Independence of Events and Conditional Probability A person tried by a 3-judge panel is declared guilty if at least 2 judges cast votes of guilty. Suppose that when the defendant is in fact guilty, each judge will independently vote guilty with probability 0.7, whereas when the defendant is in fact innocent, this probability drops to 0.2. Assume 70 percent of defendants are guilty. 


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*Let Ei, i = 1, 2, 3 denote the event that judge i casts a guilty vote. Are these events independent? Explain 

*Are Ei’s, i = 1, 2, 3, conditionally independent? Explain. 

*Compute the conditional probability that judge number 3 votes guilty given that judges 1 and 2 vote guilty.


I am honestly stuck on where to even start with this problem. I would assume that the events are independent because the problem states it as such.
 A: I will just do the first part, because the other parts are similar.
You want to find the probability that $J_3$ votes guilty given the info that $J_1$ and $J_2$ votes guilty, and for this, we first need the probability that $J_1$ and $J_2$ even votes guilty in the first place.
Let $J_{ig}$ denote the event that judge $i$ votes guilty.
\begin{align}
P(J_{1g}, J_{2g})&=P(J_{1g}, J_{2g}|\text{defendant is guilty})P(\text{defendant is guilty})\\
&+P(J_{1g}, J_{2g}|\text{defendant is innocent})P(\text{defendant is innocent}) \\
&=\left(\frac{7}{10}\right)^2\left(\frac{7}{10}\right)+\left(\frac{3}{10}\right)^2\left(\frac{2}{10}\right)\\
&=\frac{71}{200}
\end{align}
Moving on to the required probability:
\begin{align}
P(J_{3g}|J_{1g}, J_{2g})&=\frac{P(J_{3g}, J_{1g}, J_{2g}|\text{def guilty})P(\text{def guilty})+P(J_{3g}, J_{1g}, J_{2g}|\text{def innocent})P(\text{def innocent})}{P(J_{1g}, J_{2g})}\\
&=\frac{\left(\frac{7}{10}\right)^3\left(\frac{7}{10}\right)+\left(\frac{2}{10}\right)^3\left(\frac{3}{10}\right)}{\frac{71}{200}}\\
&=\frac{97}{142}
\end{align}
A: The answer to 1 is no.
$$P(E_i)=P(E_i|G)P(G)+P(E_i|I)P(I)=\frac{7}{10}\frac{7}{10}+\frac{2}{10}\frac{3}{10}=\frac{55}{100}$$
$$P(E_1E_2E_3)=P(E_1E_2E_3|G)P(G)+P(E_1E_2E_3|I)P(I)=\left(\frac{7}{10}\right)^3\frac{7}{10}+\left(\frac{2}{10}\right)^3\frac{3}{10}=\frac{2425}{10000}$$
And by the definition of independence, we have
$$P(E_1E_2E_3)\ne P(E_1)P(E_2)P(E_3)$$
The answer to 2 is no as the definition of conditional independence is
$$P((E_1\cap E_2\cap E_3)|G)=P(E_1|G)P(E_2|G)P(E_3|G)$$
We don't have
$$P(E_i|G)=P(E_i)$$
so, except for coincidence
$$P(E_1|G)P(E_2|G)P(E_3|G)\ne P(E_1)P(E_2)P(E_3)$$
and therefore
$$P((E_1\cap E_2\cap E_3)|G)\ne P(E_1\cap E_2\cap E_3)$$
For 3, $P(E_1E_2)=\frac{7}{10}\frac{7}{10}\frac{7}{10}+\frac{3}{10}\frac{2}{10}\frac{2}{10}=\frac{355}{1000}$.
Also $P(E_1E_2|E_3)P(E_3)=P(E_1)+P(E_2)-P(E_1+E_2)$, by conditional probability.
$P(E_1+E_2)=P(E_1E_2)+P(E_1E_2')+P(E_1'E_2)=\frac{355}{1000}+2\frac{195}{1000}=\frac{745}{1000}$.
$$P(E_1E_2|E_3)P(E_3)=\frac{355}{1000}$$
So
$$P(E_3|E_1E_2)=\frac{P(E_1E_2|E_3)P(E_3)}{P(E_1E_2)}=\frac{355}{745}=\frac{71}{149}$$
