The probability P(A) of an event A is a measure of the likelihood that the event will occur on any trial. Sometimes partial information determines that an event C has occurred. Given this information, it may be necessary to reassign the likelihood for each event A. This leads to the notion of conditional probability. For a fixed conditioning event C, this assignment to all events constitutes a new probability measure which has all the properties of the original probability measure. In addition, because of the way it is derived from the original, the conditional probability measure has a number of special properties which are important in applications.
The original or prior probability measure utilizes all available information to make
probability assignments 
, etc., subject to the defining conditions (P1),
(P2), and (P3). The probability P(A) indicates the likelihood that event A
will occur on any trial.
Frequently, new information is received which leads to a reassessment of the likelihood of event A. For example
An applicant for a job as a manager of a service department is being interviewed. His résumé shows adequate experience and other qualifications. He conducts himself with ease and is quite articulate in his interview. He is considered a prospect highly likely to succeed. The interview is followed by an extensive background check. His credit rating, because of bad debts, is found to be quite low. With this information, the likelihood that he is a satisfactory candidate changes radically.
A young woman is seeking to purchase a used car. She finds one that appears to be an excellent buy. It looks “clean,” has reasonable mileage, and is a dependable model of a well known make. Before buying, she has a mechanic friend look at it. He finds evidence that the car has been wrecked with possible frame damage that has been repaired. The likelihood the car will be satisfactory is thus reduced considerably.
A physician is conducting a routine physical examination on a patient in her seventies. She is somewhat overweight. He suspects that she may be prone to heart problems. Then he discovers that she exercises regularly, eats a low fat, high fiber, variagated diet, and comes from a family in which survival well into their nineties is common. On the basis of this new information, he reassesses the likelihood of heart problems.
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 AC occurs. Effectively, this makes C a new basic space. The new unit of probability mass is P(C). How should the new probability assignments be made? One possibility is to make the new assignment to A proportional to the probability P(AC). These considerations and experience with the classical case suggests the following procedure for reassignment. Although such a reassignment is not logically necessary, subsequent developments give substantial evidence that this is the appropriate procedure.
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 
satisfies
the three defining properties (P1),
(P2), and (P3) for probability, so that for fixed C,
we have a new probability measure, with all
the properties of an ordinary probability measure.
Remark. When we write P(A|C) we are evaluating the likelihood of event A when it is known that event C has occurred. This is not the probability of a conditional event A|C. Conditional events have no meaning in the model we are developing.
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 
. Once the first
engine fails, added load is placed on the second, so that 
.
Now the second engine can fail only if the other has already failed. Thus
F2⊂F1 so 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:
Respond with an honest answer to the question.
Respond “yes” to the question, regardless of the truth in the matter.
Respond “no” regardless of the true answer.
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 P(A), P(B), and P(C) are determined by a chance mechanism (i.e., a fraction P(A) selected randomly are told to answer honestly, etc.). Let E be the event the reply is “yes.” We wish to calculate P(E|A), the probability the answer is “yes” given the response is honest.
SOLUTION
Since E=EA⋁B, we have
which may be solved algebraically to give
Suppose there are 250 subjects. The chance mechanism is such that P(A)=0.7, P(B)=0.14 and P(C)=0.16. There are 62 responses “yes,” which we take to mean P(E)=62/250. According to the pattern above
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 Ai,1≤i≤5 be the event the ith of these is hired (A1 is the event
Jim is hired, A3 is the event Richard is hired, etc.). Now 
(for each i) is
the probability that finalist i is in one of the combinations of three from five. Thus,
Jim's probability of being hired, before receiving the information about Richard, is
The information that Richard is one of those hired is information that the event A3 has occurred. Also, for any pair i≠j the number of combinations of three from five including these two is just the number of ways of picking one from the remaining three. Hence,
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 P(⋅). The following are easily derived from the definition of conditional probability and basic properties of the prior probability measure, and prove useful in a variety of problem situations.
(CP1) Product rule If P(ABCD)>0, then P(ABCD)=P(A)P(B|A)P(C|AB)P(D|ABC).
Derivation
The defining expression may be written in product form: P(AB)=P(A)P(B|A). Likewise
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 Gi,1≤i≤3 be the event the ith unit selected is good. Then G1G3=G1G2G3⋁G1G2cG3. By the product rule
