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#1 2011-04-18 17:38:28

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

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!

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#2 2011-04-18 18:14:43

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,258

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2011-04-25 20:47:35

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

Re: First time licensees!

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

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