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
$$P\left({H}_{1}\right)=0.2,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({H}_{2}\right)=0.5,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({H}_{3}\right)=0.3,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(F\right{H}_{1})=0.10,\phantom{\rule{10pt}{0ex}}P\left(F\right{H}_{2})=0.02,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(F\right{H}_{3})=0.06$$
A student passes the exam. Determine for each
i the conditional
probability
$P\left({H}_{i}\right{F}^{c})$ that the student is from high school
i .
SOLUTION
$$P\left({F}^{c}\right)=P\left({F}^{c}\right{H}_{1})P\left({H}_{1}\right)+P\left({F}^{c}\right{H}_{2})P\left({H}_{2}\right)+P\left({F}^{c}\right{H}_{3})P\left({H}_{3}\right)=0.90\cdot 0.2+0.98\cdot 0.5+0.94\cdot 0.3=0.952$$
Then
$$P\left({H}_{1}\right{F}^{c})=\frac{P\left({F}^{c}{H}_{1}\right)}{P\left({F}^{c}\right)}=\frac{P\left({F}^{c}\right{H}_{1})P\left({H}_{1}\right)}{P\left({F}^{c}\right)}=\frac{180}{952}=0.1891$$
Similarly,
$$P\left({H}_{2}\right{F}^{c})=\frac{P\left({F}^{c}\right{H}_{2})P\left({H}_{2}\right)}{P\left({F}^{c}\right)}=\frac{590}{952}=0.5147\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{10pt}{0ex}}P\left({H}_{3}\right{F}^{c})=\frac{P\left({F}^{c}\right{H}_{3})P\left({H}_{3}\right)}{P\left({F}^{c}\right)}=\frac{282}{952}=0.2962$$
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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
$$P\left({A}_{i}\rightE)=\frac{P\left({A}_{i}E\right)}{P\left(E\right)}=\frac{P\left(E\right{A}_{i})P\left({A}_{i}\right)}{P\left(E\right)}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1\le i\le n\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{The}\phantom{\rule{4.pt}{0ex}}\text{law}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}\text{total}\phantom{\rule{4.pt}{0ex}}\text{probability}\phantom{\rule{4.pt}{0ex}}\text{yields}\phantom{\rule{4.pt}{0ex}}P\left(E\right)$$
Such reversals are desirable in a variety of practical situations.
A compound selection and reversal
Begin with items in two lots:
 Three items, one defective.
 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
$P\left({G}_{1}\right{G}_{2})$ .
Now the data are interpreted as
$$P\left({G}_{1}\right)=2/3,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({G}_{2}\right{G}_{1})=4/5,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({G}_{2}\right{G}_{1}^{c})=3/5$$
By the law of total probability
(CP2) ,
$$P\left({G}_{2}\right)=P\left({G}_{1}\right)P\left({G}_{2}\right{G}_{1})+P\left({G}_{1}^{c}\right)P\left({G}_{2}\right{G}_{1}^{c})=\frac{2}{3}\cdot \frac{4}{5}+\frac{1}{3}\cdot \frac{3}{5}=\frac{11}{15}$$
By Bayes' rule
(CP3) ,
$$P\left({G}_{1}\right{G}_{2})=\frac{P\left({G}_{2}\right{G}_{1})P\left({G}_{1}\right)}{P\left({G}_{2}\right)}=\frac{4/5\times 2/3}{11/15}=\frac{8}{11}\approx 0.73$$
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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
$P\left(T\right{D}^{c})$ of a
false positive , and probability
$P\left({T}^{c}\rightD)$ of a false negative. The desired
probabilities are
$P\left(D\rightT)$ and
$P\left({D}^{c}\right{T}^{c})$ .

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
$P\left(D\right)$ ,
$P\left(T\right{D}^{c})$ , and
$P\left({T}^{c}\rightD)$ , or equivalently (e.g.,
$P\left({T}^{c}\right{D}^{c})$ and
$P\left(T\rightD)$ ). Again, the desired probabilities are that the safety
alarms signals correctly,
$P\left(D\rightT)$ and
$P\left({D}^{c}\right{T}^{c})$ .

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
$P\left(E\rightH)$ and
$P\left(E\right{H}^{c})$ . The desired probability is
$P\left(H\rightE)$ .

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),
$P\left(E\rightH)$ , and
$P\left(E\right{H}^{c})$ .
The desired probability is
$P\left(H\rightE)$ .

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
$P\left(E\rightH)$ and
$P\left(E\right{H}^{c})$ indicating
the reliability of the test market. What is desired is
$P\left(H\rightE),$ the likelihood the
national market is favorable, given the test market is favorable.
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