Math Is Fun Forum

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!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: First time licensees!

Hi;

D is the answer. Use the multinomial distribution.

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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.

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.