Probability using Bayes formula

Question

Two masked robbers try to rob a crowded bank during the lunch hour but the teller presses a button that sets off an alarm and locks the front door. The robbers, realizing they are trapped, throw away their masks and disappear into the chaotic crowd. Confronted with $40$ people claiming they are innocent, the police give everyone a lie detector test. Suppose that guilty people are detected with probability $0.95$, and innocent people appear to be guilty with probability $0.01$. What is the probability that Mr. Jones is guilty given that the lie detector says he is?

The answer is $\frac{5}{6}$, but i don't know how they arrived at that.

Answer 1

Let $P$ denote the answer.

We have $$P= \frac {\frac 2{40}\times .95}{ \frac {38}{40}\times .01+\frac 2{40}\times .95}=\frac {2\times 95}{38+2\times 95}=\frac 56$$

Answer 2

Using Bayes's rule, we can write $$ \mathbb{P}(B\mid A)=\frac{\mathbb{P}(A\mid B)\mathbb{P}(B)}{\mathbb{P}(A)}=\frac{\mathbb{P}(A\mid B)\mathbb{P}(B)}{\mathbb{P}(A\mid B)\mathbb{P}(B)+\mathbb{P}(A\mid B^c)\mathbb{P}(B^c)}$$ where $B^c$ is the complement of $B$.

In your problem, let $A$ be the event that the lie detector reports that Mr. Jones is guilty, and $B$ the event that Mr. Jones is in fact guilty. Based on the information in the problem, what are $\mathbb{P}(A\mid B), \mathbb{P}(A\mid B^c), \mathbb{P}(B)$, and $\mathbb{P}(B^c)$?