My Math Forum Bayes theorem

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 September 21st, 2012, 05:35 AM #1 Senior Member   Joined: Feb 2010 Posts: 324 Thanks: 0 Bayes theorem Alright, So, I got an answer and my friend got a totally different answer... so be a tie breaker. p(d)=4% p(e|d)= 95% p(e|d')=5% p(e'|d')=95% and, p(e'|d)=5% I worked it out and got .33%, then I re-did it and got .44% then tried again and got .04%.... heeeelllpppp.
September 21st, 2012, 08:57 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Bayes theo

Hello, FreaKariDunk!

Are there typos?
The numbers don't tally up.

Quote:
 So, I got an answer and my friend got a totally different answer ... so be a tie breaker. $P(D)\,=\,4\%$ $P(E\,|\,D)\,=\,95\% \;\;\;P(E\,|\,D')\,=\,95\%\;\;\;P(E'\,|\,D ') \,=\,0.95 \;\;\;P(E#39;\,|\,D)\,=\,5\%$ I worked it out and got 33%, then I re-did it and got 44% then tried again and got 4% ... [color=beige]. . [/color][color=blue]What is the question?[/color]

If the first one is: $P(D) \,=\,40\%$,
[color=beige]. . [/color]and the question is: Find $P(E)$,
[color=beige]. . . . [/color]the problem can be solved.

 September 21st, 2012, 09:23 AM #3 Senior Member   Joined: Feb 2010 Posts: 324 Thanks: 0 Re: Bayes theo P(d|e)=p(e|d)*p(d)/p(e|d)*p(d) + p(d')* p(e|d') soooo.... .95*.04/.04*.95 + .96* .05 then .44 = 44% right?

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