car insurance stats
Math/Stats?
A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers living on one particular street, 3 were involved in a car accident last year. If 15 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year? What is the probability of getting exactly 3 who were involved in a car accident last year?
Help me solve this, Please!!!!!!
Question 1) What is the probability of getting exactly 3
who were involved in a car accident last year?
For this problem, use the Binomial Probability Formula:
P(N) = N! / [k! * (N - k)!] * p^k * q^(N – k)
where:
N = Number of opportunities for event x to occur (15)
k = Number of times that event x should happen (3)
p = Probability of a success (.08)
q = Probability of failure (.92)
! = Factorial
P(k) = N! / [k! * (N - k)!] * p^k * q^(N – k)
P(4) = 15! / [3! * (15 - 3)!] * (.08)^3 * (.92)^(15 – 3)
P(4) = .085652 or 8.57%
Yeah, the math gets really messy,
but you can use an online Binomial Probability Calculator:
http://faculty.vassar.edu/lowry/ch5apx.html
Or you can do it on your TI-83/84 Calculator:
http://mathbits.com/MathBits/TISection/Statistics2/bernoulli.htm
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Question 2) What is the probability of getting 3 or more
who were involved in a car accident last year?
1 – [P(0) + P(1) + P(2)]
P(0) = (.92)^15 = 0.286297
P(1) = 15! / [1! * (15 - 1)!] * (.08)^1 * (.92)^(15 – 1)
P(1) = .373431 or 37.34%
P(2) = 15! / [2! * (15 - 2)!] * (.08)^2 * (.92)^(15 – 2)
P(2) = .227306 or 22.73%
P(x?3) = 1 – (0.286297 + .373431 + .227306)
P(x?3) = 1 – (0.887034) = .112966 or 11.3%
The above was computed using the 8% as given in the question.
Good luck in your studies,
~ Mitch ~
P.S – I hope my explanation is clear, if not,
you’re are welcome to e-mail me with any questions.
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