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?


AGiven@ - linguistic cue that there is a conditional probability

What is the probability, conditional upon some other thing being true?


(Don=t confuse conjunction fallacy [Linda problem] with conditional probabilities)

 

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:

What is probability of being HIV + ?

1 in 500 = .2%

Need overall probability of target event NOT occurring:

What is probability of being HIV- ?

499 in 500 = 99.8%

Need probability of cue event, given target event has occurred:

What is probability of TESTING pos., given you ARE pos.?

95%

Need prob of cue event, given target even has NOT occurred:

What is probability of testing pos., given you are really HIV-

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 of me home

probability = 50% (12 hours)

2) probability of cue, given target event:

prob of radio, given I=m home

sleep 8 hours, so off then -- 4/12, 33% chance radio on

3) probability of target event not occurring:

prob me NOT home:

= 50%

4) probability of cue, given target event not occurring:

probability of radio on, given I=m not home

= 100% (discourage burglars!)

p(home, given radio) =

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