# Conditional probability

 Page 6 / 6

The calculations, as in [link] , are simple but can be tedious. We have an m-procedure called bayes to perform the calculations easily. The probabilities $P\left({A}_{i}\right)$ are put into a matrix PA and the conditional probabilities $P\left(E|{A}_{i}\right)$ are put into matrix PEA. The desired probabilities $P\left({A}_{i}|E\right)$ and $P\left({A}_{i}|{E}^{c}\right)$ are calculated and displayed

>>PEA = [0.10 0.02 0.06];>>PA = [0.2 0.5 0.3];>>bayes Requires input PEA = [P(E|A1) P(E|A2) ... P(E|An)]and PA = [P(A1) P(A2) ... P(An)] Determines PAE = [P(A1|E) P(A2|E) ... P(An|E)]and PAEc = [P(A1|Ec) P(A2|Ec) ... P(An|Ec)] Enter matrix PEA of conditional probabilities PEAEnter matrix PA of probabilities PA P(E) = 0.048P(E|Ai) P(Ai) P(Ai|E) P(Ai|Ec) 0.1000 0.2000 0.4167 0.18910.0200 0.5000 0.2083 0.5147 0.0600 0.3000 0.3750 0.2962Various quantities are in the matrices PEA, PA, PAE, PAEc, named above

The procedure displays the results in tabular form, as shown. In addition, the various quantities are in the workspace in the matrices named, so that theymay be used in further calculations without recopying.

The following variation of Bayes' rule is applicable in many practical situations.

(CP3*) Ratio form of Bayes' rule $\frac{P\left(A|C\right)}{P\left(B|C\right)}=\frac{P\left(AC\right)}{P\left(BC\right)}=\frac{P\left(C|A\right)}{P\left(C|B\right)}\cdot \frac{P\left(A\right)}{P\left(B\right)}$

The left hand member is called the posterior odds , which is the odds after knowledge of the occurrence of the conditioning event. The second fraction in the right hand member is the prior odds , which is the odds before knowledge of the occurrence of the conditioning event C . The first fraction in the right hand member is known as the likelihood ratio . It is the ratio of the probabilities (or likelihoods) of C for the two different probability measures $P\left(\phantom{\rule{0.166667em}{0ex}}\cdot \phantom{\rule{0.166667em}{0ex}}|A\right)$ and $P\left(\phantom{\rule{0.166667em}{0ex}}\cdot \phantom{\rule{0.166667em}{0ex}}|B\right)$ .

## A performance test

As a part of a routine maintenance procedure, a computer is given a performance test. The machine seems to be operating so well that the prior odds it is satisfactory are takento be ten to one. The test has probability 0.05 of a false positive and 0.01 of a false negative. A test is performed. The result is positive. What are the posterior odds the device isoperating properly?

SOLUTION

Let S be the event the computer is operating satisfactorily and let T be the event the test is favorable. The data are $P\left(S\right)/P\left({S}^{c}\right)=10$ , $P\left(T|{S}^{c}\right)=0.05$ , and $P\left({T}^{c}|S\right)=0.01$ . Then by the ratio form of Bayes' rule

$\frac{P\left(S|T\right)}{P\left({S}^{c}|T\right)}=\frac{P\left(T|S\right)}{P\left(T|{S}^{c}\right)}\cdot \frac{P\left(S\right)}{P\left({S}^{c}\right)}=\frac{0.99}{0.05}\cdot 10=198\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(S|T\right)=\frac{198}{199}=0.9950$

The following property serves to establish in the chapters on "Independence of Events" and "Conditional Independence" a number of important properties for the concept of independence and of conditional independence of events.

(CP4) Some equivalent conditions If $0 and $0 , then

$P\left(A|B\right)*P\left(A\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(B|A\right)*P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(AB\right)*P\left(A\right)P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}$
$P\left(AB\right)*P\left(A\right)P\left(B\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({A}^{c}{B}^{c}\right)*P\left({A}^{c}\right)P\left({B}^{c}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{iff}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$

where $*\text{is}<,\le ,=,\ge ,\text{or}>$ and $\diamond \text{is}>,\ge ,=,\le ,\text{or}<$ , respectively.

Because of the role of this property in the theory of independence and conditional independence, we examine the derivation of these results.

VERIFICATION of (CP4)

1. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(A|B\right)*P\left(A\right)$ (divide by $P\left(B\right)$ — may exchange A and A c )
2. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(B|A\right)*P\left(B\right)$ (divide by $P\left(A\right)$ — may exchange B and B c )
3. $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $\left[P\left(A\right)-P\left(A{B}^{c}\right)\right]*P\left(A\right)\left[1-P\left({B}^{c}\right)$ iff $-P\left(A{B}^{c}\right)*-P\left(A\right)P\left({B}^{c}\right)$ iff $P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$
4. We may use c to get $P\left(AB\right)*P\left(A\right)P\left(B\right)$ iff $P\left(A{B}^{c}\right)\diamond P\left(A\right)P\left({B}^{c}\right)$ iff $P\left({A}^{c}{B}^{c}\right)*P\left({A}^{c}\right)P\left({B}^{c}\right)$

$\square$

A number of important and useful propositons may be derived from these.

1. $P\left(A|B\right)+P\left({A}^{c}|B\right)=1$ , but, in general, $P\left(A|B\right)+P\left(A|{B}^{c}\right)\ne 1$ .
2. $P\left(A|B\right)>P\left(A\right)$ iff $P\left(A|{B}^{c}\right) .
3. $P\left({A}^{c}|B\right)>P\left({A}^{c}\right)$ iff $P\left(A|B\right) .
4. $P\left(A|B\right)>P\left(A\right)$ iff $P\left({A}^{c}|{B}^{c}\right)>P\left({A}^{c}\right)$ .

VERIFICATION — Exercises (see problem set)

$\square$

## Repeated conditioning

Suppose conditioning by the event C has occurred. Additional information is then received that event D has occurred. We have a new conditioning event $CD$ . There are two possibilities:

1. Reassign the conditional probabilities. ${P}_{C}\left(A\right)$ becomes
${P}_{C}\left(A|D\right)=\frac{{P}_{C}\left(AD\right)}{{P}_{C}\left(D\right)}=\frac{P\left(ACD\right)}{P\left(CD\right)}$
2. Reassign the total probabilities: $P\left(A\right)$ becomes
${P}_{CD}\left(A\right)=P\left(A|CD\right)=\frac{P\left(ACD\right)}{P\left(CD\right)}$

Basic result : ${P}_{C}\left(A|D\right)=P\left(A|CD\right)={P}_{D}\left(A|C\right)$ . Thus repeated conditioning by two events may be done in any order, or may be done in one step. This result extends easilyto repeated conditioning by any finite number of events. This result is important in extending the concept of "Independence of Events" to "Conditional Independence" . These conditions are important for many problems of probable inference.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
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?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive