Chapter 4. Probability Rules and Conditional Probability

General Methods

In the mathematical definition of probability, an arbitrary event $A$ is merely some subset of the sample space $S$. The following rules hold:

  1. $P(S) = 1$

  2. For any event MATH

It is also obvious from our definitions in Chapter 2 that if $A$ and $B$ are two events with $A \subseteq B$ (that is, all of the simple events in $A$ are also in $B$), then $P(A) \leq P(B)$.

It is often helpful to use elementary ideas of set theory in dealing with probability; as we show in this chapter, this allows certain rules or propositions about probability to be proved. Before going on to specific rules, we'll review Venn diagrams for sets. In the drawings below, think of all points in $S$ being contained in the rectangle, and those points where particular events occur being contained in circles. We begin by considering the union $(A\cup B)$, intersection $(A\cap B)$ and complement $(\bar{A})$ of sets (see Figure aunionb). At the URL http://stat-www.berkeley.edu/users/stark/Java/Venn.htm, there is an interesting applet which allows you to vary the area of the intersection and construct Venn diagrams for a variety of purposes.
aunionb.eps
Top panel: $A\cup B$ means $A$ OR $B$ (or possibly both) occurs.$A\cup B$ is shaded.
Middle panel: $A\cap B$ (usually written as $AB$ in probability) means $A$ and $B$ both occur. $A\cap B$ is shaded
Lower panel: $\bar{A}$ means $A$ does not occur. $\bar{A}$ is shaded

Example:

Suppose for students finishing 2A Math that 22% have a math average $\geq$ 80%, 24% have a STAT 230 mark $\geq$ 80%, 20% have an overall average $\geq$ 80%, 14% have both a math average and STAT 230 $\geq$ 80%, 13% have both an overall average and STAT 230 $\geq$ 80%, 10% have all 3 of these averages $\geq$ 80%, and 67% have none of these 3 averages $\geq$ 80%. Find the probability a randomly chosen math student finishing 2A has math and overall averages both $\geq$ 80% and STAT 230 $<$ 80%.


Solution: When using rules of probability it is generally helpful to begin by labeling the events of interest.

MATH $A$ = MATH
$B$ = MATH
$C$ = MATH

In terms of these symbols, we are given MATH, and MATH. We are asked to find $P(AB\bar{C})$, the shaded region in Figure maexample Filling in this information on a Venn diagram, in the order indicated by (1), (2), (3), etc.


maexample.jpg
Venn Diagram for Math Averages Example

(1) given
(2) $P(AC)-P(ABC)$
(3) $P(BC)-P(ABC)$
(4) $P(C)-P(AC)-.03$
(5) unknown
(6) $P(A)-P(AC)-x$
(7) $P(B)-P(BC)-x$
(8) given

(Usually, we start filling in at the centre and work our way out.)
Adding all probabilities and noting that $P(S)=1$, we can solve to get $x=.06=P(AB\bar{C})$.

Problems:

  1. In a typical year, 20% of the days have a high temperature $>22^{o}$C. On 40% of these days there is no rain. In the rest of the year, when the high temperature $\leq22^{o}$C, 70% of the days have no rain. What percent of days in the year have rain and a high temperature $\leq22^{o}$C?

  2. According to a survey of people on the last Ontario voters list, 55% are female, 55% are politically to the right, and 15% are male and politically to the left. What percent are female and politically to the right? Assume voter attitudes are classified simply as left or right.

Rules for Unions of Events

In addition to the two rules which govern probabilities listed in Section 4.1, we have the following

  1. (probability of unions)MATH
    This can be obtained by using a Venn diagram. Each point in $A\cup B$ must be counted once. Since points in $AB$ are counted twice - once in $P(A)$ and once in $P(B)$ - they need to be subtracted once.

MATH (see Figure union)
aunionbunionc.eps
The union $A\cup B\cup C$

MATH (where the subscripts are all different)
This generalization is seldom used in Stat 230.

Definition

Events $A$ and $B$ are mutually exclusive if $AB=\phi$ (the null set)

Since mutually exclusive events $A$ and $B$ have no common points, $P(AB)=0$.

In general, events MATH are mutually exclusive if $A_{i}A_{j}=\phi$ for all $i\neq j$. This means that there is no chance of 2 or more of these events occurring together. For example, if a die is rolled twice, the events
MATH are mutually exclusive. In the case of mutually exclusive events, rule 3 above simplifies to rule 4 below.


Exercise:

Think of some pairs of events and classify them as being mutually exclusive or not mutually exclusive.

  1. (unions of mutually exclusive events)

    1. Let $A$ and $B$ be mutually exclusive events. Then MATH

    2. In general, let MATH be mutually exclusive.
      Then MATH
      Proof: Use rule 3 above

  2. (probability of complements) MATH

    Proof:

    $A$ and $\bar{A}$ are mutually exclusive so
    MATH
    But MATH
    MATH
    This result is useful whenever $P(\bar{A})$ is easier to obtain than $P(A)$.

Example: Two ordinary dice are rolled. Find the probability that at least one of them turns up a 6.

Solution 1: Let $A$ = { 6 on the first die }, $B$ = { 6 on the second die } and note (rule 3) that MATH Solution 2: MATH


Example: Roll a die 3 times. Find the probability of getting at least one 6.


Solution 1:

Let MATH. Then MATH.

Using counting arguments, there are 6 outcomes on each roll, so $S$ has MATH points. For $\bar{A}$ to occur we can't have a 6 on any roll. Then $\bar{A}$ can occur in MATH ways.

MATH

Solution 2: Can you spot the flaw in this?
Let $A$ = {6 occurs on MATH roll}
$B$ = {6 occurs on MATH roll}
$C$ = {6 occurs on MATH roll}.
\mbox{Then} MATH
= $P(A)+P(B)+P(C)$
= MATH

You should have noticed that $A,B$, and $C$ are not mutually exclusive events, so we should have used
MATH
Each of $AB,AC$, and $BC$ occurs only once in the 36 point sample space for those two rolls.
MATH


Note: Rules 3, 4, and (indirectly) 5 link the concepts of addition, unions and complements. The next segment will consider intersection, multiplication of probabilities, and a concept known as independence. Making these linkages will make problem solving and the construction of probability models easier.

Problems:

  1. Let $A,B,$ and $C$ be events for which

MATH

(a) Find the largest possible value for $P(A\cup B\cup C)$

(b) For this largest value to occur, are the events $A$ and $C$ mutually exclusive, not mutually exclusive, or is this unable to be determined?

  1. Prove that MATH $\overline{B})$ for arbitrary events $A$ and $B$ in $S$.

Intersections of Events and Independence

Dependent and Independent Events:

Consider these two groups of pairs of events.

Group 1
$A$ = {airplane engine fails in flight}
$B$ = {airplane reaches its destination safely}
or (when a fair coin is tossed twice)
$A$ = {$H$ is on 1st toss}
$B$ = {$H$ on both tosses}.
Group 2
$A$ = {a coin toss shows heads}
$B$ = {a bridge hand has 4 aces}.
or (when a fair coin is tossed twice)
$A$ = {$H$ on 1st toss}
$B$ = {$H$ on 2nd toss}

What do the pairs $A,B$ in each group have in common? In group 1 the events are related so that the occurrence of $A$ affects the chances of $B$ occurring. In group 2, whether $A$ occurs or not has no effect on $B$'s occurrence.

We call the pairs in group 1 dependent events, and those in group 2 independent events. We formalize this concept in the mathematical definition which follows.

Definition

Events $A$ and $B$ are independent if and only if $P(AB)=P(A)P(B)$. If they are not independent, we call the events dependent.

If two events are independent, then the ``size'' of their intersection as measured by the probability measure is required to be the product of the individual probabilities. This means, of course, that the intersection must be non-empty, and so the events are not mutually exclusive. For example in the Venn diagram depicted in Figure independent, $P(A)=0.3,P(B)=0.4$ and $P(AB)=0.12$ so in this case the two events are independent.


independent.jpg
Independent events $A,B$

For another example, suppose we toss a fair coin twice. Let $A$ = {head on 1st toss} and $B$ = {head on 2nd toss}. Clearly $A$ and $B$ are independent since the outcome on each toss is unrelated to other tosses, so MATH.

However, if we roll a die once and let $A$ = {the number is even} and $B$ = {number $>3$} the events will be dependent since MATH (Rationale: $B$ only happens half the time. If $A$ occurs we know the number is 2, 4, or 6. So $B$ occurs $\frac{2}{3}$ of the time when $A$ occurs. The occurrence of $A$ does affect the chances of $B$ occurring so $A$ and $B$ are not independent.)

When there are more than 2 events, the above definition generalizes to:

Definition

The events MATH are independent if and only ifMATH for all sets MATH of distinct subscripts chosen from $(1,2,\cdots,n)$

For example, for $n=3$, we need
MATH
and MATH

Technically, we have defined ``mutually independent'' events, but we will shorten the name to ``independent'' to reduce confusion with ``mutually exclusive.''

The definition of independence works two ways. If we can find $P(A),P(B)$, and $P(AB)$ then we can determine whether $A$ and $B$ are independent. Conversely, if we know (or assume) that $A$ and $B$ are independent, then we can use the definition as a rule of probability to calculate $P(AB)$. Examples of each follow.


Example: Toss a die twice. Let $A$ = {first toss is a 3} and $B$ = {the total is 7}. Are $A$ and $B$ independent? (What do you think?) Using the definition to check, we get MATH (points (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) give a total of 7) and $P(AB)=\frac{1}{36}$ (only the point (3,4) makes $AB$ occur).
Therefore, $\ P(AB)=P(A)P(B)$ and so $A$ and $B$ are independent events.

Now suppose we change $B$ to the event {total is 8}.
Then MATH
and consequently $A$ and $B$ are dependent events.


This example often puzzles students. Why are they independent if $B$ is a total of 7 but dependent for a total of 8? The key is that regardless of the first toss, there is always one number on the 2nd toss which makes the total 7. Since the probability of getting a total of 7 started off being MATH, the outcome of the 1st toss doesn't affect the chances. However, for any total other than 7, the outcome of the 1st toss does affect the chances of getting that total (e.g., a first toss of 1 guarantees the total cannot be 8).



Example: A (pseudo) random number generator on the computer can give a sequence of independent random digits chosen from $S=\{0,1,\dots,9\}$. This means that (i) each digit has probability of $\frac{1}{10}$ of being any of $0,1,\dots,9$, and (ii) the outcomes for the different trials are independent of one another. We call this type of setting an "experiment with independent trials". Determine the probability that

  1. in a sequence of 5 trials, all the digits generated are odd

  2. the number 9 occurs for the first time on trial 10.


Solution:

  1. Define the events $A_{i}$ = {digits from trial $i$ is odd}, $i=1,\dots,5$.
    Then MATH since the $A_{i}$'s are mutually independent. Since $P(A_{i})=.5$, we get $P$(all digits are odd) = $.5^{5}$.

  2. Define events $A_{i}$ = {9 occurs on trial $i$}, $i=1,2,\dots$. Then we want MATH because the $A_{i}$'s are independent, and MATH.

Note: We have used the fact here that if $A$ and $B$ are independent events, then so are $\bar{A}$ and $B$. To see this note that MATH


Note: We have implicitly assumed independence of events in some of our earlier probability calculations. For example, suppose a coin is tossed 3 times, and we consider the sample space MATH Assuming that the outcomes on the three tosses are independent, and that MATH on any single toss, we get that MATH Similarly, all the other simple events have probability $\frac{1}{8}$. Note that in earlier calculations we assumed this was true without thinking directly about independence. However, it is clear that if somehow the 3 tosses were not independent then it might be a bad idea to assume each simple event had probability $\frac{1}{8}$. (For example, instead of heads and tails, suppose $H$ stands for "rain" and $T$ stands for "no rain" on a given day; now consider 3 consecutive days. Would you want to assign a probability of $\frac{1}{8}$ to each of the 8 simple events?)


Note: The definition of independent events can thus be used either to check for independence or, if events are known to be independent, to calculate $P(AB)$. Many problems are not obvious, and scientific study is needed to determine if two events are independent. For example, are the events $A$ and $B$ independent if, for a random child living in a country,
$A$ = {live within 5 km. of a nuclear power plant}
$B$ = {a child has leukemia}?
Such problems, which are of considerable importance, can be handled by methods in later statistics courses.





Problems:

  1. A weighted die is such that MATH and $P(6)=0.3.$

  2. If the die is thrown twice what is the probability the total is 9?

  3. If a die is thrown twice, and this process repeated 4 times, what is the probability the total will be 9 on exactly 1 of the 4 repetitions?

    4.3.2

  1. Suppose among UW students that 15% speaks French and 45% are women. Suppose also that 20% of the women speak French. A committee of 10 students is formed by randomly selecting from UW students. What is the probability there will be at least 1 woman and at least 1 French speaking student on the committee?

  2. Prove that $\overline{A}$ and $\overline{B}$ are independent events if and only if $\overline{A}$ and $B$ are independent.

Conditional Probability

In many situations we may want to determine the probability of some event $A$, while knowing that some other event $B$ has already occurred. For example, what is the probability a randomly selected person is over 6 feet tall, given that they are female? Let the symbol $P(A|B)$ represent the probability that event $A$ occurs, when we know that $B$ occurs. We call this the conditional probability of $A$ given $B$. While we will give a definition of $P(A|B)$, let's first consider an example we looked at earlier, to get some sense of why $P(A|B)$ is defined as it is.


Example: Suppose we roll a die once. Let $A$ = {the number is even} and $B$ = {number $>3$}. If we know that $B$ occurs, that tells us that we have a 4, 5, or 6. Of the times when $B$ occurs, we have an even number $\frac{2}{3}$ of the time. So $P(A|B)=\frac{2}{3}$. More formally, we could obtain this result by calculating $\frac{P(AB)}{P(B)}$, since MATH and $P(B)=\frac{3}{6}$.

Definition

the conditional probability of event $A$, given event $B$, isMATH

Note: If $A$ and $B$ are independent,MATH This makes sense, and can be taken as an equivalent definition of independence; that is, $A$ and $B$ are independent iff $P(A|B)=P(A)$. You should investigate the behaviour of the conditional probabilities as we move the events around on the web-site http://stat-www.berkeley.edu/%7Estark/Java/Venn3.htm.


Example: If a fair coin is tossed 3 times, find the probability that if at least 1 Head occurs, then exactly 1 Head occurs.



Solution: Define the events $A$ = {1 Head}, $B$ = {at least 1 Head}. What we are being asked to find is $P(A|B)$. This equals $P(AB)/P(B)$, and so we find MATH and MATH using either the sample space with 8 points, or the fact that the 3 tosses are independent. Thus, MATH


Example: The probability a randomly selected male is colour-blind is .05, whereas the probability a female is colour-blind is only .0025. If the population is 50% male, what is the fraction that is colour-blind?



Solution: Let
$C$ = {person selected is colour-blind}
$M$ = {person selected is male}
$F$ = {person selected is female}

We are asked to find $P(C)$. We are told that MATH To get $P(C)$ we can therefore use
MATH

Multiplication and Partition Rules

The preceding example suggests two more probability rules, which turn out to be extremely useful. They are based on breaking events of interest into pieces.

  1. Multiplication Rules
    Let     $A,B,C,D,\dots $ be arbitrary events in a sample space. ThenMATH
    and so on.

    Proof:
    The first rule comes directly from the definition $P(B|A)$. The right hand side of the second rule equals (assuming $P(AB)\neq0$) MATH and so on.

Partition Rule
Let $A_{1},\dots,A_{k}$ be a partition of the sample space $S$ into disjoint (mutually exclusive) events such thatMATH Let $B$ be an arbitrary event in $S$. Then MATH Proof: Look at a Venn diagram to see that MATH are mutually exclusive, with MATH.

Example: In an insurance portfolio 10% of the policy holders are in Class $A_{1}$ (high risk), 40% are in Class $A_{2}$ (medium risk), and 50% are in Class $A_{3}$ (low risk). The probability a Class $A_{1}$ policy has a claim in a given year is .10; similar probabilities for Classes $A_{2}$ and $A_{3}$ are .05 and .02. Find the probability that if a claim is made, it is for a Class $A_{1}$ policy.
Solution: For a randomly selected policy, let
$B$ = {policy has a claim }
$A_{i}$ = {policy is of Class $A_{i}$}, $i=1,2,3$

We are asked to find $P(A_{1}|B)$. Note that MATH and that MATH We are told thatMATH and that MATH ThusMATH

Therefore $P(B)=.04$ and MATH.

Tree Diagrams

Tree diagrams can be a useful device for keeping track of conditional probabilities when using multiplication and partition rules. The idea is to draw a tree where each path represents a sequence of events. On any given branch of the tree we write the conditional probability of that event given all the events on branches leading to it. The probability at any node of the tree is obtained by multiplying the probabilities on the branches leading to the node, and equals the probability of the intersection of the events leading to it.

For example, the immediately preceding example could be represented by the tree in Figure treetest. Note that the probabilities on the terminal nodes must add up to 1.
treetest.eps

Here is another example involving diagnostic tests for disease. See if you can represent the problem by a tree.
Example. Testing for HIV
Tests used to diagnose medical conditions are often imperfect, and give false positive or false negative results, as described in Problem 2.6 of Chapter 2. A fairly cheap blood test for the Human Immunodeficiency Virus (HIV) that causes AIDS (Acquired Immune Deficiency Syndrome) has the following characteristics: the false negative rate is 2% and the false positive rate is 0.5%. It is assumed that around .04% of Canadian males are infected with HIV.

Find the probability that if a male tests positive for HIV, he actually has HIV.

Solution: Suppose a male is randomly selected from the population, and define the events
$A$ = {person has HIV}
$B$ = {blood test is positive}

We are asked to find $P(A|B)$. From the information given we know that MATH Therefore we can find MATH and MATH Thus, if a randomly selected male tests positive, there is still only a small probability (.0727) that they actually have HIV!

Exercise: Try to explain in ordinary words why this is the case.

Note: Bayes Theorem

By using the definition of conditional probability and the multiplication rule, we get that MATH This result is called Bayes Theorem, after a mathematician Note_1 who proved it in the 1700's. It is a very trivial theorem, but it has inspired approaches to problems in statistics and other areas such as machine learning, classification and pattern recognition. In these areas the term "Bayesian methods" is often used.

Problems:

  1. If you take a bus to work in the morning there is a 20% chance you'll arrive late. When you go by bicycle there is a 10% chance you'll be late. 70% of the time you go by bike, and 30% by bus. Given that you arrive late, what is the probability you took the bus?

  2. A box contains 4 coins -- 3 fair coins and 1 biased coin for which $P$(heads) = .8. A coin is picked at random and tossed 6 times. It shows 5 heads. Find the probability this coin is fair.

  3. At a police spot check, 10% of cars stopped have defective headlights and a faulty muffler. 15% have defective headlights and a muffler which is satisfactory. If a car which is stopped has defective headlights, what is the probability that the muffler is also faulty?

Problems on Chapter 4

  1. If $A$ and $B$ are mutually exclusive events with $P(A)=0.25$ and $P(B)=0.4$, find the probability of each of the following events: MATH

  2. Three digits are chosen at random with replacement from $0,1,...,9 $; find the probability of each of the following events.

    $C$: ``the digits are all nonzero'';
    $A$: ``all three digits are the same''; $D$: ``the digits all exceed 4'';
    $B$: ``all three digits are different''; $E$ ``digits all have the same parity (all odd or all even)''.

    Then find the probability of each of the following events, which are combinations of the previous five events: MATH Show the last two of these events in Venn diagrams.

  3. Let $A$ and $B$ be events defined on the same sample space, with $P(A)=0.3$, $P(B)=0.4$ and $P(A|B)=0.5$. Given that event $B$ does not occur, what is the probability of event $A$?

  4. A die is loaded to give the probabilities:

    number 1 2 3 4 5 6
    probability .3 .1 .15 .15 .15 .15

    The die is thrown 8 times. Find the probability

    1. 1 does not occur

    2. 2 does not occur

    3. neither 1 nor 2 occurs

    4. both 1 and 2 occur.

  5. Events $A$ and $B$ are independent with $P(A)=.3$ and $P(B)=.2$. Find $P(A\cup B)$.

  6. Students $A,B$ and $C$ each independently answer a question on a test. The probability of getting the correct answer is .9 for $A$, .7 for $B$ and .4 for $C$. If 2 of them get the correct answer, what is the probability $C$ was the one with the wrong answer?

  7. 70% of the customers buying at a certain store pay by credit card. Find the probability

    1. 3 out of 5 customers pay by credit card

    2. the 5th customer is the 3rd one to pay by credit card.

  8. Let $E$ and $F$ be independent with $E=A\cup B$ and $F=AB$. Prove that either $P(AB)=0$ or else MATH.

  9. In a large population, people are one of 3 genetic types $A,B$ and $C$: 30% are type $A$, 60% type $B$ and 10% type $C$. The probability a person carries another gene making them susceptible for a disease is .05 for $A$, .04 for $B$ and .02 for $C$. If ten unrelated persons are selected, what is the probability at least one is susceptible for the disease?

  10. Two baseball teams play a best-of-seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on $A$'s field, the next three games on $B$'s field, and the last two on $A$'s field. The probability that $A$ wins a game is 0.7 at home and 0.5 away. Find the probability that:

  11. A population consists of $F$ females and $M$ males; the population includes $f$ female smokers and $m$ male smokers. An individual is chosen at random from the population. If $A$ is the event that this individual is female and $B$ is the event he or she is a smoker, find necessary and sufficient conditions on $f$, $m$, $F$ and $M$ so that $A$ and $B$ are independent events.

  12. An experiment has three possible outcomes $A$, $B$ and $C$ with respective probabilities $p$, $q$ and $r$, where $p+q+r=1$. The experiment is repeated until either outcome $A$ or outcome $B$ occurs. Show that $A$ occurs before $B$ with probability $p/(p+q)$.

  13. In the game of craps, a player rolls two dice. They win at once if the total is 7 or 11, and lose at once if the total is 2, 3, or 12. Otherwise, they continue rolling the dice until they either win by throwing their initial total again, or lose by rolling 7.
    Show that the probability they win is 0.493.
    (Hint: You can use the result of Problem 4.12)

  14. A researcher wishes to estimate the proportion $p$ of university students who have cheated on an examination. The researcher prepares a box containing 100 cards, 20 of which contain Question A and 80 Question B.
    Question A: Were you born in July or August?
    Question B: Have you ever cheated on an examination?
    Each student who is interviewed draws a card at random with replacement from the box and answers the question it contains. Since only the student knows which question he or she is answering, confidentiality is assured and so the researcher hopes that the answers will be truthful Note_2 . It is known that one-sixth of birthdays fall in July or August.

  15. Diagnostic tests. Recall the discussion of diagnostic tests in Problem 2.6 for Chapter 2. For a randomly selected person let $D=$ `person has the disease' and $R=$ `the test result is positive'. Give estimates of the following probabilities: $P(R|D)$, $P(R|\bar{D})$, $P(R)$.

  16. Slot machines. Standard slot machines have three wheels, each marked with some number of symbols at equally spaced positions around the wheel. For this problem suppose there are 10 positions on each wheel, with three different types of symbols being used: flower, dog, and house. The three wheels spin independently and each has probability 0.1 of landing at any position. Each of the symbols (flower, dog, house) is used in a total of 10 positions across the three wheels. A payout occurs whenever all three symbols showing are the same.

  17. Spam detection 1. Many methods of spam detection are based on words or features that appear much more frequently in spam than in regular email. Conditional probability methods are then used to decide whether an email is spam or not. For example, suppose we define the following events associated with a random email message.
    Spam = "Message is spam"
    Not Spam = "Message is not spam ("regular")"
    A = "Message contains the word Viagra"


    If we know the values of the probabilities $P$(Spam), $P(A|$ Spam) and $P(A|$ Not Spam), then we can find the probabilities $P$(Spam$|A)$ and $P$(Not Spam$|A)$.

  18. Spam detection 2. The method in part (b) of the preceding question would only filter out 20% of Spam messages. (Why?) To increase the probability of detecting spam, we can use a larger set of email "features"; these could be words or other features of a message which tend to occur with much different probabilities in spam and in regular email. (From your experience, what might be some useful features?) Suppose we identify $n$ binary features, and define events

    $A_{i}$ = feature $i$ appears in a message.

    We will assume that $A_{1},\dots,A_{n}$ are independent events, given that a message is spam, and that they are also independent events, given that a message is regular.

    Suppose $n=3$ and that
    $P(A_{1}|$Spam) = .2 $P(A_{1}|$Not Spam) = .005
    $P(A_{2}|$Spam) = .1 $P(A_{2}|$Not Spam) = .004
    $P(A_{3}|$Spam) = .1 $P(A_{3}|$Not Spam) = .005

    Assume as in the preceding question that $P$(Spam) = .5.

  19. Online fraud detection. Methods like those in problems 4.17 and 4.18 are also used in monitoring events such as credit card transactions for potential fraud. Unlike the case of spam email, however, the fraction of transactions that are fraudulent is usually very small. What we hope to do in this case is to "flag" certain transactions so that they can be checked for potential fraud, and perhaps to block (deny) certain transactions. This is done by identifying features of a transaction so that if $F$ = "transaction is fraudulent", thenMATH is large.