Question

g A car insurance company has determined that 8% of all drivers 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

Answer 1

`P(x \geq 3)=1- [P(X=0)+P(X=1) +P(X=2)]`

And we can find the individual probabilities like this:

`P(X=0)=(15C0)(0.08)^0 (1-0.08)^(15-0)=0.286`

`P(X=1)=(15C1)(0.08)^1 (1-0.08)^(15-1)=0.373`

`P(X=2)=(15C2)(0.08)^1 (1-0.08)^(15-2)=0.227`

`P(x \geq 3)=1- [0.286+0.373+0.227]=0.114`

Explanation:

Let X the random variable of interest "number of cars involved in an accident", on this case we now that:

`X \sim Binom(n=15, p=0.08)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

And we want to find this probability:

`P(x \geq 3)=1- [P(X=0)+P(X=1) +P(X=2)]`

And we can find the individual probabilities like this:

`P(X=0)=(15C0)(0.08)^0 (1-0.08)^(15-0)=0.286`

`P(X=1)=(15C1)(0.08)^1 (1-0.08)^(15-1)=0.373`

`P(X=2)=(15C2)(0.08)^1 (1-0.08)^(15-2)=0.227`

`P(x \geq 3)=1- [0.286+0.373+0.227]=0.114`