# Conditional Probability [duplicate]

A judge is 35% sure that X has committed a crime. A and B are two witnesses who know whether X is innocent or guilty. However, A is X’s friend and will lie with probability 0.25 if X is guilty. He tells the truth if he is innocent. B is X's enemy and will lie with probability 0.30 if X is innocent. He will tell the truth if X is guilty. What is the probability that, in the course of the trial, A and B will give conflicting testimony?

L=lie
T=tell the truth
G=guilty
I=innocent


I have attached my solution. I have a feeling that my step: $P(C|G)=P(L|G)\times P(T|G)$ is wrong but I'm not too sure if it is.

I am not quite sure how you got some probabilities, like 0.1 for the second branch for A. Also your probability statement $P(C|G) = P(L|G) \times P(T|G)$ is not quite clear to me.

Here is how I would approach to this interesting question.

I use "$A = \{X = G\}$" denote that $A$'s testimony is that $X$ is guilty. Other notations have the analogous meaning. According to the context, we have following statements. $$P\left(A = \{X = G\} \mid X=G \right) = 0.75$$ $$P\left(A = \{X = I\} \mid X=G \right) = 0.25$$ $$P\left(A = \{X = G\} \mid X=I \right) = 0$$ $$P\left(A = \{X = I\} \mid X=I \right) = 1$$ Similarly, $$P\left(B = \{X = G\} \mid X=G \right) = 1$$ $$P\left(B = \{X = I\} \mid X=G \right) = 0$$ $$P\left(B = \{X = G\} \mid X=I \right) = 0.7$$ $$P\left(B = \{X = I\} \mid X=I \right) = 0.3$$

What is the probability that, in the course of the trial, A and B will give conflicting testimony?

This question is answered by the following probability statement, \begin{align*} \tag{1} P\left(A = \{X = G\} \cap B = \{X = I\} \right) + P\left(A = \{X = I\} \cap B = \{X = G\} \right). \end{align*} Now we need some assumptions to compute the above probability,

1. The prior probability $P(X = G) = 0.35$, according to the judge;
2. $A$ and $B$ would make their own independently, solely depending on whether $X$ is guilty or not. That said, $A$'s testimony would not affect $B$'s judgement, vice verse.

Then for the first piece in (1), \begin{align*} & \quad P\left(A = \{X = G\} \cap B = \{X = I\} \right) \\ & = P\left(A = \{X = G\}\right) \cdot P\left(B = \{X = I\} \right) \end{align*} Furthermore, for example, \begin{align*} & \quad P\left(A = \{X = G\}\right)\\ & = P\left(A = \{X = G\} \mid X=G \right) P\left(X = G \right) + P\left(A = \{X = G\} \mid X=I \right) P\left(X = I \right) \\ & = 0.75 \times 0.35 + 0 \times (1-0.35) \\ & = 0.2625 \end{align*} And you can calculate other pieces analogously. Based on my approach and calculation, the final probability to your question is $$0.0511875 + 0.5936875 = 0.644875$$ Correct me if I did wrong.

(Since this is a self-study question, I removed the other part of the solution. The above hints should be enough since the idea is straightforward. I can provide the full answer if you still have questions.)