# Math Help - Conditional Probabilty

1. ## Conditional Probabilty

Suppsoe it takes at least 10 votes of guilty from a 12-member jury to convice a defendant. Suppose that the probability an individual juror votes a guilty person innocent is 0.2, whereas the probability a juror votes an innocent person guilty is 0.1, and assume jurors reach their decisions independently.

(a) What is the distribution of the number of votes of guilty if the defendant is innocent, and if they are guilty?
(b) Find the probability a defendant is convicted given that he is guilty, and find the probability a defendant is convicted given that he is innocent.
(c) If 65% of all defendants are guilty, what is the probability the jury reaches the correct decision on a randomly selected defendant?
(d) Again, if 65% of all defendants are guilty, what is the probability that a convicted defendant is actually innocent?

Does anyone know how to do this? I don't understand part (a) and think i may need that for the other parts?

2. Originally Posted by Revilo
(b) Find the probability a defendant is convicted given that he is guilty,
He can be convicted guilty or he can be convited innocent. (We are assuming he is really guilty).

Each person has a .8 chance of convicting him guilty. So the probability that he is convicted guilty is $\sum_{n=10}^{12}C_{12,n}(.8)^n(.2)^{12-n}$.

Each person has a .2 chance of convicting him innocent. So the probability that he is convicted innocent is $\sum_{n=10}^{12}C_{12,n}(.2)^n(.8)^{12-n}$.

3. Originally Posted by Revilo
(c) If 65% of all defendants are guilty, what is the probability the jury reaches the correct decision on a randomly selected defendant?
Let $C$ be the event that they make a correct decision. Let $I$ be the event that he is innocent and $G$ be that he is guilty (so $G=I^C$).

Then,
$P(C) = P(C|I)P(I)+P(G|I^C)P(I^C) = P(C|I)P(I)+P(C|G)P(G)$

In this case $P(I)=.35$ and $P(G) = .65$.
Thus, $P(C) = (.35)P(C|I)+(.65)P(C|G)$.

Now $P(C|I)$ is the probability that the jury reaches a correct decision given that he is innocent. So we need 10 or 11 or 12 to convict him innocent. The probability that an individual member says innocent is .9. Thus, the overal probability that we seek is $\sum_{n=10}^{12} C_{12,n}(.9)^n (.1)^{12-n}$. Do the same with the other one.

4. Thanks for the replys. I can follow your reasoning, but i don't understand what the $C_{12,n}$ stands for in the following equations:

Originally Posted by ThePerfectHacker
$\sum_{n=10}^{12}C_{12,n}(.8)^n(.2)^{12-n}$.
$\sum_{n=10}^{12}C_{12,n}(.2)^n(.8)^{12-n}$.

5. Originally Posted by Revilo
Thanks for the replys. I can follow your reasoning, but i don't understand what the $C_{12,n}$ stands for in the following equations:
It is a binomial coefficient:

$C_{12.n} = \frac{12!}{(12-n)!n!}$

Though having both parameters as subscripts is not standard notation.

RonL

6. What he means:

there are 12 jurors. Of these 12 jurors, at least 10 are going to vote GUILTY if he is to be convicted guilty. But it could be any permutation of the 12 jurors. So we do ${12 \choose 10}+{12 \choose 11} + {12 \choose 12}=\sum_{n=10}^{12}{12 \choose n}$

Now $P(C|I)$ is the probability that the jury reaches a correct decision given that he is innocent. So we need 10 or 11 or 12 to convict him innocent. The probability that an individual member says innocent is .9. Thus, the overal probability that we seek is $\sum_{n=10}^{12} C_{12,n}(.9)^n (.1)^{12-n}$. Do the same with the other one.
$\sum_{n=3}^{12} C_{12,n}(.9)^n (.1)^{12-n}$