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)
I am not getting any answers right!!!!
I dont know whats going to be my plight in the exams!
D is the answer. Use the multinomial distribution.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.