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Hi can some one please explain why the conditional probability P(c/g^c) is 0.2. Shouldn't it be p(c)=0.2 where c is the event that a person has the characteristic?

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    $\begingroup$ If the total population with the characteristic is large then both will have almost the same value $\endgroup$
    – Henry
    Jul 13, 2021 at 1:38

1 Answer 1

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Hi can some one please explain why the conditional probability $P(C\mid G^{\small\complement})$ is $0.2$. Shouldn't it be $P(C)=0.2$ where $C$ is the event that a person has the characteristic?

No; not any person. Our $C$ is the event that the suspect has the characteristic. This is not representative of the general population; their selection was biased by the old evidence.

Should the suspect be actually the guilty party, then they would certainly have the characteristic of the guilty party. Thus $P(C\mid G)=1.00$

Should the suspect actually be innocent, and so merely coincidentally fit the profile of the old evidence, then we may consider them to be a representative of the general population (for the new evidence). Thus $P(C\mid G^{\small\complement}) = 0.2$

$\mathbf{Edit:}$

The reason the word "may" is used here in the last line "when suspect is innocent..,then we may consider them to be a representative of general population is because of the fact that the actual guilty person also belongs to the general population who possess the characteristic.

Let $x$ be the total population. Then for calculating the probability that the suspect possess the characteristic when he is not guilty, the sample space would consists of number of innocent people which is equal to $x-1$ and the number of outcomes favourable to the event will be equal to $0.2x-1$ (=number of people who are not guilty but possess the characteristic, since the guilty one belongs to those $0.2x$ people).

$$P(C\mid G^{\small\complement})=\frac{P(C\cap G^{\small\complement})}{P(G^{\small\complement})} = \frac{N(C\cap G^{\small\complement})}{N(G^{\small\complement})}$$

$$=\frac{0.2x-1}{x-1}=\frac{0.2-\frac{1}{x}}{1-\frac{1}{x}} \approx 0.2$$ (since total population $x$ is very large $1\over x$ is neglected)

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