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**getback0****Member**- Registered: 2010-10-10
- Posts: 25

A large pool of adults earning their first drivers 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 drivers 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!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,475

Hi;

D is the answer. Use the multinomial distribution.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**authentication****Member**- Registered: 2011-04-05
- Posts: 2

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

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