Hi again

This one is kind of tough. If anybody have a clue please let me know

Attachment 1742

Thanks

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- Feb 16th 2007, 03:31 AMmauro21plProbability
Hi again

This one is kind of tough. If anybody have a clue please let me know

Attachment 1742

Thanks - Feb 16th 2007, 08:20 AMCaptainBlack
the proportion of people that have accidents in a fixed year is the sum

of the proportion in each category times the probability of accident in a year for that category:

In this case this is:

0.2*0.05+0.5*0.15+0.3*0.3 = 0.175

For the second part we can use Bayes' theorem:

P(good|no accidents) = P(no accidents|good)P(good)/P(no accidents)

..........................= (1-0.05)*0.2/(1-0.175) = 0.230

RonL