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View Full Version : Compute the PROBABILITY that Broderick is GUILTY.


Statisticallychallenged
May 6, 2007, 09:21 PM
:eek: B. Broderick was tried for multiple crimes. A forensic expert testified that based on blood samples taken from the crime scene, the assilant had type B blood characteristics and a unique genetic marker. She reported that only 0.32 % of the population of the county had these blood characteristics and that Broderick had blood type B and the genetic marker.

If I define events G and E as following:
G= Broderick is guilty of the crimes
E= Evidence at crime scenes is type B and has the genetic marker

Circumstantial evidence about this case makes me think that P(G) is 0.10 (subjective probability). Given the forensic eveidence, what is the probability that Broderick is guilty?

While a 10% probability of guilt is not "beyond a reasonable doubt" what happens to the probabilty after accounting the forensic eveidence?

HINTS:
1. P(E,given G)=1. If Broderick did the crime then the evidence would certainly be type B blood with genetic marker

2. Argue that P9E, given not G)= 0.0032

3. Use the subjective quilt probability P(G)=0.10 and the probabilities in the previous hints to compute P(G, given E), the probability of guilt given the evidence.


YIKES!! HOPEFULLY SOMEONE OUT THERE KNOWS HOW TO MAKE SENSE OF THIS MONSTER OF A QUESTION!!

ebaines
May 7, 2007, 12:32 PM
Not sure what question 2 is, but as for question 3, it's clearly a Bayes Theorem problem to me...

Using your terminology, Bayes Theorem is stated like this:

P(G|E) = P(E|G)*P(G)/P(E).

Here P(E|G) is 1.0, as you stated. We know P(G) = 0.1. So what is P(E)? Stated in english: what is the probability that the blood evidence would be found at the scene of the crime? It's the following:

P(E) = P(E|G)*P(G) + P(E|~G)*P(~G)

Here the ~ symbol means "not." The probability that the evidence of the blood markers would be found is equal to the sum of the probabilities that Broiderick did it and the blood was found and the probability that he didn't do it and the blood with the marker was found anyway. Do the math, and you find that P(E) = 1*.1 + .9*.0032*.9 = 0.102592.

So P(G|E) = 1*.1/0.102592 = 97.47%.

Lowtax4eva
May 7, 2007, 12:47 PM
I would have said roughly 76 percent

This:

https://www.askmehelpdesk.com/math-sciences/announcement-u-b-read-first-expectations-homework-help-board-b-u.html

ebaines
May 7, 2007, 01:21 PM
Based on what, Lowtax?