Conditional Probability - Derivation of Probability Unknown

Question

Here's a problem from Sheldon Ross's A First Course in Probability that I don't understand:

At a certain stage of a criminal investigation, the inspector in charge is 60 percent convinced of the guilt of a certain suspect. Suppose, however, that a new piece of evidence which shows that the criminal has a certain characteristic (such as left-handedness, baldness, or brown hair) is uncovered. If 20 percent of the population possesses this characteristic, how certain of the guilt of the suspect should the inspector now be if it turns out that the suspect has the characteristic?

The solution is:

Letting $G$ denote the event that the suspect is guilty and $C$ the event that he possesses the characteristic of the criminal, we have $$ P(G\mid C) = \frac{P(GC)}{P(C)} = \frac{P(C\mid G)P(G)}{P(C\mid G)P(G) + P(C\mid G^c)P(G^c)} = \frac{1(0.6)}{1(0.6) + (0.2)(0.4)} = 0.882 $$

What I don't understand is why $P(C) = P(C\mid G)P(G) + P(C\mid G^c)P(G^c)$. I feel that it should be either 1, if the suspect has the characteristic, else 0.2 if it's unknown. If someone could explain how this is derived, I'd really appreciate it.

Answer 1

$$\begin{align}P(C) & = P(C\mid G)P(G) + P(C\mid G^c)P(G^c) \\ & = P(C \cap G) + P(C \cap G^c)\end{align}$$

$P(C)$ is the measure of the inspector's belief that the suspect would possess the characteristic prior to examination. It is equal to the inspector's belief they could possess the characteristic and be guilty, plus the inspector's belief that they could possess the characteristic and be innocent.

Your instinct that the value is either $1$ for someone who is guilty, or $0.20$ for someone who is not, is why we assign those values to $P(C\mid G) = 1$ and $P(C\mid G^c) = 0.2$.

We then multiply each of these by the inspector's prior belief that the particular suspect is guilty or innocent (as appropriate) and add these together to obtain the inspector's prior belief that the suspect could have the characteristic.

$\begin{align}P(C) & = P(C\mid G)P(G) + P(C\mid G^c)P(G^c) \\ & = (1)(0.6)+(0.2)(0.4) \\ ~ & = 0.68 \end{align}$

This is then used to update the inspector's belief in the suspect's guilt posterior to discovering that the suspect does have that characteristic.

$$P(G\mid C) = \frac{P(G)P(C\mid G)}{P(C)}$$

Answer 2

$$ \Pr(C) = \Pr( (C\ \&\ G)\text{ or }(C\ \&\ \text{not }G)). $$

That much is true for reasons having nothing to do with probability. It's true because $C$ is logically the same as $( (C\ \&\ G)\text{ or }(C\ \&\ \text{not }G))$.

But then we can add \begin{align} & = \Pr(C\ \&\ G) + \Pr(C\ \&\ \text{not }G) \\[10pt] & = \Pr(C\mid G)\Pr(G) + \Pr(C\mid \text{not }G)\Pr(\text{not }G). \end{align}

Answer 3

i have another approach, i don't know if it's correct

a normal person in a population of N will be (1/N)100% guilty

initially in inspectors mind, suspect was 60% guilty

let the population be N, we can assume that the suspect is equivalent to 'set of X people' such that new population is N+X-1 and event G (that suspect is guilty) is equivalent to H, any of the set of X people is guilty (i have just distributed guilt weightage of suspect to X people equally)

therefore P(H) = X/(N+X-1) = 60%

for larger N, N+X-1 = N+X

=> N= 4*X/6

now we know that suspect belongs to 20%(=N/5) of the population that have some certain characteristic, So the set of X people will also possess the same characteristic such that now there are X+(N/5-1) people with the characteristic, all are same in inspectors eyes, also criminal belongs to these X+(N/5-1) people

probability of suspect to be guilty, P(G) = P(H) = X/(X+N/5-1)

for larger N, N/5-1 = N/5

so we have to find X/(X+N/5)

that is X/(X+4*X/6*5) = 1/(1+4/30) = 30/34 = 0.8823