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At a certain stage of a criminal investigation, the inspector in charge is 60% certain that the suspect is guilty. Suppose however that a new piece of evidence shows that the suspect has a certain characteristic that only 20% of the population has. If 20% of the population possess this characterstic how certain of guilt should the inspector now be if that suspect has the characteristic. This could translated to what is the probability that the suspect is guilty given that they have the characteristic?

$E=$ the suspect is guilty $.60$

$E^c=$ the suspect is not guilty $.40$ or $1-.60$

$F=$ the suspect has the characteristic, .20

$F^c$= the suspect doesnt have the characteristic $.80$

$P(E \vert F)=\frac{P(EF)P(F)}{P(F)}$

$=\frac{P(F \vert E)(PE)}{P(F \vert E)P(E)+P(F \vert E^C)P(E^c)}$

$=\frac{P(F \vert E).60}{.60P(F \vert E)+P(F \vert E^c)(.40)}$

My question is how in hell am I supposed to obtain $1$ for $P(F \vert E)$ and $.2$ for $P(F \vert E^c)$. Is there any way to obtain $P(F \vert E)$ with the formula $\frac{P(EF)}{P(F)}$ How on earth do I calcualte $P(FE)$? which is calcualted as Probability that the person has the charactersitic and is guilty?

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    How is $E^c$ $1-.4$? – 2017-02-06
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    Because it 1-E and E is .60 It is the probability that they are not guilty – 2017-02-06
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    Please check your question again. Quote:"$.40$ or $1-.40$" This doesn´t match. – 2017-02-06
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    So what is the probability that the suspect is guilty knowing that the suspect fits the characteristic requirements? In other words, $P(E|F)$. – 2017-02-06
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    @callculus that's what I was wondering as well. – 2017-02-06
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    @Oliver821 I have read your comment. I´m curious if the OP will notice the flaw. Btw your answer is nice. – 2017-02-06
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    @callculus I hope so too and thanks. Just pure manipulation of the conditional probability is all we need to know. – 2017-02-06
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    I just dont know how to get$ P(E \cap F)$. @Oliver821. Your answer didnot clear that up for me. I corrected my mistake it was 1-.60 – 2017-02-06
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    You haven't had much exposure to conditional probability problems have you? – 2017-02-06
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    No I have not. I am trying to read through the problems on my own and it is a real struggle we haven't done chapter 3 (conditional) But my teacher is an adjunct and hes not very good nor does he have office hours. I am screwed without help on here – 2017-02-06

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Assuming you know the equation for the conditional probability of two independent variables, we have

$$P(F|E) = \frac {P(F \cap E)}{P(E)}$$ but note that $$P(F\cap E) = P(E\cap F)$$ And $$P(E|F) = \frac{P(E\cap F)}{P(F)}$$ Solving for $P(E\cap F)$ in the latter equation yields $$P(E\cap F)= P(E|F)*P(F)$$ Lastly, we have $$P(F|E) = \frac {P(E|F)*P(F)}{P(E)}$$ I'll leave it up to you to fill in the missing information, most of which is given in the problem already.

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    You wrote down your equation wrong for $P(E|F)$. You need to use the cap feature since we arre comparing two events. – 2017-02-06