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?

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:

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

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:

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

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

i believe the correct answer is b.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation: