Catalogue of Useful Matlab FilesThe probability
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Summary: The probability P(A) of an event A is a measure of the likelihood that the event will occur on any trial. New, but partial, information determines a conditioning event C , which may call for reassessing the likelihood of event A. For a fixed conditioning event C, this new assignment to all events constitutes a new probability measure. In addition, because of the way it is derived from the original, or prior, probability, the conditional probability measure has a number of special properties which are important in applications. Determination of the conditioning event is key.
The probability
The original or prior probability measure utilizes all available information to make
probability assignments
Frequently, new information is received which leads to a reassessment of the likelihood of event A. For example
New, but partial, information determines a conditioning
event C , which may call for reassessing the likelihood of event A. For one thing,
this means that A occurs iff the event
Definition. If C is an event having positive probability, the conditional probability of A, given C is
For a fixed conditioning event C, we have a new likelihood assignment to the event A. Now
Thus, the new function
Remark. When we write
A survey of student opinion on a proposed national health care program included 250 students, of whom 150 were undergraduates and 100 were graduate students. Their responses were categorized Y (affirmative), N (negative), and D (uncertain or no opinion). Results are tabulated below.
| Y | N | D | |
| U | 60 | 40 | 50 |
| G | 70 | 20 | 10 |
Suppose the sample is representative, so the results can be taken as typical of the student body. A student is picked at random. Let Y be the event he or she is favorable to the plan, N be the event he or she is unfavorable, and D is the event of no opinion (or uncertain). Let U be the event the student is an undergraduate and G be the event he or she is a graduate student. The data may reasonably be interpreted
Then
Similarly, we can calculate
We may also calculate directly
Conditional probability often provides a natural way to deal with compound trials carried out in several steps.
An aircraft has two jet engines. It will fly with only one engine operating. Let
F1 be the event one engine fails on a long distance flight, and F2 the event
the second fails. Experience indicates that
Thus reliability of any one engine may be less than satisfactory, yet the overall reliability may be quite high.
The following example is taken from the UMAP Module 576, by Paul Mullenix, reprinted in UMAP Journal, vol 2, no. 4. More extensive treatment of the problem is given there.
In a survey, if answering “yes” to a question may tend to incriminate or otherwise embarrass the subject, the response given may be incorrect or misleading. Nonetheless, it may be desirable to obtain correct responses for purposes of social analysis. The following device for dealing with this problem is attributed to B. G. Greenberg. By a chance process, each subject is instructed to do one of three things:
Let A be the event the subject is told to reply honestly, B be the event the
subject is instructed to reply “yes,” and C be the event the answer is to be “no.”
The probabilities
SOLUTION
Since
which may be solved algebraically to give
Suppose there are 250 subjects. The chance mechanism is such that
The formulation of conditional probability assumes the conditioning event C is well defined. Sometimes there are subtle difficulties. It may not be entirely clear from the problem description what the conditioning event is. This is usually due to some ambiguity or misunderstanding of the information provided.
Five equally qualified candidates for a job, Jim, Paul, Richard, Barry, and Evan, are identified on the basis of interviews and told that they are finalists. Three of these are to be selected at random, with results to be posted the next day. One of them, Jim, has a friend in the personnel office. Jim asks the friend to tell him the name of one of those selected (other than himself). The friend tells Jim that Richard has been selected. Jim analyzes the problem as follows.
ANALYSIS
Let
The information that Richard is one of those hired is information that the event A3
has occurred. Also, for any pair
The conditional probability
This is consistent with the fact that if Jim knows that Richard is hired, then there are two to be selected from the four remaining finalists, so that
Discussion
Although this solution seems straightforward, it has been challenged as being incomplete. Many feel that there must be information about how the friend chose to name Richard. Many would make an assumption somewhat as follows. The friend took the three names selected: if Jim was one of them, Jim's name was removed and an equally likely choice among the other two was made; otherwise, the friend selected on an equally likely basis one of the three to be hired. Under this assumption, the information assumed is an event B3 which is not the same as A3. In fact, computation (see Example 5, below) shows
Both results are mathematically correct. The difference is in the conditioning event, which corresponds to the difference in the information given (or assumed).
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In addition to its properties as a probability measure, conditional probability has
special properties which are consequences of the way it is related to the original
probability measure
(CP1) Product rule If
Derivation
The defining expression may be written in product form:
and
This pattern may be extended to the intersection of any finite number of events. Also, the events may be taken in any order.
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An electronics store has ten items of a given type in stock. One is defective. Four successive customers purchase one of the items. Each time, the selection is on an equally likely basis from those remaining. What is the probability that all four customes get good items?
SOLUTION
Let Ei be the event the ith customer receives a good item. Then the first chooses one of the nine out of ten good ones, the second chooses one of the eight out of nine goood ones, etc., so that
Note that this result could be determined by a combinatorial argument: under the assumptions, each combination of four of ten is equally likely; the number of combinations of four good ones is the number of combinations of four of the nine. Hence
Three items are to be selected (on an equally likely basis at each step) from ten, two of which are defective. Determine the probability that the first and third selected are good.
SOLUTION
Let
(CP2) Law of total probability
Suppose the class
Five cards are numbered one through five. A two-step selection procedure is carried out as follows.
Let Ai be the event the ith card is drawn on the first selection and let Bi
be the event the card numbered i is drawn on the second selection (from the box).
Determine
SOLUTION
From Example 4, we have
Now we can draw card five on the second selection only if it is selected on the
first drawing, so that
Also, since
We thus have
Occurrence of event B1 has no affect on the likelihood of the occurrence of A1. This condition is examined more thoroughly in the chapter on "Independence of Events".
Often in applications data lead to conditioning with respect to an event but the problem calls for “conditioning in the opposite direction.”
Students in a freshman mathematics class come from three different high schools. Their
mathematical preparation varies. In order to group them appropriately in class
sections, they are given a diagnostic test. Let Hi be the event that a
student tested is from high school i,
A student passes the exam. Determine for each i the conditional
probability
SOLUTION
Then
Similarly,
The basic pattern utilized in the reversal is the following.
(CP3) Bayes' rule
If
Such reversals are desirable in a variety of practical situations.
Begin with items in two lots:
One item is selected from lot 1 (on an equally likely basis); this item is added to lot 2; a selection is then made from lot 2 (also on an equally likely basis). This second item is good. What is the probability the item selected from lot 1 was good?
SOLUTION
Let G1 be the event the first item (from lot 1) was good, and G2 be the event the
second item (from the augmented lot 2) is good. We want to determine
By the law of total probability (CP2),
By Bayes' rule (CP3),
The calculations, as in Example 8, are simple but can be tedious. We have an m-procedure
called bayes to perform the calculations easily. The probabilities
>> PEA = [0.10 0.02 0.06];
>> PA = [0.2 0.5 0.3];
>> bayes
Requires input PEA = [P(E|A1) P(E|A2) ... P(E|An)]
and PA = [P(A1) P(A2) ... P(An)]
Determines PAE = [P(A1|E) P(A2|E) ... P(An|E)]
and PAEc = [P(A1|Ec) P(A2|Ec) ... P(An|Ec)]
Enter matrix PEA of conditional probabilities PEA
Enter matrix PA of probabilities PA
P(E) = 0.048
P(E|Ai) P(Ai) P(Ai|E) P(Ai|Ec)
0.1000 0.2000 0.4167 0.1891
0.0200 0.5000 0.2083 0.5147
0.0600 0.3000 0.3750 0.2962
Various quantities are in the matrices PEA, PA, PAE, PAEc, named above
The procedure displays the results in tabular form, as shown. In addition, the various quantities are in the workspace in the matrices named, so that they may be used in further calculations without recopying.
The following variation of Bayes' rule is applicable in many practical situations.
(CP3*) Ratio form of Bayes' rule
The left hand member is called the posterior odds, which is the odds after
knowledge of the occurrence of the conditioning event. The second fraction in the right
hand member is the prior odds, which is the odds before knowledge of the
occurrence of the conditioning event C. The first fraction in the right hand member
is known as the likelihood ratio. It is the ratio of the probabilities (or likelihoods)
of C for the two different probability measures
As a part of a routine maintenance procedure, a computer is given a performance test. The machine seems to be operating so well that the prior odds it is satisfactory are taken to be ten to one. The test has probability 0.05 of a false positive and 0.01 of a false negative. A test is performed. The result is positive. What are the posterior odds the device is operating properly?
SOLUTION
Let S be the event the computer is operating satisfactorily and let T be the
event the test is favorable. The data are
The following property serves to establish in the chapters on "Independence of Events" and "Conditional Independence" a number of important properties for the concept of independence and of conditional independence of events.
(CP4) Some equivalent conditions
If
where
Because of the role of this property in the theory of independence and conditional independence, we examine the derivation of these results.
VERIFICATION of (CP4)
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A number of important and useful propositons may be derived from these.
VERIFICATION — Exercises (see problem set)
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Suppose conditioning by the event C has occurred. Additional
information is then received that event D has occurred. We have a new conditioning event
Basic result:
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This work is licensed by Paul E Pfeiffer under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Daniel Williamson on Sep 18, 2009 12:18 pm -0500.