Question

Empirical evidence suggests that 25% of Florida drivers are uninsured.
If four random Florida drivers are involved in an accident, what is the probability that more than one of them are uninsured?

Answer 1

If the question is asking if only one isnt insured, the chances are 1/12.
If the question is asking if at least one isnt insured, the chances are 1/4.

Answer 2

Answer:

The probability is 0.2617

Step-by-step explanation:

The variable that said the number of drivers that are uninsured follow a binomial distribution, so the probability that x drivers are uninsured is calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

Where n is the number of drivers involved in an accident and p is the probability that any driver is uninsured, then:

$P(x)=\frac{4!}{x!(4-x)!}*0.25^{x}*(1-0.25)^{4-x}$

So, the probability that more than one of them are uninsured is:

P(x>1) = P(2) + P(3) + P(4)

Therefore, the values of P(2), P(3) and P(4) are:

$P(2)=\frac{4!}{2!(4-2)!}*0.25^{2}*(1-0.25)^{4-2}=0.2109$

$P(3)=\frac{4!}{3!(4-3)!}*0.25^{3}*(1-0.25)^{4-3}=0.0469$

$P(x)=\frac{4!}{4!(4-4)!}*0.25^{4}*(1-0.25)^{4-4}=0.0039$

Finally, the probability that more than one of them are uninsured is:

P(x>1) = 0.2109 + 0.0469 + 0.0039 = 0.2617