# 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) ,

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

• 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.

find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant hi vedant can u help me with some assignments Solomon find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive