# Insurance Company accident probability

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• May 18th 2007, 06:50 PM
hello there
Insurance Company accident probability
I'm a little unshure on how to solve this problem please help.

The clayton Automobile Insurance Company only insures drivers that are 18 years of age. It has been determined that the probability that an 18 year old driver has an accident in a one year span is 24.65%. If the company insures twenty 18 year old drivers for a one year contract, determine the following.

a) What is the probability that none of the drivers have an accident during their contract?

b) What is the probability that at least 2 drivers have an accident during their contract?
• May 18th 2007, 07:14 PM
solace
I think i figured this one out

A)
P(x=0)= 20C0(.2465)^0(.7535)^20
P(x=0)= .00348

B)
P(x at least 2)= 1 - P(0) + P(1)
1 - (20C0(.2465)^0(.7535)^20 + 20C1(.2465)^1(.7535)^19
1- (.00348 + .0228) = .974
P(x at least2) = .974

Can some verify that these answers are correct please?

We are both reviewing for a final exam and we need to know that we are prepared.
• May 19th 2007, 01:00 AM
CaptainBlack
Quote:

I'm a little unshure on how to solve this problem please help.

The clayton Automobile Insurance Company only insures drivers that are 18 years of age. It has been determined that the probability that an 18 year old driver has an accident in a one year span is 24.65%. If the company insures twenty 18 year old drivers for a one year contract, determine the following.

a) What is the probability that none of the drivers have an accident during their contract?

This one we can do straight off. All twenty must have no accident, so

p=(1-0.2465)^20 ~= 0.003481

But we may as well observe that the number of drivers who do have
accidents has a binomial distrition with single case probability 0.2465,
that is P(N drivers with accidents) ~ B(20, 0.2465).

Then prob none have accidents is:

b(0; 20, 0.2465) = 20!/[0! (20-0)!] 0.2465^0 (1-0.2465)^{20-0}

............................. = (1-0.2465)^20 ~= 0.003481

Quote:

b) What is the probability that at least 2 drivers have an accident during their contract?
The probability that at least 2 drivers have an accident is the probability
that 2 or more drivers have accidents which is 1 minus the probability that
onle 1 or 0 drivers have accidents, so here:

p(n>=2) = 1 - (p(0) + p(1))

........... = 1 - (b(0; 20, 0.2465) + b(1; 20, 0.2465))

........... ~= 1 - 0.003481 - 20!/[1! 19!] 0.2465^1 (1-0.2465)^19

........... ~= 1 - 0.003481 - 0.022773 ~= 0.9737

RonL