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July 26th, 2011, 07:03 AM  #1 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  probability question
It is estimated that onequarter of the drivers on the road between 11pm and midnight have been drinking during the evening.If a driver has not been driving,the probability that he will have an accident at that time of night is 0.004%;if he has been drinking,the probability of a accident goes up to 0.02%. a)What is the probability that a car selected at random at that time of night will have an accident b)A policeman on the beat at 11.30pm see a car run into a lamppost,and jumps to the conclusion that driver has been drinking.What is the probability that he is right? my solution a) P(car selected at random at night will have accident) D=driver have been drinking,A=Accident happen =P(DA)+P(D'A) =(1/4 x 0.0002)+ (3/4 x 0.00004) =8 x 10^5 Probability that car selected at random at that time of night will have an accident is 8x 10^5 x 100% = 0.008% b)i think it is a probability of event.But i try several way but still can't get the answer.I need help. 
July 26th, 2011, 08:08 AM  #2 
Newbie Joined: Jul 2011 Posts: 1 Thanks: 0  Re: probability question
b) the probability that the policeman is right is the probability that the driver was drinking conditional on that he had an accident. Then we get from bayes theory P(DA)=P(AD)P(D)/P(A) where P(AD)=0.02%, p(D)=1/4 and P(A) is what you found in a) 
July 27th, 2011, 05:36 AM  #3 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  Re: probability question
Thank you very much.I get it now.


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