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A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a four. Find the probability that it is actually a four. |
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Answer» Solution :In a throw of die, let `E_1` = event of getting a FOUR `E_2` = event of getting not four and E = event that the man REPORTS that it is a four Thus,`P(E_1)=1/6 " and " P(E_2)=1-1/6=5/6` `P(E|E_1)` = probability that the man reports that four occurs, when four has actually occurred. = probability that the man speaks the truth `=4/5` `P(E|E_2)` = probability that the man reports that four occurs, when four has not actually occurred. = probability that the man does not SPEAK the truth `=(1-4/5)=1/5` Probability of getting a four, given that the man reports it to be a four `=P(E_1|E)` `=(P(E|E_1)P(E_1))/(P(E|E_1)P(E|E_2)P(E_2))` `=((4/5)(1/6))/((4/5)(1/6)+(1/5)(5/6))` `=4/(4+5)=4/9` |
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