A probability problem about a thief? Imagine that a detective is 60 percent sure that Mr.X is the thief in an investigation.
2 days later, They find new information about the real thief. The real thief is left-handed. Mr.X is left-handed too !
They know that 20 percent of the people are left-handed in the city.
Now, How many percents should detective be sure that Mr.X is the thief?
 A: You have been given new and apparently certain evidence that a guilty suspect will certainly be left handed, but that anyone in the public will only be left-handed with probability $20\%$.
You wish to evaluate the probability that a particular suspect is guilty given that the suspect is left handed and has a $60\%$ prior probability of being guilty.
Let $G$ be the event that this suspect is guilty, and $L$ be the event that a suspect is left handed.
$$\begin{array}{|l|l|}\hline \mathsf P(G) = 0.60 &\textsf{ the prior probability of guilt.} \\ \hline \mathsf P(L\mid G) = 1.00 &\textsf{ the probability of a guilty suspect being left handed.} \\ \hline \mathsf P(L) = 0.20 &\textsf{ the probability of anyone being left handed.} \\\hline
 \end{array} \\ \mathsf P(G\mid L)=\mathsf P(G)~\mathsf P(L\mid G) = 0.60\cdot 1.00 = 0.60$$
Thus the new evidence cannot influence the estimate that a left handed suspect is guilty; it can only clear right handed suspects.
(And you cannot assume someone is more guilty because you now have fewer suspects. Sherlock Homes was wrong. When you have eliminated the impossible what is left is still only the possible.)
