Question

At a certain stage in a trial, the judge feels the odds are two to one the defendant is guilty. It is determined that the defendant is left handed. An investigator convinces the judge this is six times more likely if the defendant is guilty than if he were not. What is the likelihood, given this evidence, that the defendant is guilty? (Use Bayes’ Theorem.)

Answer 1

here from above probability(guilty)+P(not guilty)=1

also probability(guilty)=2*P(not guilty)

hence P(guilty)=2/3 and P(not guilty)=1/3

P(left handed|guilty)=6x/7

P(left handed|not guilty)=x/7

hence from Bayes theorem:

P(guilty given left handed)=P(guilty)*P(left handed|guilty)/(P(guilty)*P(left handed|guilty)+P(not guilty)*P(left handed|not guilty))

=(2/3)*(6/7)/((2/3)*(6/7)+(1/3)*(1/7))=12/13