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P1.T2.20.2. More probabilities and Bayes rule


New Member
Hi @David Harper CFA FRM Can you give more explanation to the example in the study notes of Probabilities, I have highlighted my doubts in yellow. :)

I don't understand on 39.13*75(which is conditional beats of first year?)


David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi Navjyot (@navjyotbirdy ) In #2.3, because we don't know whether the manager is a star/not, the probability she beats next year is a weighted probability: 39.13% probability that she is a star, which we induced via Bayes Pr[S|3B], multiplied by 75% probability that a star beats; plus 60.87% probability that she is not a star, which is given by (1-60.87%), multiplied by the 50% probability that a non-star beats (i.e., 50%). Put is this way, if we only used the a priori (unconditional) probabilities of star/not, which are 16%/84%, then what is the probability of next year's beat? It is 16%*75% + 84%*50%. But uses Bayes (because we have observed 3 beats) we updated the probabilities of start/not to 39.13%/(1-39.13%); i.e., after 3 beats in a row, it is more likely this is a star!

Re GARP'2 #4, it's just SHOWING the answer to the question. Although "Neutral" looks like a typo. Should be Pr(Passage | Constant) = 6.0%/27.0% = 22.22%. That should be the to the question, which is a conditional probability. Dividing the blue joint probability (6.0%) by the orange unconditional probability (27.0%) gives us the conditional probability that answer's the question. Thanks,