# Cambridge Question - The Prosecutor's Fallacy.

• Oct 1st 2013, 08:14 AM
zikcau25
Cambridge Question - The Prosecutor's Fallacy.
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EXERCISE 4B Question 13
- Statistics 1 (Steve Dobbs & Jane Miller)

"The Prosecutor's Fallacy. An accused prisoner is on trial. The defense lawyer asserts that, in the absence of further evidence, the probability that the prisoner is guilty is 1 in a million. The prosecuting lawyer produces a further piece of evidence and asserts that, if the prisoner were guilty, the probability that this evidence would be obtained is 999 in 1000, and if he were not guilty it would be only 1 in 1000; in other words, $P(evidence|guilty)= 0.999$, and $P(evidence|not guilty)= 0.001$. Assuming that the court admits the legality of the evidence, and that both lawyers' figures are correct, what is the probability that the prisoner is guilty?"

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This is one of the last few challenging and out of syllabus question in my textbooks. I am confused about ways to tackle it, whether this need Bayes' theorem, some linear solution or simply by Tree diagram.

I have been desperately listening Alexandra Stan (Million) all day, atleast I won't forget the question and you can do same, trying to find a solution to that problem and hoping someone to clearly show a solution. Thank you.

Alexandra Stan feat Carlprit - Million - Official Video - YouTube
• Oct 1st 2013, 08:56 PM
chiro
Re: Cambridge Question - The Prosecutor's Fallacy.
Hey zikcau25.

Hint: P(Evidence|guilty) + P(Evidence|not guilty) = P(Evidence). P(Guilty) = 1/1000000. P(Evidence and Guilty) = P(Evidence|Guilty)*P(Guilty).

What is P(Guilty|Evidence) equal to? (Remember also that P(A|B) = (P(B|A)*P(A))/P(B))
• Oct 2nd 2013, 04:26 AM
zikcau25
Re: Cambridge Question - The Prosecutor's Fallacy.
Master Chiro, thank you so much for guiding me to the correct answer. All credits goes to you but it contradicts two of your proposals.

Firstly,
$P(Evidence|Guilty) + P(Evidence|not Guilty) \neq P(Evidence)$
(Checked using numerical values, IF P(A)=0.75, P(B|A)=0.8, P(B|A')=0.6, P(B)=0.75, P(A|B)=0.8 then P(B|A) + P(B|A')P(B) as 0.8 + 0.6 ≠ 0.75)

Secondly,
The probability of the prisoner is guilty is interpreted by P(Guilty|Evidence) instead of P(Evidence and Guilty)

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Here my solution,

(expressed in terms of G for Guilty, G' for not Guilty, E for Evidence, E' for no Evidence).

Bayes' Theorem says:

1. Simple form:

$P(E|G)=\frac{P(E)\cdot P(G|E)}{{\color{DarkBlue} P(G)}}$

$P(G|E)=\frac{P(G)\cdot P(E|G)}{{\color{DarkBlue} P(E)}}$

2. Extended form:

$P(E|G)=\frac{P(E)\cdot P(G|E)}{{\color{DarkBlue} P(E)\cdot P(G|E)+P(E')\cdot P(G|E')}}$

$P(G|E)=\frac{P(G)\cdot P(E|G)}{{\color{DarkBlue} P(G)\cdot P(E|G)+P(G')\cdot P(E|G')}}$

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Now, referring back to our initial question,

Given that:

$P(G)=\frac{1}{10^{6}}$, so $P(G')=\frac{999999}{10^{6}}$

$P(E|G)=0.999$

$P(E|G')=0.001$

Evidently, the second equation of the extended form $P(G|E)=\frac{P(G)\cdot P(E|G)}{{\color{DarkBlue} P(G)\cdot P(E|G)+P(G')\cdot P(E|G')}}$ requires just all the values that we have got.

Substituting the values,

$P(G|E)=\frac{\frac{1}{10^{6}}\times 0.999}{\frac{1}{10^{6}}\times 0.999+\frac{999999}{10^{6}}\times 0.001}$= 0.000 998 (3 sig.)
• Oct 2nd 2013, 10:20 AM
HallsofIvy
Re: Cambridge Question - The Prosecutor's Fallacy.
I really hate using long, complicated, formulas like that! Here is how I would do it:

Imagine a population of 1,000,000,000 people (chosen to avoid fractions). Since "in the absence of further evidence, the probability that the prisoner is guilty is 1 in a million", 1000 of those people are guilty. Of those 1000 people, 999 would have this particular "evidence" against them. Of the 999,991,000 who are not guilty, 1 in a thousand, 999,991 would have this evidence. That is, of the total 999+ 999,991= 1000990 people against whom we could find this evidence, 999 are guilty. The probability that a person, against whom we have that evidence, is guilty is [tex]\frac{999}{1000990}= 0.000998[/quote] as you say.