Answer:
There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.
Step-by-step explanation:
For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And
is the probability of X happening.
In this problem
15 drivers are randomly selected, so
.
A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that
.
What is the probability of getting 3 or more who were involved in a car accident last year?
This is
.
Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:
![P(X < 3) + P(X \geq 3) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%2B%20P%28X%20%5Cgeq%203%29%20%3D%201)
![P(X \geq 3) = 1 - P(X < 3)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%203%29%20%3D%201%20-%20P%28X%20%3C%203%29)
In which
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
![P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B15%2C0%7D.%280.08%29%5E%7B0%7D.%280.92%29%5E%7B15%7D%20%3D%200.2863)
![P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B15%2C1%7D.%280.08%29%5E%7B1%7D.%280.92%29%5E%7B14%7D%20%3D%200.3734)
![P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B15%2C2%7D.%280.08%29%5E%7B2%7D.%280.92%29%5E%7B13%7D%20%3D%200.2273)
So
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.2863%20%2B%200.3734%20%2B%200.2273%20%3D%200.887)
Finally
![P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%203%29%20%3D%201%20-%20P%28X%20%3C%203%29%20%3D%201%20-%200.887%20%3D%200.113)
There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.