Eating my lunch break is gonna you wanna I was like a a little ninja but you didn’t want me to come to pick y’all lol I gotta is a time to go eat some lunch or later in my day or just let y’all go bye see y’all later love y’all and I have a a lot going on in my house I don’t have any hw I want y’all I want you can you guys come over and play it for like three minutes and I play ooooo you want me a big big house I don’t wanna is that iiiiiioooo was the night you got me a big play bye I love y’all so bye love y’all I love y’all so bye love y’all so I love you so bye love you too love y’all so bye love you too love y’all so bye love you bye love

6 0

3 years ago

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

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.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

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 $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

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:

$P(X < 3) + P(X \geq 3) = 1$