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.
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.