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# PROBABILITY FOR ENGR APPL ECSE 4500

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This 6 page Class Notes was uploaded by Miss Damien Crooks on Monday October 19, 2015. The Class Notes belongs to ECSE 4500 at Rensselaer Polytechnic Institute taught by John Woods in Fall. Since its upload, it has received 39 views. For similar materials see /class/224760/ecse-4500-rensselaer-polytechnic-institute in ELECTRICAL AND COMPUTER ENGINEERING at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

ECSE 4500 Lecture 3 John W Woods 942008 Bayes Theorem Theorem Let the Als be a mutually exclusive and collectively exhaustive decomposition off then if PlB gt 0 and for all i PlAi gt 0 then PlBlA39lPlA39l P A B l 1 l 271PiBiAiiPiAii Proof By the Total Probability Theorem the denominator of the righthand side of this result isjust PlBl Then the result follows from the de nition of the conditional probability PlAlel ll Example Test for Virus Let the sample space 0 00011011 with i being the logical variable for a test result andj being the logical variable for the presence of a virus The string thus will be our outcomes g for this problem For instance 01 is the outcome where the test is negative and the virus is present Correspondingly 01gt is its elementary event Besides the four elementary events let us de ne two more A 3 1011gt corresponding to a positive test and B 3 0111gt correspondng to the virus being present We are given the following observed information PlAlB PlAClBC 095 and PB 0005 Based on this we na39I39vely say the test looks good But is it really What is PlBlA In words this says What is the probability that the person has the virus given that the test was positive To answer this we use Bayes Therem with the decomposition B U BC 0 First we rewrite Bayes Theorem for this case with n 2 PlAlBlPlBl PlAlB PlB PlAchlPlBC 095 X 0005 095 X 0005 10 7 09510 7 0005 0087 PlBlAl So the test is bad It has what is called a high false positive rate because PlBClA 10 7 0087 0913 is way to high Combinations and Permutations The number of arrangements permutations of n objects taken r at a time equals nr nn 71n 7 r l nlrl The number of combinations of n objects taken r at a time equals C e nwr nl n 7 rlrll since there are rl arrangements of a set of r objects We note that this is also the number of ways of dividing up a population of n objects into two sets of size r and n 7 r Application All the subsets of a set of n objects C0 2 Eg Four things taken two at a time We start with the set 1 234gt and n 4 and r 2 unordered subsets 1 2 2 3 3 4 1 3 1 4 2 4 Check 9 6 ordered strings 12 and 21 23 and 32 34 and 43 13 and 31 14 and 41 24 and 42 Check 42 12 ordered Wrepacement All strings Check 42 164 11 s s 21 22 unordered Wrepacement 31 32 33 It turns out to 41 42 43 44 be 0391 Here equal to 104

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