Math 124: Lecture for Week 6 of 17online.sfsu.edu/meredith/statistics/S2008/Lecture_6_slides.pdf · Week 6 Review Tests Questions and Answers Probability Theory Basic Terms and Deﬁnitions

Math 124:Week 6

Review Tests

Questions andAnswers

ProbabilityTheoryBasic Terms andDefinitions

Rules for ProbabilityCalculations

ConditionalProbability

Math 124: Lecture for Week 6 of 17

David MeredithDepartment of Mathematics

San Francisco State University

March 4, 2008

Math 124:Week 6

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What we will do tonight

1 Review Tests

2 Questions and Answers

3 Probability TheoryBasic Terms and DefinitionsRules for Probability CalculationsConditional Probability

Math 124:Week 6

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1 Review Tests



Math 124:Week 6

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1 Review Tests



Math 124:Week 6

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Review tests

• Midterm Review• Essay Question• Distribution of Answers

• Midterm

• Quiz www.cmu.edu/oli

www.cmu.edu/oli

Math 124:Week 6

Review Tests





Review tests

• Midterm Review• Essay Question• Distribution of Answers

• Midterm• Quiz www.cmu.edu/oli

www.cmu.edu/oli

Math 124:Week 6

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Questions and Answers

Review My Response questions for the week

Math 124:Week 6

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Determine the sample space of an experiment

Suppose you want to apply probability theory to somesituation or experiment.

• The set of possible outcomes in your situation is calledthe sample space

• The sample space for tomorrow’s weather is {sunny,cloudy, rainy}

• The sample space for tossing a coin is {H,T}• The sample space for tossing a nickel and a dime is:

Nickel H H T TDime H T H T

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Suppose you want to apply probability theory to somesituation or experiment.• The set of possible outcomes in your situation is called

the sample space

• The sample space for tomorrow’s weather is {sunny,cloudy, rainy}



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the sample space• The sample space for tomorrow’s weather is {sunny,

cloudy, rainy}



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cloudy, rainy}• The sample space for tossing a coin is {H,T}

• The sample space for tossing a nickel and a dime is:Nickel H H T TDime H T H T

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cloudy, rainy}• The sample space for tossing a coin is {H,T}• The sample space for tossing a nickel and a dime is:


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• An event in your situation is a subset of a samplespace

• For tomorrow’s weather, “it doesn’t rain” = {sunny,cloudy} is an event

• For two coins, nickel comes up H is an event• For two coins, one H and one T is an event

• Two special events• the empty subset of the sample space = “nothing

happens”• the whole sample space = “something happens”

• Two events are disjoint if they don’t contain anycommon outcome

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• For two coins, nickel comes up H is an event

• For two coins, one H and one T is an event• Two special events

• the empty subset of the sample space = “nothinghappens”

• the whole sample space = “something happens”


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• Two special events

• the empty subset of the sample space = “nothinghappens”



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happens”



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Question 1:

1 Find the sample space associated with drawing arandom card from an ordinary deck of cards.

2 How many outcomes are there?3 Are the events “draw a spade” and “draw a face card”

disjoint?4 Give an example of two disjoint events with at least four

outcomes in each event.

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• In the context of a probability model, every event has aprobability, which is a decimal number between 0 and1.

• If A is an event, a subset of the sample space S, thenthe probability of A is denoted P(A)

• P(∅) = 0The probability that nothing happens is 0

• P(S) = 1The probability that something happens is 1

• If A and B and disjoint events, then

P(A ∪ B) = P(A) + P(B)

The probability that either A or B happens is the sum ofthe probabilities of A and B

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P(A ∪ B) = P(A) + P(B)


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P(A ∪ B) = P(A) + P(B)


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P(A ∪ B) = P(A) + P(B)


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P(A ∪ B) = P(A) + P(B)


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P(A ∪ B) = P(A) + P(B)


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Find probability of events in the case when all outcomes are

equally likely.

A consequence of these rules is:

•∑

o∈S P(o) = 1• The sum of the probabilities of all the individual

outcomes is 1• In the special case that there are n equally likely

outcomes, then the probability of each outcome is1n

.

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equally likely.

A consequence of these rules is:•

∑o∈S P(o) = 1

• The sum of the probabilities of all the individualoutcomes is 1

• In the special case that there are n equally likely


.

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equally likely.


∑o∈S P(o) = 1




.

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equally likely.


∑o∈S P(o) = 1




.

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equally likely.

Question 2: Assuming that you are equally likely to drawany card, what is the probability of:

1 drawing the ace of spades2 drawing an ace3 drawing a face card4 drawing a spade

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equally likely.

Question 3: If you toss a nickel and a dime, what is theprobability that you get one H and one T?

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Translate real world problems into probabilistic

representation.

• Math 124.16 consists of

Fr So Jr Sr Uncl Cl Fac TotalF 9 5 4 8 3 1 0 30M 5 5 2 1 2 1 1 17

Total 14 10 6 9 5 2 1 47

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representation.

• The situation under consideration is choosing a personat random from our class

• The sample space is the whole class• Each person is an outcome• “Choosing a junior” is an event• “Choosing a senior” is an event• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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representation.


• The sample space is the whole class

• Each person is an outcome• “Choosing a junior” is an event• “Choosing a senior” is an event• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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representation.


• The sample space is the whole class• Each person is an outcome

• “Choosing a junior” is an event• “Choosing a senior” is an event• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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representation.


• The sample space is the whole class• Each person is an outcome• “Choosing a junior” is an event

• “Choosing a senior” is an event• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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representation.


• The sample space is the whole class• Each person is an outcome• “Choosing a junior” is an event• “Choosing a senior” is an event

• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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representation.


• The sample space is the whole class• Each person is an outcome• “Choosing a junior” is an event• “Choosing a senior” is an event• These events are disjoint

• “Choosing an upper-division student” is the union ofthese events

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representation.


• The sample space is the whole class• Each person is an outcome• “Choosing a junior” is an event• “Choosing a senior” is an event• These events are disjoint• “Choosing an upper-division student” is the union of

these events

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equally likely.

• The probability of choosing any person is the same asthe probability of choosing any other person.

• So the probability of choosing any particular person is1

47= 0.021.

• If we choose somebody• The probability of choosing someone is 1• The probability of choosing nobody is 0

• The probability of choosing a junior is647

= 0.13

• The probability of choosing a senior is9

47= 0.19

• The probability of choosing an upper-division student is0.13 + 0.19 = 0.32

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equally likely.



47= 0.021.



= 0.13


47= 0.19


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equally likely.



47= 0.021.

• If we choose somebody

• The probability of choosing someone is 1• The probability of choosing nobody is 0


= 0.13


47= 0.19


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equally likely.



47= 0.021.

• If we choose somebody• The probability of choosing someone is 1

• The probability of choosing nobody is 0


= 0.13


47= 0.19


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equally likely.



47= 0.021.



= 0.13


47= 0.19


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equally likely.



47= 0.021.



= 0.13


47= 0.19


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equally likely.



47= 0.021.



= 0.13


47= 0.19


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equally likely.



47= 0.021.



= 0.13


47= 0.19


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Apply probability rules in order to find the likelihood of an

event

Question 4: What is the probability of choose a malefreshman or a female senior?

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event

• Next we need to develop some rules for calculatingnon-obvious probabilities

• There are two rules.• Master them and you can do any probability problem.

• The rules are based on three little words: “and”, “or”and “not”

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event


• There are two rules.

• Master them and you can do any probability problem.


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event




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event




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event

The Negation Rule

• If A is an event, then not A is the complement of A, allthe outcomes that aren’t in A.

• If A = choose a junior, thennot A = choose someone who is not a junior

• P(not A) = 1− P(A) because not A and A are disjoint,so

P(A) + P(not A) = P(A ∪ not A)

= P(S)

= 1

• The probability that a student is not a junior is1− 0.13 = 0.87

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event

The Negation Rule





= P(S)

= 1


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event

The Negation Rule





= P(S)

= 1


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event

The Negation Rule





= P(S)

= 1


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event

The Negation Rule





= P(S)

= 1


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event

Question 5: What is the probability that a student is anundergraduate?

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event

• If A and B are events, then A or B = A ∪ B, all theoutcomes in A or B or both.

• If A = junior and B = male thenA or B = junior or male or both

• A and B = A ∩ B, all the outcomes in A and in B• In the example, A and B = male junior

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event




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event



• A and B = A ∩ B, all the outcomes in A and in B

• In the example, A and B = male junior

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event




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event

• The Addition Rule

P(A and B) + P(A or B) = P(A) + P(B)

• If you know any three of these probabilities, you canfind the fourth.

• In the example, P(junior) = 0.13, P(male) = 0.36 andP(junior male) = 0.043

• Thus

P(male or junior) = P(male) + P(junior)− P(male and junior)

= 0.36 + 0.13− 0.043= 0.45

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event





• Thus


= 0.36 + 0.13− 0.043= 0.45

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event





• Thus


= 0.36 + 0.13− 0.043= 0.45

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event





• Thus


= 0.36 + 0.13− 0.043= 0.45

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event

Question 6: What is the probability that a student is a maleor a senior?

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event

• The addition rule has one special case that you havealready seen

• A and B are disjoint events if they have no outcomes incommon

• That means A ∩ B = ∅ or P(A and B) = 0.• In that case our general addition rule simplifies to the

rule for disjoint events:

P(A or B) = P(A) + P(B)

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event





P(A or B) = P(A) + P(B)

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event



• That means A ∩ B = ∅ or P(A and B) = 0.

• In that case our general addition rule simplifies to therule for disjoint events:

P(A or B) = P(A) + P(B)

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event





P(A or B) = P(A) + P(B)

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event

• Example: A = choose a male freshman andB = choose a female senior

• P(A) =547

= 0.11 and P(B) =8

47= 0.17

• A and B are disjoint events

P(A and B)

= P(student is male freshman and a female senior)= 0

P(choose a male freshman or a female senior)= P(A or B)

= P(A) + P(B)

= 0.11 + 0.17= 0.28

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event


• P(A) =547

= 0.11 and P(B) =8

47= 0.17


P(A and B)



= P(A) + P(B)

= 0.11 + 0.17= 0.28

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event


• P(A) =547

= 0.11 and P(B) =8

47= 0.17


P(A and B)



= P(A) + P(B)

= 0.11 + 0.17= 0.28

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event


• P(A) =547

= 0.11 and P(B) =8

47= 0.17


P(A and B)



= P(A) + P(B)

= 0.11 + 0.17= 0.28

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event


• P(A) =547

= 0.11 and P(B) =8

47= 0.17


P(A and B)



= P(A) + P(B)

= 0.11 + 0.17= 0.28

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event

Question 7: If you draw a card at random from an ordinarydeck, what is the probability that you get:

1 a nine or a jack2 a club or a jack

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event

• We say that A and B are independent events ifP(A and B) = P(A)P(B)

• in that case the occurrence of one of the events doesnot affect the probability that the other will occur.

• Example: If I toss a nickel and a dime, the probabilitythat the nickel comes up H is not affected by how thedime falls.

P(N = H and D = H) = 0.25= 0.5× 0.5= P(N = H)P(D = H)

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event




P(N = H and D = H) = 0.25= 0.5× 0.5= P(N = H)P(D = H)

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event




P(N = H and D = H) = 0.25= 0.5× 0.5= P(N = H)P(D = H)

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event




P(N = H and D = H) = 0.25= 0.5× 0.5= P(N = H)P(D = H)

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event

• Example: If choose a card and put it back, the value ofthe card has no effect on the value of the next carddrawn. The probability that I will draw a spade and thena 9 is

P(card 1 = spade) = 0.25P(card 2 = nine) = 0.077

P(card 1 = space and card 2 = nine) = (0.25)(0.077)

= 0.019

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event




= 0.019

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event




= 0.019

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event

Question 8: If you choose two students at random from ourclass with replacement,

1 what is the probability that you get two women?2 what is the probability that you get a man and a

woman?

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Break

Please return in 15 minutes.

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Find conditional probabilities and interpret them

Question 9:1 If you choose a woman from our class, what is the

probability that she would be a junior?2 If you choose a junior from our class, what is the

probability that the person would be a woman?

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• If you want to find the probability that somethinghappens, and you already know something about thesituation, we call that a conditional probability—theprobability is conditional on the knowledge

• Example: if you know that the person you chose fromour class was a woman, what would the probability bethat she was a junior?

•4

30= 0.13

• Example: if you know that the person you chose fromour class was a junior, what would the probability bethat the person was a woman?

•46

= 0.67

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Question 10:1 If you choose a man from our class, what is the

probability that he would be a senior?2 If you choose someone from our class, what is the

probability that the person would be a male senior?

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• The probability that B is true, given that A is true, isdenoted P(B|A)

• P(junior | woman) = 0.13• P(woman | junior) = 0.67

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Question 11:1 Find P(man)2 Find P(man and sophomore)3 Find P(sophomore | man)

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Explain the reasoning behind conditional probability, and

how this reasoning is expressed by the definition of

conditional probability

• We have seen that P(B|A) =P(B and A)

P(A)

• P(sophomore|man) =P(sophomore and man)

P(man)

• In other words, P(B|A)P(A) = P(B and A)

• P(sophomore|man)P(man) = P(sophomore and man)

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Determine whether two events are independent or not

• A and B are independent events ifP(B)P(A) = P(B and A)

• It is always true that P(B|A)P(A) = P(B and A)

• Therefore A and B are independent events ifP(B) = P(B|A)

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Question 12:1 Show that the probability of drawing a spade from an

ordinary deck of cards is independent of the probabilityof drawing a face card.

2 If you remove an ace from the deck, what is theprobability that a card drawn at random from theremaining cards is a king?

3 Use the formula

P(card 1 = ace)P(card 2 = king|card 1 = ace)

= P(card 1 = ace and card 2 = king)

to calculate the probability of drawing an ace followedby a king.

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Question 13:1 If you selected two people at random from our class,

what is the probability that the first person is a man andthe second person is a woman?

2 If you selected two people at random from our class,what is the probability that one person is a man and theother person is a woman?

Documents

Math 124: Lecture for Week 6 of 17online.sfsu.edu/meredith/statistics/S2008/Lecture_6_slides.pdf · Week 6 Review Tests Questions and Answers Probability Theory Basic Terms and Deﬁnitions