of the balls are yellow. The balls are numbered 1 through 200. 150 of the balls are even or red.What is the probability of choosing a ball that is even and red?

At a college, the following percentages of freshmen students take these classes in their first semester.

English (E)78%Math(M)81%History (H)46%English and Math61%English and History38%Math and History32%English, Math, and History25%

What is the probability that a student will not take any of these classes in the first semester of college?

Use the figure on p. 36 to show that

P (A B) = 1 P (A B). Let the gray area be region d.

Chapter 2: ProbabilitySection 2.6: Conditional Probability P (A|B) is the probability that A occurs given that B will occur.

Definintion 2.1. If A and B are any two events in a sample space S and P(A) 0, the conditional probability of B given A is

P (A B)P (B|A) = -------------- P (A)

Use the Venn diagram on p. 36 to find P(B|A).

P(B|A) = __________

C: rolling a doubleD: rolling at least a 9

P (C|D) = ______________

A mechanic has just ordered new and used (refurbished) parts for the industrial machines that she maintains. She knows from experience that some of the parts are defective.

DefectiveNot defective

New694

Used1175

Given that a new part is used, the probability is ____________ that it will not be defective.

For a part that will be found to be defective, the probability is ___________ that it will be used.

Theorem 2.9: If A and B are any two events in a sample space S and P (A) 0, then

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

Theorem 2.10: If A, B, and C are any three events in a sample space S such that P (A B) 0, then

P (A B C) = P (A) P (B|A) P (C|A B)

An urn contains 20 red, 30 blue, and 40 green balls.

1. Find the probability of choosing 3 red balls. The balls are not put back in the urn. This is an example of without replacement.

2. Find the probability of choosing 2 green balls. After each ball is picked, it is returned to the urn. This is an example of with replacement .

3. Find the probability of choosing 3 green balls and two blue balls. The chosen balls are not returned to the urn after each pick.

Chapter 2: ProbabilitySection 2.7: IndependenceTwo events are said to be independent if the occurrence (or nonoccurrence) of one event does not affect the probability of the other event occurring.

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

Definition 2.2. Two events A and B are independent if and only if

P(A B) = P(A) P(B)

Determine if the two events are independent. The experiment is rolling two dice.

A: rolling a multiple of 3B: rolling a double

Theorem 2.11. If A and B are independent, then A and B are also independent.

See Proof on p. 46.

Definition 2.3. Events A1, A2, . . . , and AK, are independent if and only if the probability of the intersection of any 2, 3, . . ., or k of these events equals the product of their respective probabilities.

For three events A, B, and C,

P (A B) = P(A) P(B)P(A C) = P(A) P(C)P(B C) = P(B) P(C)andP(A B C) = P(A) P(B) P(C)

must occur for the three events to be independent.

Note: Pairwise independence does not mean that there will be the three events will be independent.

Events A, B, and C are independent with their probabilities given below.P (A) = 0.24P (B) = 0.35P(C) = 0.08

P(A B C) = _________________

P(A B C) = ______________________

P(A|B C) = ______________________

John can hit the target 45% of the time. Letitia is able to hit the target 82% of the time while Mario hits the target 91% of the time. Assuming each attempt by each person is independent, answer the following for when each person takes a shot at the target.

1. What is the probability that all three people hit the target?

2. What is the probability that John and Letitia hit the target but Mario misses the target?

3. What is the probability that only two of the people hit the target?

Extra Topic: Odds

The odds that an event will occur are the number of successes compared to the number of failures.

The odds that an event will not occur are the number of failures compared to the number of successes.

1. What are the odds in favor of choosing an ace from a standard deck of cards?

2. What are the odds against choosing a face card?

3. The odds against Born to Run winning the race are 8 to 3. What is the probability that Born to Run will win the horse race?

4. A horse owner has one horse whose odds in favor of winning a race are 1 to 2 and another horse whose odds in favor of winning are 2 to 9. What are the odds in favor of one the owners horses winning the race?

Chapter 2: ProbabilitySection 2.8: Bayes Theorem

Exercise 2.106The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one car accident is incorrectly attributed to faulty brakes is 0.03.a. What is the probability that a car accident will be attributed to faulty brakes?

A: The probability that a car accident will be attributed to faulty brakes.B: The probability that a car accident is due to faulty brakes.

b. What is the probability that a car accident attributed to faulty brakes was actually due to faulty brakes?

Part a. is an example of the rule of total probability.P(A) = P(B) * P(A|B) + P(B)*P(A| B) and can be generalized byTheorem 2.12. If the events B1, B2, . . . , Bk constitute a partition of the sample space S and P(Bi) 0 for i = 1, 2, . . . k, then for event A in S

P(A) =

A charity sells pink, blue, and purple bags for $10 each: 35% of the bags are pink, 25% of the bags are blue, and 40% of the bags are purple. In 7% of the pink bags, 6% of the blue bags, and 8% of purple bags are gift cards for $100. Find the probability that any given bag contains a $100 gift card.

Theorem 2.13. If the events B1, B2, . . . , Bk constitute a partition of the sample space S and P(Bi) 0 for r = 1, 2, . . . k, then for event A in S such that P(A) 0

r = 1, 2, . . . k.

See the previous example. Knowing that a bag containing a gift card was chosen, what is the probability that the bag is purple? Chapter 1: IntroductionSection 1.2: Combinatorial MethodsTheorem 1.1 p.2If an operation consists of two steps, of which the first can done in n1 ways and the second can be done in n2 ways, then the whole operation can be done in n1 n2 ways.

Theorem 1.2p.4If an operation consists of k steps, of which the first can be done in n1 ways, for each of these the second step can be done in n2 ways, for each of the first two, the third step can be done in n3 ways, and so forth, the whole operation can be done in n1 n2 nk ways.

Example: At a food truck you have a choice of 2 types of bread, 4 kinds of meat and three varieties of cheese.

How many kinds of sandwiches can be made? Make the tree diagram.

Example: In how many ways can five multiple-choice questions be answered if each question has four choices?

Combination for n objects, a combination is the set or collection of those objects. The collection of objects is in no particular order. ABC and CAB would be the same combination of letters.Permutation for n objects, a permutation is an arrangement of the n objects. ABC, ACB, BAC, BCA, CAB, and CBA are the six permutations for the letters A, B, and C. The combination is ABC in no particular order.Theorem 1.3p. 5The number of permutations of n distinct objects is n!.

5. In how many ways can five tasks be assigned to five people, if each person is assigned exactly one task?

6. In how many ways can two members of a committee of six be chosen to fill the positions of chair and secretary?

7. In how many ways can three people be chosen from a group of 12 to be on a committee?

8. In how many ways can we get 6 people to stand in a straight line? In how many ways can they be seated around a table?

Theorem 1.4p. 5The number of permutations of n distinct objects taken r at a time is

for r = 0, 1, 2, . . . n.Theorem 1.5p. 7The number of permutations of n distinct objects arranged in a circle is (n 1)!Theorem 1.6p. 7The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . nk are of a kth kind and n1 + n2 + . . . + nk = n is

Theorem 1.7p. 8The number of combinations of n distinct objects taken r at a time is

for r = 0, 1, 2, . . . n.

Theorem 1.8p. 10The number of ways in which a set of n distinct objects can be partitioned into k subsets with n1 objects in the first subset, n2 objects in the second subset, . . . , nk objects in the kth subset is

5. How many words can be made using all and only the letters a, a, b, b, d, d, d, d, and r?

6. In how many ways can 8 people be assigned to three committees needing 1, 1, and 6 new members?

7. Find the expression for how many ways that r indistinguishable objects can be distributed among n cells. First, look at how 7 loaves of bread be can distributed among 4 customers?

8. In how many ways can 12 loaves of bread be distributed among 5 customers, if each customer gets at least one loaf of bread?

Chapter 2: ProbabilitySection 2.1: Introduction Classical ProbabilityAll outcomes are equally likely. Number of SuccessesP(Success) = ------------------------------------ Total Number of Outcomes

nP(Success) = ---- N

Example: An urn contains balls that are numbered from 1 to 40. Find the probability of a multiple of 7 being drawn.

Frequency InterpretationProbability is defined as the proportion of time that some outcome occurs.Example. We might say that a person is at her home 40% of the time or there is a 0.40 probability that the person will be home.

Axiomatic Approach . . . the axiomatic approach in which probabilities are defined as mathematical objects that behave according to certain well defined rules. P. 24This is the approach used in this chapter.

Chapter 2: ProbabilitySection 2.2: Sample SpacesExperiment: This refers to the process of observation or measurement.This should not be confused with the term experiment as used in research.Flipping a coin, rolling a die, and picking a card from a deck are all experiments.

Outcome: This is the most basic measurement. In rolling two pair of dice, an outcome could be defined in more than one way.Outcome: The sum of the dots on the sides facing up.Sample Space: This is the set of all possible outcomes.For the previous example:Outcome: The sum of the dots on the sides facing up.Sample Space: S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Outcome: The pairs of the sums of the dots on the red die and green die.Sample Space: S = {(x, y)| x = 1, 2, . . . , 6, y = 1, 2, . . . , 6}

Both of these sample spaces have a finite number of elements.

For rolling the experiment of rolling dice, which type of outcome do we usually use?

Do we have sample spaces that have an infinite number of elements or outcomes?

Suppose an outcome is how far the length of a towel made at a factory? In this case, the sample space is continuous.

A sample space is countable if can be matched up one-to-one with some subset of the set of whole numbers. Example: The experiment of choosing a card from standard deck of 52 cards.

It can even be a set with an infinite number of elements. Example: A good example is given in the book. The experiment is flipping a coin until you get heads.{H, TH, TTH, TTTH, TTTTH, . . . . }

A sample space that contains a countable number of elements is said to be discrete.Chapter 2: ProbabilitySection 2.3: EventsEvent: A subset of the sample space. Consider rolling a red die and green die.See figure 2.1 on p. 27The sample space could be considered as a set of the sample points. These could be represented by ordered pairs.Some possible events:A: both die show the same number of dots.B: The sum of both die is even.

Find the sample space for the experiment of tossing a coin three times or tossing three coins.

A: two headsB: at least two headsC: no headsD: 3 tails.

P(A) = __________

P(B) = ___________

P(C) = ___________

P (D) = ____________

Suppose A is a subset of sample space S.The complement of A is A, the set of elements not included in but in S.A A = ________A A = ________P (A) + P (A) = _________P (A) = _____________P (A) = ____________Two events are said to be mutually exclusive if they no outcomes in common.For rolling two dice when events are mutually exclusive.A: Rolling an 8B: Rolling an odd numberC: Rolling the same number on each dieD: Rolling a sum of at most 5E: Rolling a 4

A and B are mutually exclusive events. P(A) = 0.3 and P(B) = 0.5.5. P(A B) = ____________________

6. P(A) = _________________

7. P[(A B)] = ________________

8.

Venn Diagrams

Mr. Valk is responsible for advising 94 freshmen. Sixty-two of the students are taking English and 70 of the students are taking mathematics. Forty seven of the students are taking both English and mathematics courses.

d. Draw the Venn diagram.

e. What is the probability of randomly choosing a student who is taking English but not mathematics?

f. What is the probability of randomly choosing a student who is taking neither English nor mathematics courses?

2.72. Ride ProbabilityJungle Cruise (J)0.74Monorail (M)0.70Matterhorn (H)0.62J and M0.52J and H0.46M and H0.44J, M, and H0.34

Draw the Venn diagram.

6. Find the probability that s/he will go on at least one ride.

7. Find the probability that s/he will not go on any of these rides

8. P (J M) = ______________

9. P[(J M)(MH)] = ______________

10. P(J MH) = _________________

Know the results from exercises 2.1 through 2.4.

Show: P[ (A B) (A B)] < P( A B)

Chapter 2: ProbabilitySection 2.4: The Probability of an Event

Postulate 1: The probability of an event is a nonnegative real number; that is, P (A) > 0 for any subset A of S.Postulate 2: P(S) = 1Postulate 3: If A1, A2, A3, . . . . Is a finite or infinite sequence of mutually exclusive events of S, then P(A1 A2 A3 ) = P(A1) + P(A2) + P(A3) + Recall, postulates are statements that require no proof. These will be used to prove theorems about probability.

Theorem 2.1. If A is an event in a discrete sample space S, then P(A) equals the sum of the probabilities of the individual outcomes comprising A.

Example: Find the probability of rolling dice and getting a sum of 6.

1Is P(x) = ------- a probability function where x = 1, 2, 3, 4, . . . ? x + 1

Theorem 2.2. If an experiment can result in any of N different equally likely outcomes, and if n of these outcomes together constitute event A, then the probability of event A is nP (A) = ----- N Find the probability of rolling two dice and getting a sum of at least 10.

Find the probability of being dealt a three of a kind in a six card hand.

Find the probability of being dealt a three of a kind and a pair (but not three of a kind) six card hand.

Find the probability of being dealt exactly two pairs (but not as part of three of a kind or four of a kinds) in a six card hand.

Suppose you roll seven dice. What is the probability you roll exactly 2 pairs but not including a three of a kind higher?

Chapter 2: ProbabilitySection 2.5: Some Rules for Probability

Theorem 2.3: If A and A are complementary events in a sample space S, then P (A) = 1 P (A)

Theorem 2.4 P () = 0 for any sample space S.

Theorem 2.5. If A and B are events in a sample space S and A B, then P (A) < P (B).

Theorem 2.6. 0 < P(A) < 1 for any event A.

Theorem 2.7. If A and B are any two events in a sample space S, P (A B) = P (A) + P (B) P (A B)See proof on p. 36.

Theorem 2.8. If A, B, and C are three events in a sample space S, then P (A B C) = P(A) + P(B) + P(C) P(A B) P(A C) P(B C) + P(A B C )

In an urn, 80 of the balls are red, 90 of the balls are blue, and 30 of the balls are yellow. The balls are numbered 1 through 200. 150 of the balls are even or red.What is the probability of choosing a ball that is even and red?

At a college, the following percentages of freshmen students take these classes in their first semester.

English (E)78%Math(M)81%History (H)46%English and Math61%English and History38%Math and History32%English, Math, and History25%

What is the probability that a student will not take any of these classes in the first semester of college?

Use the figure on p. 36 to show that

P (A B) = 1 P (A B). Let the gray area be region d.

Chapter 2: ProbabilitySection 2.6: Conditional Probability P (A|B) is the probability that A occurs given that B will occur.

Definintion 2.1. If A and B are any two events in a sample space S and P(A) 0, the conditional probability of B given A is

P (A B)P (B|A) = -------------- P (A)

Use the Venn diagram on p. 36 to find P(B|A).

P(B|A) = __________

C: rolling a doubleD: rolling at least a 9

P (C|D) = ______________

A mechanic has just ordered new and used (refurbished) parts for the industrial machines that she maintains. She knows from experience that some of the parts are defective.

DefectiveNot defective

New694

Used1175

Given that a new part is used, the probability is ____________ that it will not be defective.

For a part that will be found to be defective, the probability is ___________ that it will be used.

Theorem 2.9: If A and B are any two events in a sample space S and P (A) 0, then

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

Theorem 2.10: If A, B, and C are any three events in a sample space S such that P (A B) 0, then

P (A B C) = P (A) P (B|A) P (C|A B)

An urn contains 20 red, 30 blue, and 40 green balls.

4. Find the probability of choosing 3 red balls. The balls are not put back in the urn. This is an example of without replacement.

5. Find the probability of choosing 2 green balls. After each ball is picked, it is returned to the urn. This is an example of with replacement .

6. Find the probability of choosing 3 green balls and two blue balls. The chosen balls are not returned to the urn after each pick.

Chapter 2: ProbabilitySection 2.7: IndependenceTwo events are said to be independent if the occurrence (or nonoccurrence) of one event does not affect the probability of the other event occurring.

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

Definition 2.2. Two events A and B are independent if and only if

P(A B) = P(A) P(B)

Determine if the two events are independent. The experiment is rolling two dice.

A: rolling a multiple of 3B: rolling a double

Theorem 2.11. If A and B are independent, then A and B are also independent.

See Proof on p. 46.

Definition 2.3. Events A1, A2, . . . , and AK, are independent if and only if the probability of the intersection of any 2, 3, . . ., or k of these events equals the product of their respective probabilities.

For three events A, B, and C,

P (A B) = P(A) P(B)P(A C) = P(A) P(C)P(B C) = P(B) P(C)andP(A B C) = P(A) P(B) P(C)

must occur for the three events to be independent.

Note: Pairwise independence does not mean that there will be the three events will be independent.

Events A, B, and C are independent with their probabilities given below.P (A) = 0.24P (B) = 0.35P(C) = 0.08

P(A B C) = _________________

P(A B C) = ______________________

P(A|B C) = ______________________

John can hit the target 45% of the time. Letitia is able to hit the target 82% of the time while Mario hits the target 91% of the time. Assuming each attempt by each person is independent, answer the following for when each person takes a shot at the target.

4. What is the probability that all three people hit the target?

5. What is the probability that John and Letitia hit the target but Mario misses the target?

6. What is the probability that only two of the people hit the target?

Extra Topic: Odds

The odds that an event will occur are the number of successes compared to the number of failures.

The odds that an event will not occur are the number of failures compared to the number of successes.

5. What are the odds in favor of choosing an ace from a standard deck of cards?

6. What are the odds against choosing a face card?

7. The odds against Born to Run winning the race are 8 to 3. What is the probability that Born to Run will win the horse race?

8. A horse owner has one horse whose odds in favor of winning a race are 1 to 2 and another horse whose odds in favor of winning are 2 to 9. What are the odds in favor of one the owners horses winning the race?

Chapter 2: ProbabilitySection 2.8: Bayes Theorem

Exercise 2.106The probability that a one-car accident is due to faulty brakes is 0.04, the probability that a one-car accident is correctly attributed to faulty brakes is 0.82, and the probability that a one car accident is incorrectly attributed to faulty brakes is 0.03.c. What is the probability that a car accident will be attributed to faulty brakes?

A: The probability that a car accident will be attributed to faulty brakes.B: The probability that a car accident is due to faulty brakes.

d. What is the probability that a car accident attributed to faulty brakes was actually due to faulty brakes?

Part a. is an example of the rule of total probability.P(A) = P(B) * P(A|B) + P(B)*P(A| B) and can be generalized byTheorem 2.12. If the events B1, B2, . . . , Bk constitute a partition of the sample space S and P(Bi) 0 for i = 1, 2, . . . k, then for event A in S

P(A) =

A charity sells pink, blue, and purple bags for $10 each: 35% of the bags are pink, 25% of the bags are blue, and 40% of the bags are purple. In 7% of the pink bags, 6% of the blue bags, and 8% of purple bags are gift cards for $100. Find the probability that any given bag contains a $100 gift card.

Theorem 2.13. If the events B1, B2, . . . , Bk constitute a partition of the sample space S and P(Bi) 0 for r = 1, 2, . . . k, then for event A in S such that P(A) 0

r = 1, 2, . . . k.

See the previous example. Knowing that a bag containing a gift card was chosen, what is the probability that the bag is purple?

Fin