3
$\begingroup$

The question I'm trying to answer is part d) of the below question:

The defendant in a court case is either guilty or innocent. Suppose the defendant is convicted if at least $10$ of the $12$ members of the jury vote that he is guilty, and acquitted otherwise. Suppose also that an individual juror votes that a guilty person is innocent with probability $0.20$ and that an innocent person is guilty with probability $0.30$. Suppose too that all jurors cast their votes independently. (a) What is the distribution of the number of votes of guilty if the defendant is innocent? (b) Find the probability that an innocent defendant is convicted. (c) Find the probability that a guilty defendant is convicted. (d) Suppose that $10$% of all defendants are guilty. Find the probability that a convicted defendant is actually innocent. Find the probability that a defendant who is acquitted is actually guilty.

The question is based on binomial distributions. For part b) and c) I obtained $0.00021$ and $0.5584$ respectively. However for part d) I'm not exactly sure how to proceed. I know that the probability of someone being convicted is $P(\text{convicted}) = 0.00021 + 0.5584 = 0.55861$. So $55.861$% of the time someone is convicted. Let $E = \text{event that someone is convicted}$, $G = \text{person is guilty}$, $I= \text{person is innocent}$. Now, the probability that a convicted defendant is actually innocent would therefore be the conditional probability that given $E$, $G$ occurs. This would be $$P(I|E) = \frac{P(I \cap E)}{P(E)} = \frac{P(\text{defendant is convicted and innocent})}{P(E)} = \frac{0.00021}{0.55861}=0.000376$$ Similarly I found $$P(G|E') = \frac{P(G \cap E')}{P(E')} = \frac{P(\text{defendant is acquitted and guilty})}{P(E')} = \frac{1-0.5584}{1-0.55861}=1$$ Now both of my answers were obviously wrong. The probability that a convicted defendant is actually innocent is $0.003373$ and the probability that a defendant who is acquitted is actually guilty is $0.04679$. I'm not exactly sure how these answers were obtained.

$\endgroup$
1
$\begingroup$

The solution is given in the image. You are correct in the first few part and the last one is almost correct with only a few mistakes. You will have to multiply P(Innocent) and P(Guilty) to respective probabilities. If you have any doubt let me know.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.