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#1 2011-04-19 15:38:28

getback0
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First time licensees!

A large pool of adults earning their first driver’s license includes 50% low-risk drivers,
30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior
driving record, an insurance company considers each driver to be randomly selected from
the pool. This month, the insurance company writes 4 new policies for adults earning their
first driver’s license. What is the probability that these 4 will contain at least two more
high-risk drivers than low-risk drivers?
(A) 0.006 (B) 0.012 (C) 0.018 (D) 0.049 (E) 0.073


I figured -

P(L) = 0.5
P(H) = 0.2
P(M) = 0.3

Now, probability (atleast 2 more high risk drivers than low risk drivers) =

p( 0 L, 2 H, 2 M) + p(0L, 3 H, 1M) + p(0L, 4H, 0M) + p(1L, 3H)

= (0.2^2)*(0.3^2) + (0.2^3*0.3) + (0.2^4) + (0.5* 0.2^3)

= .0116


I am not getting any answers right!!!!

I dont know whats going to be my plight in the exams!

 

#2 2011-04-19 16:14:43

bobbym
Administrator

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Re: First time licensees!

Hi;

D is the answer. Use the multinomial distribution.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#3 2011-04-26 18:47:35

authentication
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Re: First time licensees!

Wow nice! this is really helpful. I'm still figuring out things right here.

 

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