MHB Wackerly/Mendenhall/Schaeffer Problem 2.74: Double the Probability?

[Ackbach]

Problem: A lie detector will show a positive reading (indicate a lie) $10\%$ of the time when a person is telling the truth and $95\%$ of the time when the person is lying. Suppose two people are suspects in a one-person crime and (for certain) one is guilty.

What is the probability that the detector shows a positive reading for both suspects?

Answer: Let $L$ be the event that a suspect is lying, and let $IL$ be the event that a lie detector indicates a lie. We are given the following probabilities:
\begin{align*}
P(IL|L)&=0.95 \\
P(IL|\overline{L})&=0.1.
\end{align*}
Moreover, we may assume that the innocent suspect is telling the truth, and the guilty suspect is lying. We do not know which suspect is guilty and which is innocent, presumably. We may infer that
\begin{align*}
P(\overline{IL}|L)&=0.05 \\
P(\overline{IL}|\overline{L})&=0.9.
\end{align*}

Here's my question: since we don't know which suspect is innocent and which is guilty, should we multiply $P(IL|L) \cdot P(IL| \overline{L})$ by two, since there are two ways to assign the guilty and innocent suspects? Or am I overthinking it here? I'm probably missing something obvious.

Thanks for your time!

 

[chisigma]

Problem: A lie detector will show a positive reading (indicate a lie) $10\%$ of the time when a person is telling the truth and $95\%$ of the time when the person is lying. Suppose two people are suspects in a one-person crime and (for certain) one is guilty.

What is the probability that the detector shows a positive reading for both suspects?

Answer: Let $L$ be the event that a suspect is lying, and let $IL$ be the event that a lie detector indicates a lie. We are given the following probabilities:
\begin{align*}
P(IL|L)&=0.95 \\
P(IL|\overline{L})&=0.1.
\end{align*}
Moreover, we may assume that the innocent suspect is telling the truth, and the guilty suspect is lying. We do not know which suspect is guilty and which is innocent, presumably. We may infer that
\begin{align*}
P(\overline{IL}|L)&=0.05 \\
P(\overline{IL}|\overline{L})&=0.9.
\end{align*}

Here's my question: since we don't know which suspect is innocent and which is guilty, should we multiply $P(IL|L) \cdot P(IL| \overline{L})$ by two, since there are two ways to assign the guilty and innocent suspects? Or am I overthinking it here? I'm probably missing something obvious.

Thanks for your time!

We have to find the probability of the union of two separate events, one with probability $P_{1}= P(IL|L)=.95$ and the other with probability $P_{2} = P(IL|\overline {L}) = .1$. The requested probability is $P = P_{1}\ P_{2} = .095$. You can easilily verify that considering that the set of all possible events has probability $P_{T} = P_{1}\ P_{2} + P_{1}\ (1 - P_{2}) + P_{2}\ (1 - P_{1}) + (1 - P_{1})\ (1 - P_{2}) = 1$...

Kind regards

$\chi$ $\sigma$

 

[polygamma]

Do we know that the guilty person will lie and the innocent person won't?

 

[Ackbach]

We have to find the probability of the union

Did you mean "intersection" here?

of two separate events, one with probability $P_{1}= P(IL|L)=.95$ and the other with probability $P_{2} = P(IL|\overline {L}) = .1$. The requested probability is $P = P_{1}\ P_{2} = .095$.

This assumes that $IL|\overline{L}$ and $IL|L$ are independent events. Is that correct?

You can easily verify that considering that the set of all possible events has probability $P_{T} = P_{1}\ P_{2} + P_{1}\ (1 - P_{2}) + P_{2}\ (1 - P_{1}) + (1 - P_{1})\ (1 - P_{2}) = 1$...

That's true of independent events, correct? Is it also true that if the equation you wrote down holds, the events must be independent?

Kind regards

$\chi$ $\sigma$

Do we know that the guilty person will lie and the innocent person won't?

I am assuming that. Presumably the guilty one would not want to go to jail, and would lie to get out of it.

 

[chisigma]

a)Did you mean "intersection" here?...

b) This assumes that $IL|\overline{L}$ and $IL|L$ are independent events. Is that correct?...

c) That's true of independent events, correct?... Is it also true that if the equation you wrote down holds, the events must be independent?...

a) I apologize but in fact set theory has ever been troublesome for me, so that I confuse 'union' and 'intersection'... 'intersection' is correct...

b) Yes!... it is implicit in the definition You have done...

c) Non necessarly!... we have simply two different events, the first with probability $P_{1}$ to be KO and $1-P_{1}$ to be OK, the second with probability $P_{2}$ to be KO and $1-P_{2}$ to be OK, and the probability of the overall set of possibilities must be 1...

It is important to note that all that is true no matter if the questioned people are 'guilty' or 'innocent'...Kind regards

$\chi$ $\sigma$