# Conditional probability  (Page 2/6)

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Definition. If C is an event having positive probability, the conditional probability of A , given C is

$P\left(A|C\right)=\frac{P\left(AC\right)}{P\left(C\right)}$

For a fixed conditioning event C , we have a new likelihood assignment to the event A . Now

$P\left(A|C\right)\ge 0,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(\Omega |C\right)=1,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(\phantom{\rule{0.166667em}{0ex}}\underset{j}{\overset{}{\bigvee }}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{A}_{j}|C\right)=\frac{P\left(\phantom{\rule{0.166667em}{0ex}},{\bigvee }_{j}^{},\phantom{\rule{0.166667em}{0ex}},\phantom{\rule{0.166667em}{0ex}},{A}_{j},C\right)}{P\left(C\right)}=\sum _{j}P\left({A}_{j}C\right)/P\left(C\right)=\sum _{j}P\left({A}_{j}|C\right)$

Thus, the new function $P\left(\phantom{\rule{0.166667em}{0ex}}\cdot \phantom{\rule{0.166667em}{0ex}}|C\right)$ 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 $P\left(A|C\right)$ 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.

## Conditional probabilities from joint frequency data

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

$P\left(G\right)=100/250,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(U\right)=150/250,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(Y\right)=\left(60+70\right)/250,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(YU\right)=60/250,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{etc.}$

Then

$P\left(Y|U\right)=\frac{P\left(YU\right)}{P\left(U\right)}=\frac{60/250}{150/250}=\frac{60}{150}$

Similarly, we can calculate

$P\left(N|U\right)=40/150,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(D|U\right)=50/150,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(Y|G\right)=70/100,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(N|G\right)=20/100,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(D|G\right)=10/100$

We may also calculate directly

$P\left(U|Y\right)=60/130,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(G|N\right)=20/60,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{etc.}$

Conditional probability often provides a natural way to deal with compound trials carried out in several steps.

## Jet aircraft with two engines

An aircraft has two jet engines. It will fly with only one engine operating. Let F 1 be the event one engine fails on a long distance flight, and F 2 the event the second fails. Experience indicates that $P\left({F}_{1}\right)=0.0003$ . Once the first engine fails, added load is placed on the second, so that $P\left({F}_{2}|{F}_{1}\right)=0.001$ . Now the second engine can fail only if the other has already failed. Thus ${F}_{2}\subset {F}_{1}$ so that

$P\left({F}_{2}\right)=P\left({F}_{1}{F}_{2}\right)=P\left({F}_{1}\right)P\left({F}_{2}|{F}_{1}\right)=3×{10}^{-7}$

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 extensivetreatment of the problem is given there.

## Responses to a sensitive question on a survey

In a survey, if answering “yes” to a question may tend to incriminate or otherwise embarrass the subject, the response given may beincorrect or misleading. Nonetheless, it may be desirable to obtain correct responses for purposes of social analysis. Thefollowing 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:

1. Respond with an honest answer to the question.
2. Respond “yes” to the question, regardless of the truth in the matter.
3. 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 $P\left(A\right)$ , $P\left(B\right)$ , and $P\left(C\right)$ are determined by a chance mechanism (i.e., a fraction $P\left(A\right)$ selected randomly are told to answer honestly, etc.). Let E be the event the reply is “yes.” We wish to calculate $P\left(E|A\right)$ , the probability the answer is “yes” given the response is honest.

SOLUTION

Since $E=EA\bigvee B$ , we have

$P\left(E\right)=P\left(EA\right)+P\left(B\right)=P\left(E|A\right)P\left(A\right)+P\left(B\right)$

which may be solved algebraically to give

$P\left(E|A\right)=\frac{P\left(E\right)-P\left(B\right)}{P\left(A\right)}$

Suppose there are 250 subjects. The chance mechanism is such that $P\left(A\right)=0.7$ , $P\left(B\right)=0.14$ and $P\left(C\right)=0.16$ . There are 62 responses “yes,” which we take to mean $P\left(E\right)=62/250$ . According to the pattern above

$P\left(E|A\right)=\frac{62/250-14/100}{70/100}=\frac{27}{175}\approx 0.154$

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20/(×-6^2)
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