# probility problem

• Feb 25th 2010, 05:28 AM
chialin4
probility problem
i think it is not easy!!

A judge is 65% sure that a suspect has committed a crime .
During the course of the trial a witness convinces the judge that there is an 85% chance that the criminal is left-handed. If 23% of the population is left-handed and the suspect is also left-handed,with this imformation,
how certain should the judge be of the guilt of the suspect?
• Feb 26th 2010, 09:46 AM
davismj
Quote:

Originally Posted by chialin4
A judge is 65% sure that a suspect has committed a crime .
During the course of the trial a witness convinces the judge that there is an 85% chance that the criminal is left-handed. If 23% of the population is left-handed and the suspect is also left-handed,with this imformation,
how certain should the judge be of the guilt of the suspect?

Let P(A) be the probability of the judges initial suspicion: P(A) = 0.65.

Let P(S) be the probability (according to the judge's thought process) that the criminal is left handed: P(S) = 0.85

Let P(L) be the proportion of the population that is left handed: P(L) = 0.23.

So, let's start with the left-handedness. We know that there's a fairly good (85%) chance that the suspect is left-handed; however, there's a slight (23%) chance that whomever they put up on the stand--regardless of guilt--would be left-handed. This means that if they picked a random person off the street and put them on the stand, there's a 23% chance that they would have an 85% chance of being wrongly accused, or about 21.25% (since it is most likely that the suspicion that the guilty part is left-handed is most-likely independent of any assumptions on the population). In other words,

P(S | L) = P(S)P(L) = (0.23)(0.85) = 0.2125

Let E be the probability that the evidence shows that he is guilty. Given the above conditional probability--the probability that a left-handed person would be wrongly accused--we know that

P(E) = 1 - (0.2125) = 0.7875

Now, the judge knows there is a fairly good chance that it is no coincidence that this person is the suspect, and he's somewhat (65%) sure that this person is guilty anyway. These are certainly independent, and so we have.

P(A | E) = P(A)P(E) = (0.65)(0.7875) = 0.5119

Therefore, there is about a 51.2% chance that this person is guilty given the evidence and the judge's intuition.

Edit: I think I've read this problem wrong. I'm going to look at it again.
• Feb 26th 2010, 02:18 PM
davismj
I'd like to take another look at what we know. Perhaps a better interpretation of our data will culminate in better results.

We have a the set of all people, from which one and only one person is guilty. 23% of the population is left-handed, and 77% of the population is right-handed. We've picked someone out of the crowd--a lefty--and we've decided there's a 65% chance we've picked the guilty person. Let's call this G:

P(G) = 0.65

Okay. Now we bring in some evidence. A witness decides that there's an 85% chance the guilty person is left-handed, which makes our suspect look more guilty. In other words, if we pick the guilty individual, there's an 85% chance he's left-handed. Thus, we have:

P(L | G) = 0.85

However, 23% of people are left-handed, which means we could easily have picked the wrong guy. But ultimately, what we want to show is:

P(G | L) = p

In other words, what is the probability that we've picked the guilty person given his left-handedness.

Okay, let's not assume independence, since we may very well have picked this individual out of a lineup specifically because he's left-handed. We know that

P(G | L) = $\displaystyle \frac{P(G \cap L)}{P(L)}$

and

P(L | G) = $\displaystyle \frac{P(L \cap G)}{P(G)} = \frac{P(G \cap L)}{0.65} = 0.85 \implies P(G \cap L) = (0.65)(0.85) = 0.5525$

Now here is the tricky part. We need to know P(L). When I first worked through this, I assumed, that P(L) = 0.23. However, with our figures for P(L | G) and $\displaystyle P(L \cap G)$, we measure L as the probability that this individual would be left-handed. Let's look at it like this: There is a dichotomy involved with our left-handed suspect. He's either left-handed and guilty, or left-handed and not guilty (since we know he is left-handed). So we have

$\displaystyle P(L) = P(L \cap G) + P(L \cap G') = P(L | G)P(G) + P(L | G')P(G')$

We know P(L | G), P(G), and P(G'), so what is P(L | G')? This figure is precisely the population figure of 23%; that is, since only one person is guilty, of all the non-guilty people, 23% are left-handed, and 77% are right-handed. Thus, we have:

P(L) = (0.85)(0.65) + (0.23)(0.35) = 0.633.

So finally, plugging this into the original equation, we have:

$\displaystyle P(G | L) = \frac{0.5525}{0.633} = 0.8728$

In other words, the additional information provided regarding is left-handedness provided us with an additional 22.3% confidence that we have the correct suspect.
• Feb 28th 2010, 06:18 AM
chialin4
Quote:

Originally Posted by davismj
I'd like to take another look at what we know. Perhaps a better interpretation of our data will culminate in better results.

We have a the set of all people, from which one and only one person is guilty. 23% of the population is left-handed, and 77% of the population is right-handed. We've picked someone out of the crowd--a lefty--and we've decided there's a 65% chance we've picked the guilty person. Let's call this G:

P(G) = 0.65

Okay. Now we bring in some evidence. A witness decides that there's an 85% chance the guilty person is left-handed, which makes our suspect look more guilty. In other words, if we pick the guilty individual, there's an 85% chance he's left-handed. Thus, we have:

P(L | G) = 0.85

However, 23% of people are left-handed, which means we could easily have picked the wrong guy. But ultimately, what we want to show is:

P(G | L) = p

In other words, what is the probability that we've picked the guilty person given his left-handedness.

Okay, let's not assume independence, since we may very well have picked this individual out of a lineup specifically because he's left-handed. We know that

P(G | L) = $\displaystyle \frac{P(G \cap L)}{P(L)}$

and

P(L | G) = $\displaystyle \frac{P(L \cap G)}{P(G)} = \frac{P(G \cap L)}{0.65} = 0.85 \implies P(G \cap L) = (0.65)(0.85) = 0.5525$

Now here is the tricky part. We need to know P(L). When I first worked through this, I assumed, that P(L) = 0.23. However, with our figures for P(L | G) and $\displaystyle P(L \cap G)$, we measure L as the probability that this individual would be left-handed. Let's look at it like this: There is a dichotomy involved with our left-handed suspect. He's either left-handed and guilty, or left-handed and not guilty (since we know he is left-handed). So we have

$\displaystyle P(L) = P(L \cap G) + P(L \cap G') = P(L | G)P(G) + P(L | G')P(G')$

We know P(L | G), P(G), and P(G'), so what is P(L | G')? This figure is precisely the population figure of 23%; that is, since only one person is guilty, of all the non-guilty people, 23% are left-handed, and 77% are right-handed. Thus, we have:

P(L) = (0.85)(0.65) + (0.23)(0.35) = 0.633.

So finally, plugging this into the original equation, we have:

$\displaystyle P(G | L) = \frac{0.5525}{0.633} = 0.8728$

In other words, the additional information provided regarding is left-handedness provided us with an additional 22.3% confidence that we have the correct suspect.

thx for ur help
• Feb 28th 2010, 06:18 AM
chialin4
Quote:

Originally Posted by davismj
Let P(A) be the probability of the judges initial suspicion: P(A) = 0.65.

Let P(S) be the probability (according to the judge's thought process) that the criminal is left handed: P(S) = 0.85

Let P(L) be the proportion of the population that is left handed: P(L) = 0.23.

So, let's start with the left-handedness. We know that there's a fairly good (85%) chance that the suspect is left-handed; however, there's a slight (23%) chance that whomever they put up on the stand--regardless of guilt--would be left-handed. This means that if they picked a random person off the street and put them on the stand, there's a 23% chance that they would have an 85% chance of being wrongly accused, or about 21.25% (since it is most likely that the suspicion that the guilty part is left-handed is most-likely independent of any assumptions on the population). In other words,

P(S | L) = P(S)P(L) = (0.23)(0.85) = 0.2125

Let E be the probability that the evidence shows that he is guilty. Given the above conditional probability--the probability that a left-handed person would be wrongly accused--we know that

P(E) = 1 - (0.2125) = 0.7875

Now, the judge knows there is a fairly good chance that it is no coincidence that this person is the suspect, and he's somewhat (65%) sure that this person is guilty anyway. These are certainly independent, and so we have.

P(A | E) = P(A)P(E) = (0.65)(0.7875) = 0.5119

Therefore, there is about a 51.2% chance that this person is guilty given the evidence and the judge's intuition.

Edit: I think I've read this problem wrong. I'm going to look at it again.

thx for ur help