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A judge is 35% sure that Mary has committed a crime. Jack and Tim are two witnesses who know whether Mary is innocent or guilty. However, Jack is Mary's friend and will lie with probability 0.25 if Mary is guilty. He tells the truth if she is innocent. Tim is Mary's enemy and will lie with probability 0.30 if Mary is innocent. He will tell the truth if Mary is guilty. What is the probability that Mary is guilty if Jack and Tim gave conflicting testimonies?

Another witness, David, convinces the judge that there is 85% chance that the criminal is left-handed. If 23% of the population is left-handed and Mary is also left-handed, with this new information, how certain should the judge be of the guilt of Mary?

By the Law of Total Probability, Let G be the event where Mary is guilty. Let C be the event where Jack and Tim gave conflicting testimonies.

P(C) = P(C|G)P(G)+P(C|G^c)P(G^c) = 0.25*0.35 + 0.30*0.65 = 0.2825.

I need help with the 2nd part of the question...

Thanks in advance!

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    $\begingroup$ Do we know specifically how Jack and Tim testified? $\endgroup$ – paw88789 Feb 15 '15 at 11:35
  • $\begingroup$ Hi, i edited the question $\endgroup$ – Stephanie Kim Feb 15 '15 at 11:45
  • $\begingroup$ How is the fact that "a judge is $35\%$ sure that Mary has committed a crime" has anything to do with the question? $\endgroup$ – barak manos Feb 17 '15 at 6:39
  • $\begingroup$ I suppose it is the question's way of stating that P(G) = 0.35 $\endgroup$ – Stephanie Kim Feb 17 '15 at 6:59
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In the first question, you are asked to compute $P(G|C)$. Let $J$ be the event "Jack said guilty". Then

$$ P(G|C) = \frac{P(C|G) P(G)}{P(C)}=\frac{P(C|G) P(G)}{P(C|G) P(G) +P(C|\bar{G}) P(\bar{G}) }$$

And $P(C|G) = P(J ,\bar M |G) + P(\bar J , M |G)$. Can you go on from here?

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  • $\begingroup$ what does P(J,M¯|G) mean? $\endgroup$ – Stephanie Kim Feb 15 '15 at 14:22
  • $\begingroup$ I fixed the letters. $P(J ,\bar T |G) =$ probability that Jack says "guilty" and Tim says "not guilty" given that Mary is guilty. The bar means the complementary event (not...). $\endgroup$ – leonbloy Feb 15 '15 at 15:38
  • $\begingroup$ Do you have any idea how to solve the next part of the question where David comes in? Thank you for your help for the previous part! $\endgroup$ – Stephanie Kim Feb 16 '15 at 12:05

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