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A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probaility that it is actually a six. |
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Answer» Solution :In a throw of a die, LET `E_1` = EVENT of getting a SIX, `E_2`= event of not gettinga six, and E = event that the man reports that it is a six. Then, `P(E_1)=1/6, and P(E_2) =(1-1/6)=5/6`. `P(E//E_1)` = probability that the manreports that six occurs, when six has actually occurred = probability that the man speaks the TRUTH `=3/4`. `P(E//E_2)` = probability that the man reports that six occurs, when six has not actually occurred = probability that the man does not speak the truth `=(1-3/4)=1/4`. Probability of getting a six, GIVEN that the man reports it to be six `=P(E_1//E)` `=(P(E//E_1).P(PE_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2))`[by Bayes's theorem) `((3/4xx1/6))/((3/4xx1/6)+(1/4xx5/6))=(1/8xx3)=3/8`. Hence,the required probability is `3/8`. |
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