# Conditional probability  (Page 5/6)

 Page 5 / 6

## Reversal of conditioning

Students in a freshman mathematics class come from three different high schools. Their mathematical preparation varies. In order to group them appropriately in classsections, they are given a diagnostic test. Let H i be the event that a student tested is from high school i , $1\le i\le 3$ . Let F be the event the student fails the test. Suppose data indicate

$P\left({H}_{1}\right)=0.2,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({H}_{2}\right)=0.5,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({H}_{3}\right)=0.3,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(F|{H}_{1}\right)=0.10,\phantom{\rule{10pt}{0ex}}P\left(F|{H}_{2}\right)=0.02,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(F|{H}_{3}\right)=0.06$

A student passes the exam. Determine for each i the conditional probability $P\left({H}_{i}|{F}^{c}\right)$ that the student is from high school i .

SOLUTION

$P\left({F}^{c}\right)=P\left({F}^{c}|{H}_{1}\right)P\left({H}_{1}\right)+P\left({F}^{c}|{H}_{2}\right)P\left({H}_{2}\right)+P\left({F}^{c}|{H}_{3}\right)P\left({H}_{3}\right)=0.90\cdot 0.2+0.98\cdot 0.5+0.94\cdot 0.3=0.952$

Then

$P\left({H}_{1}|{F}^{c}\right)=\frac{P\left({F}^{c}{H}_{1}\right)}{P\left({F}^{c}\right)}=\frac{P\left({F}^{c}|{H}_{1}\right)P\left({H}_{1}\right)}{P\left({F}^{c}\right)}=\frac{180}{952}=0.1891$

Similarly,

$P\left({H}_{2}|{F}^{c}\right)=\frac{P\left({F}^{c}|{H}_{2}\right)P\left({H}_{2}\right)}{P\left({F}^{c}\right)}=\frac{590}{952}=0.5147\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{10pt}{0ex}}P\left({H}_{3}|{F}^{c}\right)=\frac{P\left({F}^{c}|{H}_{3}\right)P\left({H}_{3}\right)}{P\left({F}^{c}\right)}=\frac{282}{952}=0.2962$

The basic pattern utilized in the reversal is the following.

(CP3) Bayes' rule If $E\subset \phantom{\rule{0.166667em}{0ex}}\underset{i=1}{\overset{n}{\bigvee }}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{A}_{i}\phantom{\rule{0.277778em}{0ex}}$ (as in the law of total probability), then

$P\left({A}_{i}|E\right)=\frac{P\left({A}_{i}E\right)}{P\left(E\right)}=\frac{P\left(E|{A}_{i}\right)P\left({A}_{i}\right)}{P\left(E\right)}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1\le i\le n\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{The}\phantom{\rule{4.pt}{0ex}}\text{law}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}\text{total}\phantom{\rule{4.pt}{0ex}}\text{probability}\phantom{\rule{4.pt}{0ex}}\text{yields}\phantom{\rule{4.pt}{0ex}}P\left(E\right)$

Such reversals are desirable in a variety of practical situations.

## A compound selection and reversal

Begin with items in two lots:

1. Three items, one defective.
2. Four items, one defective.

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 itemis good. What is the probability the item selected from lot 1 was good?

SOLUTION

Let G 1 be the event the first item (from lot 1) was good, and G 2 be the event the second item (from the augmented lot 2) is good. We want to determine $P\left({G}_{1}|{G}_{2}\right)$ . Now the data are interpreted as

$P\left({G}_{1}\right)=2/3,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({G}_{2}|{G}_{1}\right)=4/5,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({G}_{2}|{G}_{1}^{c}\right)=3/5$

By the law of total probability (CP2) ,

$P\left({G}_{2}\right)=P\left({G}_{1}\right)P\left({G}_{2}|{G}_{1}\right)+P\left({G}_{1}^{c}\right)P\left({G}_{2}|{G}_{1}^{c}\right)=\frac{2}{3}\cdot \frac{4}{5}+\frac{1}{3}\cdot \frac{3}{5}=\frac{11}{15}$

By Bayes' rule (CP3) ,

$P\left({G}_{1}|{G}_{2}\right)=\frac{P\left({G}_{2}|{G}_{1}\right)P\left({G}_{1}\right)}{P\left({G}_{2}\right)}=\frac{4/5×2/3}{11/15}=\frac{8}{11}\approx 0.73$

## Additional problems requiring reversals

• Medical tests . Suppose D is the event a patient has a certain disease and T is the event a test for the disease is positive. Data are usually of the form: prior probability $P\left(D\right)$ (or prior odds $P\left(D\right)/P\left({D}^{c}\right)$ ), probability $P\left(T|{D}^{c}\right)$ of a false positive , and probability $P\left({T}^{c}|D\right)$ of a false negative. The desired probabilities are $P\left(D|T\right)$ and $P\left({D}^{c}|{T}^{c}\right)$ .
• Safety alarm . If D is the event a dangerous condition exists (say a steam pressure is too high) and T is the event the safety alarm operates, then data are usually of the form $P\left(D\right)$ , $P\left(T|{D}^{c}\right)$ , and $P\left({T}^{c}|D\right)$ , or equivalently (e.g., $P\left({T}^{c}|{D}^{c}\right)$ and $P\left(T|D\right)$ ). Again, the desired probabilities are that the safety alarms signals correctly, $P\left(D|T\right)$ and $P\left({D}^{c}|{T}^{c}\right)$ .
• Job success . If H is the event of success on a job, and E is the event that an individual interviewed has certain desirable characteristics, thedata are usually prior $P\left(H\right)$ and reliability of the characteristics as predictors in the form $P\left(E|H\right)$ and $P\left(E|{H}^{c}\right)$ . The desired probability is $P\left(H|E\right)$ .
• Presence of oil . If H is the event of the presence of oil at a proposed well site, and E is the event of certain geological structure (salt dome or fault), the data are usually $P\left(H\right)$ (or the odds), $P\left(E|H\right)$ , and $P\left(E|{H}^{c}\right)$ . The desired probability is $P\left(H|E\right)$ .
• Market condition . Before launching a new product on the national market, a firm usually examines the condition of a test market as anindicator of the national market. If H is the event the national market is favorable and E is the event the test market is favorable, data are a prior estimate $P\left(H\right)$ of the likelihood the national market is sound, and data $P\left(E|H\right)$ and $P\left(E|{H}^{c}\right)$ indicating the reliability of the test market. What is desired is $P\left(H|E\right),$ the likelihood the national market is favorable, given the test market is favorable.

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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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