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Conditional Probability

STA281 Fall 2005

1 Definition

Often we are only interested in particular rows or columns of the table. Consider the newspaperexample, and the question Of those that receive the morning paper, what proportion receivethe afternoon paper? This question does not concern the entire population of households, itonly concerns those who receive a morning paper. Probabilities that refer only to subsets of thepopulation are called conditional probabilities. Recall the probability table we constructed

M Mc

A 0.10 0.20 0.30

Ac 0.50 0.20 0.70

0.60 0.40 1.00

The question asked concerns only those who receive a morning paper, which is 60% of the entirepopulation. We want to know, out of that 60%, what proportion receive the afternoon paper. Toprovide an intuitive feel for how this question is answered, suppose we sampled 100 people fromthe population. On average 60 of those people would receive the morning paper. Looking at the Mcolumn of the table, we see that of those 60, on average 50 receive only the morning paper while 10receive the morning and afternoon paper. So 10 of the 60 people who receive the morning paper

also receive the afternoon paper, so the conditional probability is 10/60 = 1/6.Usually we dont go through the argument concerning sampling a set of people and just divide

the probabilities directly. There are 60% of the people who receive a morning paper, with 50%receiving only the morning paper and 10% receiving the afternoon paper. So 0.10/0.60 = 1/6 isthe conditional probability of receiving an afternoon paper given one receives a morning paper.

Mathematically, a conditional probability has two parts. First, a conditional probability onlyasks about a subset of the population, not the entire population. Second, a conditional probabilityasks some property of that subset. In our example question, the subset of interest was those whoreceive the morning paper, while the property we are interested in was receiving an afternoon paper.In general, we have a question of those who are in subset A, what is the probability they are in B.This question is translated into mathematical symbols as P(B|A), which is read the probability

of B given A.Notice how we solved the problem. First, we found which of the individuals were in the subset

of interest. This involved finding P(A), the unconditional probability of the subset. Then, withinthat subset, we found how many individuals had the property we were interested in. The result was

P(B|A) =people in subset A that have property B

people in subset A

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Instead of writing this fraction in words, we can just use symbols. For the denominator, the peoplein subset A refers to P(A). For the numerator, the people must be subset A, but they must alsohave property B. Since both criteria must be satisfied (in subset A and have property B), thenumerator is P(A

B), resulting in the formula

P(B|A) = P(B

A)P(A)

Mathematically, this formula is often used as the definition of conditional probability.

2 Recognizing Conditional Probabilities

Remember the key point about conditional probability is that we are only interested in a subset ofthe population. The first step in evaluating a conditional probability is determining which outcomeswe are interested in. In our example we were interested in people who received a morning paper.The second step is to identify the property of interest (in our example it was receiving an afternoon

paper). After that you use the definition of conditional probability P(B|A) = P(B

A)/P(A).In our example, the subset was receiving a morning paper while the property was receiving anafternoon paper. Thus, we were interested in finding P(A|M).

Whenever you see a probability stated or are asked for a probability, ask yourself two questions.First, Who is this statement about. In a conditional probability we are only interested in a subsetof the population. It is important to determine which subset as soon as possible so we can proceed.Second, What are they asking about. That is, what property of the subset is the question orstatement about.

For example, suppose our population is all registered voters. Constrast the two questions Whatproportion of women are Democrats? and What proportion of voters are women Democrats.The first statement does not ask about all registered voters, it only asks about women. Therefore

it is a conditional probability. After deciding it only asks about women, we must that determinewhat exactly it wants to know about women. In this example, the property of interest is beingDemocratic. IfW is the event voter is female and D is the event voter is a Democrat, the firstquestion asks for P(D|W). The second question does not place any restriction on the populationsince it asks for the proportion of voters, not the proportion of any particular subset. The propertyit is interested in is whether a voter is a Democratic women. To be a democratic woman, a votermust be both a woman AND a democrat. Thus, the second question is asking for P(W

D). These

are separate questions, so you must recognize which one you are being asked.Of course, in any language there are multiple ways to ask the same question. The following

questions are equivalent, all ask for the probability of receiving an afternoon paper in the set ofoutcomes where a morning paper was received.

What proportion of those that receive a morning paper receive an afternoon paper?

Given a person receives a morning paper, what is the probability they receive an afternoonpaper.

If someone receives a morning paper, what is the probability they receive an afternoon paper?

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There are a number of ways to report the result P(A|M) = 1/6 as well, such as

1/6 of those that receive a morning paper receive an afternoon paper.

Given someone receives a morning paper, there is a 1/6 probability of receiving an afternoonpaper.

If someone receives a morning paper there is a 1/6 probability of receiving an afternoon paper.

The probability table allows us to compute conditional probabilities such as P(A|M) fairlyeasily, since we are just involved with one row or column of the table. More complicated conditionalprobabilities may be computed as well. For example, given that someone receives at least one ofthe papers, what is the probability they receive at most one of the papers? This is a conditionalprobability since we are interested only in those who receive at least one paper, not everyone. Thissubset of individuals includes 3 cells of our table, since receiving at least one paper involves the cellsP(M

A) = 0.10, P(M

Ac) = 0.50 and P(Mc

A) = 0.20 for a total of 80% of the population.

Within that subset, we are interested in the property receive both papers. As with allconditional probabilities, we look at the individuals within the subset and try to determine whichobey the property. Just looking at the cells which compose the subset, we find only the cellsP(M

Ac) and P(Mc

A) satisfy the property receive at most on the papers. These two boxes

total 70% of the overall probability. The conditional probability is thus 0.70/0.80 = 7/8.

3 Using Conditional Probabilities

The definition of conditional probability is that P(B|A) = P(B

A)/P(A). With some algebra,we find P(B

A) = P(B|A)P(A). This formula allows us to take conditional probabilities as given

information and complete a probability table.Suppose in a small university, students work in either the dorms or the library. 70% of the

students do some work in the library while 10% do work only in the dorms. Of those who work inthe library, 30% also do work in the dorms. Use this information to construct a probability table.

Students may either work in dorms or the library. Since they may either work or not work inthe library and work or not work in dorms, we may construct a probability table

L Lc

D

Dc

1.00

We are given the information that 70% of the students do some work in library. Since this infor-

mation was given without any reference to whether or not those students work in the dorms, it isthe overall proportion of students that work in library, including students who work in both andstudents who work only in the library. It is therefore placed in the margins of the table. We arealso given that 10% work only in the dorms. This is one of the 4 core cells of the table, since itinvolves students who work in the dorms and also do not work in the library. We fill those valuesin the table.

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L Lc

D 0.10

Dc

0.70 1.00

Before working with the conditional probability, we may use some arithmetic to fill in some cellsof the table. If 70% work in the library, then the other 30% do not work in library. We may alsoconclude that the proportion who work in neither is 20%.

L Lc

D 0.10

Dc 0.20

0.70 0.30 1.00

To complete the table, we must use the conditional probability given in the problem. We aregiven the information that of those who work in the library, 30% do work in the dorms. Thisis the conditional probability, not the probability of those that work in both. It is a conditionalprobability because it only discusses those who work in the library, not everyone.

We are conditioning on people who work in the library (70% of the students). We are giventhat 30% of those 70% also work in the dorm. 30% of 70% may be found by multiplying theprobabilities, so (0.30)(0.70) = 0.21 work in the dorm and the library. Mathematically, we havebeen given P(D|L) = 0.3, and we have found P(D

L) = P(D|L)P(L) = (0.30)(0.70) = 0.21. We

may complete the table.

L Lc

D 0.21 0.10 0.31

Dc 0.49 0.20 0.69

0.70 0.30 1.00

Since we have computed the probabilities of all the outcomes, we may compute any probabilitiesfrom the table.

What is the probability a student works only in the library? 0.49

What is the probability a student works in the dorms? 0.31

What is the probability a student works either in the dorms or the library? (Remember thatin probability or means one or the other or both) 0.21 + 0.49 + 0.10 = 0.80

What is the probability a student works in neither the library nor the dorms? 0 .20

What is the probability a student that does not work in the library works in the dorms?0.10/0.30 = 1/3

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4 More difficult problems

In the dorm and library example, the conditional probability provided allowed the direct compu-tation of a cell probability. Problems involving conditional probability may be more difficult.

4.1 Applicants Example

Suppose a company is looking at applicants for a position. The position requires (A) experienceand (B) a masters degree. Suppose that 90% of the applicants have at least one of (A) or (B).Suppose further that, of those applicants with a least one of (A) or (B), 50% have both. Of thoseapplicants with exactly one of (A) or (B), 2/3 have (A).

We can construct a probability table and fill in one of the cells directly. Since 90% of theapplicants have at least one of (A) or (B), we may determine by the complement rule that 10%have neither. This results in the table

A Ac

B

Bc

0.101.00

To fill in the remainder of the table, we have to use the conditional probabilities. We are given that50% of the people with at least one have both. That is, 50% of the 90% that have at least one. Wecan therefore solve 50% of 90% is (0.5)(0.9) = 0.45 is the probability an individual has both.

A Ac

B 0.45

Bc 0.10

1.00

The remaining cell must be found be utilizing that 2/3 of the applicants with exactly one of (A)or (B) have (A). Although we are not given the proportion of individuals with exactly one directly,we can derive it to be 1 0.45 0.10 = 0.45 (the four core cells must sum to 1, and we have beengiven all cells except those corresponding to exactly one). Since 2/3 of those 0.45% have (A), weconclude P(A

Bc) = 0.30 (A

Bc is the only cell in exactly one that has A) and complete the

table through arithmetic.

A Ac

B 0.45 0.15 0.60

Bc 0.30 0.10 0.400.75 0.25 1.00

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4.2 Professors and their Stories Example

Here is another example. Suppose that 10% of professors can think of decent stories to go withtheir exam problems. Of those that can, 60% can think of funny stories. Suppose also that, ofthose professors whose stories are at least one of funny/decent, 75% of the professors write funnystories. Construct a probability table.

The first two probabilities allow us to complete part of the table.

D Dc

F 0.06

Fc 0.04

0.10 0.90 1.00

To fill in the remaining entries, we must use the information that of those professors whose storiesare at least one of funny/decent, 75% of the professors write funny stories. Unfortunately, withthe table so far we cannot compute the probability being at least one of funny or decent, because

we are missing the cell P(D

cF). Lets just fill in this unknown quantity with x and see if we canmake progress in solving for x.

D Dc

F 0.06 x

Fc 0.04

0.10 0.90 1.00

In terms of the unknown x, the proportion of professors whose stories are at least one of funny ordecent is 0.06 + 0.04 + x = 0.10 + x. We are given

P(F|at least one) = P(F

at least one)P(at least one)

= 0.06 + x0.10 + x

= 0.75

Solving for x, we find x = 0.06, so we may complete the table.

D Dc

F 0.06 0.06 0.12

Fc 0.04 0.84 0.88

0.10 0.90 1.00

As with all probability tables, once the table is completed we may compute any probability.

What proportion of professors write both decent and funny stories? 0.06

Of the professors with funny stories, what proportion write decent stories? 0 .06/0.12 = 0.5

Of those professors whose stories are exactly one of funny or decent, what proportion writefunny stories? 0.06/0.10 = 0.6

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5 Independence

If P(A|B) is different than P(A|Bc), than A and B are related. Return the newspaper example.We find P(A|M) = 1/6 while P(A|Mc) = 1/3. People who receive the morning paper are lesslikely to receive the afternoon people than people who do not receive the morning paper. Sincethe probability the event A occurs depends on whether or not M occurred, we call these eventsdependent. There are many events of this form. Any events with a causal relationship, forexample, will be of this form. Individuals with a Ph.D. are much more likely to get a facultyposition at a research university than individuals who do not have a Ph.D., the simple reason beinga Ph.D. is a requirement for such a faculty position. Indirect links between variables also createdependence. An elementary school student with large feet is more likely to read well. Foot size andreading ability have no causal link, but children with big feet tend to be older, and older childrentend to read better than younger children. All these types of relationships between variables resultin dependent events. Much of science is concerned with dependence. In medicine, for example, wecould like to know whether administering a treatment to a patient increases the patients chanceof recovery.

Occasionally two variables are unrelated. If you flip a fair coin twice, the coin has no memory.

Getting a head on the first flip has no effect on whether a head appears in the second flip. Fur-thermore, the probability of heads is always 0.5. IfH1 is the event heads on the first flip and H2is the event heads on the second flip, then P(H2|H1) = 0.5 and P(H2|Hc1) = 0.5. They are thesame probability. The result of the first flip is not affecting the probability of heads on the secondflip.

Events A and B where P(A|B) = P(A|Bc) are called independent events. Essentially, whetheror not B occurred does not affect the probability A occurred. The purpose of this section is toderive some special properties of independent events. With the information P(A|B) = P(A|Bc),we can derive some equivalent definitions of independence. Recall the Law of Total Probability

P(A) = P(A

B) + P(A

Bc)

Using the rule for intersections

P(A) = P(A|B)P(B) + P(A|Bc)P(Bc)

Now using the assumed information P(A|B) = P(A|Bc)

P(A) = P(A|B)P(B) + P(A|B)P(Bc)

P(A) = P(A|B) [P(B) + P(Bc)]

P(A) = P(A|B)

This equivalent definition of independence states that the overall probability of A, P(A), is thesame as the conditional probability of A given B, P(A|B) (one could also show P(A) = P(A|Bc).If you know the overall probability A, knowing whether or not B occurred does not change thatprobability. Another equivalent definition may be found by using the rule for intersections

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P(A

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

This third definition is the one used for mathematical purposes. The reason is that the conditionalprobability is a fraction, and thus there is the possibility of dividing by zero. The equation above

present no such problems.If P(A) and P(B) are both greater than 0, then all three definitions are equivalent. However,

if P(A) = 0 or P(B) = 0, then the first two definitions (P(A|B) = P(A|Bc) and P(A|B) = P(A))might result in division by zero problems. To verify independence in this course, you must showthe third definition, that P(A

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

For an example, reconsider the newspaper example. If you were asked are the events M andA independent?, you should check

0.10 = P(M

A) = P(M)P(A) = (0.6)(0.3) = 0.18

Since the equality required for independence does not hold, the events are dependent, not indepen-

dent. Alternatively, suppose a probability table has

A Ac

B 0.08 0.12 0.20

Bc 0.32 0.48 0.80

0.40 0.60 1.00

In this example,

0.08 = P(A

B) = P(A)P(B) = (0.40)(0.20) = 0.08

and thus the events are independent.P(A|Mc) are largely unrelated.

6 The Relationship between Disjoint and Independent

The short answer is that there is no relationship between two disjoint events and two independentevents. They are separate definitions, and they are useful in different scenarios.

Disjoint events are defined as events A and B such that A

B = . Disjoint events are usefulin that they simplify the rule for unions. In general the rule for unions states

P(A

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

B)

If A and B are disjoint, then P(A

B) = P() = 0 and thus the union rule simplifies to

P(A

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

Remember, you need the disjoint assumption to make the simplification (technically you only needP(A

B) = 0, which you can derive from the disjoint assumption).

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Independence is defined as P(A

B) = P(A)P(B), and thus results in a simplification of therule for intersections. In general,

P(A

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

If A and B are independent, this simplifies to

P(A

B) = P(A)P(B)

Remember you have to have to have the independence assumption to make this simplication.Independent events are sometimes disjoint and sometimes not, while disjoint events are some-

times independent and sometimes not. You have to know your purpose and check the appropriatedefinition.

7 Sample Questions

The following questions are divided into three levels of difficulty (all relative). The easiest

problems contain no conditional probabilities in the given information, but do ask you to computeconditional probabilities from the table. The medium problems have conditional probabilities inthe given information as well, so you must use the conditional probabilities to construct the table.The hardest problems often require some type of algebra to construct the probability table.

1. (Easiest) In a study of 3756 court cases, Kalven and Zeisell (1966) recorded the jury panelsdecision. In addition, they separately asked the judge how he or she would have decided thesame case if there were no jury panel. They found that

the judge would not have convicted in 17% of the cases

both the judge and the jury would have convicted in 64% of the cases

the judge and jury disagreed on whether to convict in 22% of the cases.a) Construct a probability table.

b) What is the probability that neither the judge nor the jury would convict?

c) Given the panel convicts, what is the probability the judge would also convict?

d) Are the events Judge convicts and Jury Panel convicts disjoint? Why or why not?

2. (Easiest) Suppose in a particular company, employees either use workstations or PCs (couldbe both or neither). The probability an employee uses a PC is 0.95 and the probability anemployee uses a workstation is 0.20. The probability an employee uses both is 0.19.

a) Construct a probability table

b) What is the probability an employee who does not use a PC uses a workstation?

c) What is the probability an employee uses exactly one of the machines?

d) Given an employee uses at least of the machines, what is the probability they use aworkstation?

e) Are the events use a PC and use a workstation disjoint? Why or why not?

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3. (Easiest) Suppose in a certain county 30% of the voters are Democrat (the remainder areRepublican). Suppose further that 50% of the voters are male, and that 40% of the votersare male Republicans.


b) Given a voter is male, what is the probability they are a Republican?

c) Given a voter is female, what is the probability they are a Republican?

d) Given a voter is Republican, what is the probability they are male?

4. (Easiest) A clinical trial explores the effect of two drugs, A and B. 50% of the patients in thestudy are assigned to drug A. 20% of the patients are assigned to drug B and are not cured.10% of the patients are assigned to drug A and are cured.

a) Construct a probability table.

b) What proportion of people in the study are cured?

c) Of those assigned to drug B, what proportion of patients are cured?

5. (Easiest) A muscle cell has 2 sites where electricity can conduct into the cell. Every time thebody intends to stimulate the muscle cell, an attempt is made to channel electricity througheach of the 2 sites.

The probability both sites conduct electricity is 0.4. The probability that exactly one siteconducts electricity is 0.4. Finally, the probability site 1 conducts electricity is 0.7.


b) What is the probability site 2 conducts electricity?

c) Given at least one of the sites conducts electricity, what is the probability both sites conductelectricity?

6. (Medium) Suppose a professor only writes two types of exams, easy or hard. Supposethat 90% of the exams are hard. There is a 80% chance that the first question on a hard examwill be difficult, and a 15% chance that the first question on an easy exam will be difficult.


b) On a given exam, what is the probability it is a hard exam that contains a non-difficultfirst question?

c) On a given exam, what is the probability that the first question will be non-difficult?

d) Suppose the first question on a given exam is non-difficult. Given this information, whatis the probability the exam is hard.

e) are the events easy exam and difficult first question independent?

7. (Medium) Suppose a couple of your friends go to a restaurant on either Tuesday or Saturdayeach week (not both). 30% of the time they go on Tuesday. On Tuesdays, the probability ofreceiving good service is 0.8. On Saturdays, the probability of receiving good service is only0.1.


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b) What is the probability that they went to the restaurant on Tuesday and received goodservice there?

c) What is the probability they received good service at the restaurant last week?

d) Suppose you dont know which day they went last week, but they tell you they receivedgood service at the restaurant. What is the probability that they went on Tuesday?

8. (Medium) Suppose at a particular park visitors can hike or raft. The probability that someonewill raft is 0.40. The probability that someone wont hike is 0.10. Given someone hikes, theprobability they raft is 0.40.


b) What is the probability of rafting?

c) What is the probability of not hiking?

d) What is the probability of someone participating in exactly one?

e) What is the probability of someone participating in at least one?

f) Of those who do not participate in both, what proportion participate in neither?9. (Medium) Suppose that 10% of all cases involving an insanity plea go to trial. Of those that

go to trial, the defendent is found guilty in 95% of the cases. Of those that do not go to trial,the defendant is found guilty (through a plea bargain) in 40% of the cases.


b) What proportion of defendants are found guilty?

c) Given a defendent is found guilty, what is the probability their case went to trial?

d) What proportion of cases that go to trial do not result in a guilty plea?

10. (Medium) A small university has an applicant pool that is 60% male. The university admits

25% of the male applicants and 75% of the female applicants.a) Construct a probability table showing the admissions at the university.

b) Of admitted students, what proportion are female?

c) Does this table provide evidence in favor of sexual discrimination in admissions at theuniversity? (this part will only be graded on effort, no answer will be marked wrong, but youhave to answer something)

11. (Hardest) Investment advisors might subscribe to the Wall Street Journal or Investors Busi-ness Daily. Suppose 0.80 subscribe to the WSJ. Of those that subscribe to the WSJ, 0.75 alsosubscribe to IBD. Also, 0.20 of those that subscribe to exactly one of the papers subscribe toIBD.


b) What is the probability an investment advisor subscribe to neither paper?

c) What proportion of investment advisors subscribe to IBD?

d) Given an investment advisor receives IBD, what is the probability they also receive theWSJ?

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e) Suppose an investment advisor receives at least one of the papers. What is the probabilitythey receive the WSJ?

12. (Hardest) Suppose there are two restaurants in a small town, Abelards Attic and BaltazarsBuffet. Suppose further that 20% of the people in the town dine at neither restaurant. Ofthose that go to at least one of the restaurants, 75% dine at both. Of those that dine at

exactly one of the restaurants, 75% dine at Abelards.


b) What proportion dine at Baltazars?

c) Of those that dine at Baltazars, what proportion dine at Abelards?

13. (Hardest) Let A and B be events. Suppose that P(A) = 0.3 and P(Bc|A) = 0.2. Supposefurther that given exactly one of two events occurs, 40% of the time it is B that occurred.


b) Given that at least one of the events occurred, what is the probability B occurred?

c) What is P(B|A)?14. (Hardest) Let A and B be events. Suppose P(A

Bc) = 0.4. Suppose also that, conditional

on exactly one of the events occuring, the probability A occurs is 0.8. Finally, suppose thatthe probability neither event occurs given B did not occur is 1/9.


b) What is the probability that at least one of the events occurs?

c) Given that at most one of the events occur, what is the probability A occurs?

15. (Hardest) Let A and B be events. Suppose that P(A) = 0.4. Given that at least one of theevents happens, there is a 1/3 chance that Ac occurs. Given that Bc occurs, there is a 1/5

chance that A occurs.a) Construct a probability table.

b) What is P(Ac)?

c) What is P(B|A)?

8 Solutions for Sample Problems

1. (Easiest) In a study of 3756 court cases, Kalven and Zeisell (1966) recorded the jury panelsdecision. In addition, they separately asked the judge how he or she would have decided thesame case if there were no jury panel. They found that

the judge would not have convicted in 17% of the cases

both the judge and the jury would have convicted in 64% of the cases

the judge and jury disagreed on whether to convict in 22% of the cases.


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J Jc

P 0.64 0.03 0.67

Pc 0.19 0.14 0.33

0.83 0.17 1.00

b) What is the probability that neither the judge nor the jury would convict? 0 .14

c) Given the panel convicts, what is the probability the judge would also convict? 0 .64/0.67 =0.9552

d) Are the events Judge convicts and Jury Panel convicts disjoint? Why or why not?They are not disjoint, they occur together with probability 0.64.

2. (Easiest) Suppose in a particular company, employees either use workstations or PCs (couldbe both or neither). The probability an employee uses a PC is 0.95 and the probability anemployee uses a workstation is 0.20. The probability an employee uses both is 0.19.


P C P C c

W S 0.19 0.01 0.20

W Sc 0.76 0.04 0.80

0.95 0.05 1.00

b) What is the probability an employee who does not use a PC uses a workstation? 0.01/0.05 =0.20

c) What is the probability an employee uses exactly one of the machines? 0.7 6 + 0.01 = 0.77

d) Given an employee uses at least of the machines, what is the probability they use aworkstation? 0.20/(0.19 + 0.01 + 0.76) = 0.2083

e) Are the events use a PC and use a workstation disjoint? Why or why not? No, theyoccur together with probability 0.19.

3. (Easiest) Suppose in a certain county 30% of the voters are Democrat (the remainder areRepublican). Suppose further that 50% of the voters are male, and that 40% of the votersare male Republicans.


Let D, R, M, and F be the events Democrat, Republican, Male and Female.

D R

M 0.10 0.40 0.50

F 0.20 0.30 0.500.30 0.70 1.00

b) Given a voter is male, what is the probability they are a Republican? 0.40/0.50=0.80

c) Given a voter is female, what is the probability they are a Republican? 0.30/0.50=0.60

d) Given a voter is Republican, what is the probability they are male? 0.40/0.70=4/7=0.5714

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4. (Easiest) A clinical trial explores the effect of two drugs, A and B. 50% of the patients in thestudy are assigned to drug A. 20% of the patients are assigned to drug B and are not cured.10% of the patients are assigned to drug A and are cured.


Assigned A Assigned BCured 0.10 0.30 0.40

Not cured 0.40 0.20 0.60

0.50 0.50 1.00

b) What proportion of people in the study are cured? 0.40

c) Of those assigned to drug B, what proportion of patients are cured? 0 .30/0.50 = 0.60

5. (Easiest) A muscle cell has 2 sites where electricity can conduct into the cell. Every time thebody intends to stimulate the muscle cell, an attempt is made to channel electricity througheach of the 2 sites.

The probability both sites conduct electricity is 0.4. The probability that exactly one siteconducts electricity is 0.4. Finally, the probability site 1 conducts electricity is 0.7.


Site 1 conducts Site 1 doesnt conduct

Site 2 conducts 0.40 0.10 0.50

Site 2 doesnt conduct 0.30 0.20 0.50

0.70 0.30 1.00

b) What is the probability site 2 conducts electricity? 0.50

c) Given at least one of the sites conducts electricity, what is the probability both sites conductelectricity? 0.40/0.80 = 0.5

6. (Medium) Suppose a professor only writes two types of exams, easy or hard. Supposethat 90% of the exams are hard. There is a 80% chance that the first question on a hard examwill be difficult, and a 15% chance that the first question on an easy exam will be difficult.


Let E be the event an exam is easy. Ec is the event that the exam is hard. Let D be theevent the first question is difficult.

E Ec

D 0.015 0.720 0.735

Dc 0.085 0.180 0.265

0.100 0.900 1.000

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b) On a given exam, what is the probability it is a hard exam that contains a non-difficultfirst question? 0.180

c) On a given exam, what is the probability that the first question will be non-difficult? 0.265

d) Suppose the first question on a given exam is non-difficult. Given this information, whatis the probability the exam is hard. 0.180/0.265 = 0.6792

e) The events are dependent, since

0.015 = P(E

D) = P(E)P(D) = (0.100)(0.735) = 0.0735

7. (Medium) Suppose a couple of your friends go to a restaurant on either Tuesday or Saturdayeach week (not both). 30% of the time they go on Tuesday. On Tuesdays, the probability ofreceiving good service is 0.8. On Saturdays, the probability of receiving good service is only0.1.


T Tc

GS 0.24 0.07 0.31

GSc 0.06 0.63 0.69

0.30 0.70 1.00

b) What is the probability that they went to the restaurant on Tuesday and received goodservice there? 0.24

c) What is the probability they received good service at the restaurant last week? 0 .31

d) Suppose you dont know which day they went last week, but they tell you they received goodservice at the restaurant. What is the probability that they went on Tuesday? 0.24/0.31 =0.7742

8. (Medium) Suppose at a particular park visitors can hike or raft. The probability that someonewill raft is 0.40. The probability that someone wont hike is 0.10. Given someone hikes, theprobability they raft is 0.40.


H Hc

R 0.36 0.04 0.40

Rc 0.54 0.06 0.60

0.90 0.10 1.00

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b) What is the probability of rafting? 0.40

c) What is the probability of not hiking? 0.10

d) What is the probability of someone participating in exactly one? 0 .54 + 0.04 = 0.58

e) What is the probability of someone participating in at least one? 0.3 6 + 0.5 4 + 0.04 = 0.94

f) Of those who do not participate in both, what proportion participate in neither? 0 .06/0.64 =0.0938

9. (Medium) Suppose that 10% of all cases involving an insanity plea go to trial. Of those thatgo to trial, the defendent is found guilty in 95% of the cases. Of those that do not go to trial,the defendant is found guilty (through a plea bargain) in 40% of the cases.


Let T be the event go to trial and G be the event found guilty.

T Tc

G 0.095 0.360 0.455

Gc 0.005 0.540 0.545

0.100 0.900 1.000

b) What proportion of defendants are found guilty? 0.455

c) Given a defendent is found guilty, what is the probability their case went to trial? 0.095/0.455=0.20

d) What proportion of cases that go to trial do not result in a guilty plea? 0.005/0.100=0.05

10. (Medium) A small university has an applicant pool that is 60% male. The university admits25% of the male applicants and 75% of the female applicants.

a) Construct a probability table showing the admissions at the university.

Male Female

Admitted 0.15 0.30 0.45

Not admitted 0.45 0.10 0.55

0.60 0.40 1.00

b) Of admitted students, what proportion are female? 0.30/0.45 = 2/3

c) Does this table provide evidence in favor of sexual discrimination in admissions at theuniversity? (this part will only be graded on effort, no answer will be marked wrong, but youhave to answer something)

Any answer will be graded as correct. Certainly a higher proportion of female applicants areadmitted than male applicants. This indicates there is some difference between how men andwomen work through way through the admission process. However, while this difference mightbe discrimination in admissions, it could also be caused by a third variable, here unobserved,that differentiates between men and women. The handout on Simpsons paradox discussesthis in more detail.

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11. (Hardest) Investment advisors might subscribe to the Wall Street Journal or Investors Busi-ness Daily. Suppose 0.80 subscribe to the WSJ. Of those that subscribe to the WSJ, 0.75 alsosubscribe to IBD. Also, 0.20 of those that subscribe to exactly one of the papers subscribe toIBD.


W SJ W SJ c

IBD 0.60 0.05 0.65

IB Dc 0.20 0.15 0.35

0.80 0.20 1.00

b) What is the probability an investment advisor subscribes to neither paper? 0.15

c) What proportion of investment advisors subscribe to IBD? 0.35

d) Given an investment advisor receives IBD, what is the probability they also receive the

WSJ? 0.60/0.65 = 0.9231e) Suppose an investment advisor receives at least one of the papers. What is the probabilitythey receive the WSJ? P(WSJ at least one) = 0.60/(0.60+0.20+0.05) = 0.60/0.85 = 0.7059

12. (Hardest) Suppose there are two restaurants in a small town, Abelards Attic and BaltazarsBuffet. Suppose further that 20% of the people in the town dine at neither restaurant. Ofthose that go to at least one of the restaurants, 75% dine at both. Of those that dine atexactly one of the restaurants, 75% dine at Abelards.


A A

c

B 0.60 0.05 0.65

Bc 0.15 0.20 0.35

0.75 0.25 1.00

b) What proportion dine at Baltazars? 0.65

c) Of those that dine at Baltazars, what proportion dine at Abelards? 0 .60/0.65 = 0.9231

13. (Hardest) Let A and B be events. Suppose that P(A) = 0.3 and P(Bc|A) = 0.2. Supposefurther that given exactly one of two events occurs, 40% of the time it is B that occurred.


A Ac

B 0.24 0.04 0.28

Bc 0.06 0.66 0.72

0.30 0.70 1.00

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b) Given that at least one of the events occurred, what is the probability B occurred?0.28/0.34 = 0.8235

c) What is P(B|A)? 0.24/0.30 = 0.8

14. (Hardest) Let A and B be events. Suppose P(A

Bc) = 0.4. Suppose also that, conditional

on exactly one of the events occuring, the probability A occurs is 0.8. Finally, suppose thatthe probability neither event occurs given B did not occur is 1/9.


A Ac

B 0.45 0.10 0.55

Bc 0.40 0.05 0.45

0.85 0.15 1.00

b) What is the probability that at least one of the events occurs? 0 .95

c) Given that at most one of the events occur, what is the probability A occurs? 0.40/0.55 =0.7273

15. (Hardest) Let A and B be events. Suppose that P(A) = 0.4. Given that at least one of theevents happens, there is a 1/3 chance that Ac occurs. Given that Bc occurs, there is a 1/5chance that A occurs.


A Ac

B 0.30 0.20 0.50

Bc 0.10 0.40 0.50

0.40 0.60 1.00

b) What is P(Ac)? 0.60

c) What is P(B|A)? 0.30/0.40 = 0.75

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