Bayes= theorem:
For making probability judgments
Does presence of cue increase or decrease probability?
Is CONDITIONAL probability different from overall probability?
EX.
If you test positive for HIV, what is the likelihood that you actually are HIV+?
What is the likelihood you are HIV+, GIVEN you test positive?
A
Given@ - linguistic cue that there is a conditional probabilityWhat is the probability, conditional upon some other thing being true?
Bayes theorem:
p(target event, given cue) =
p(cue, given target event)*p(target event)
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p(cue, given target)*p(target event) + p(cue, given no target event)*p(no target event)
Need OVERALL probability of target event:
1 in 500 = .2%
Need overall probability of target event NOT occurring:
499 in 500 = 99.8%
Need probability of cue event, given target event has occurred:
95%
Need prob of cue event, given target even has NOT occurred:
2%
So, GIVEN you TEST pos, how likely you ARE pos?
Probability being HIV+, given positive test =
p(positive test, given HIV+)*p(HIV+)
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p(positive test, given HIV+)*p(HIV+) + p(positive test, given HIV-)*p(HIV-)
(.95)*(.002)
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(.95)*(.002) +(.02) *(.998)
(.0019)/(.0019) + (.01996) = .0869
With no HIV test, best estimate that you are HIV pos?
(pop. percent -- .2 %, .002, 1 in 500)
After testing pos., best guess?
.0869 -- 8.69 out of 100
Odds increase from less than 1 in 100 to close to 1 in 10
Likelihood of ACTUALLY being HIV+, given testing HIV+ = 8.69 out of 100
Likelihood of testing HIV+, given that you are HIV+ = 95 out of 100
False positive: probability of testing pos, given you really are negative
p(pos. test, given HIV-) = .02
What is the probability I=m home, given my radio is on?
In probability terms: p(home, given radio)
1) prob of target event:
probability = 50% (12 hours)
sleep 8 hours, so off then -- 4/12, 33% chance radio on
= 50%
= 100% (discourage burglars!)
p(radio, given home)*p(home)
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p(radio, given home) * p (home) + p(radio, given not home)* p (not home)
(.33*.50)/(.33*.50)+(1.0)*(.50) = .165/.665 = .248
What about probability I=m NOT home given the radio is on?
p(not home, given radio) =
p(radio, given not home) * p (not home)
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p(radio, given not home)* p (not home) + p(radio, given home)*p (home)
.5/(.5)+(.165) = .752
Probability (I=m home, given the radio is on) is .248
Probability (the radio is on, given I=m home) is .33