Question

Suppose you have the following information. 12% of all drivers do not have a valid driver’s license, 6% of all drivers have no insurance, and 4% have neither (i.e., do not have driver’s license and do not have insurance). The probability that a randomly selected driver either fails to have a valid license or fails to have insurance is about
(a) 0.18

(b) 0.2

(c) 0.22

(d) 0.072

(e) 0.14

Answer 1

Answer:

e) 0.14

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a driver does not have a valid driver's license.

B is the probability that a driver does not have insurance.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a driver does not have a valid driver's license but has insurance and $A \cap B$ is the probability that a driver does not have any of these things.

By the same logic, we have that:

$B = b + (A \cap B)$

We start finding these values from the intersection.

4% have neither

This means that $(A \cap B) = 0.04$

6% of all drivers have no insurance

This means that $B = 0.06$. So

$B = b + (A \cap B)$

$0.06 = b + 0.04$

$b = 0.02$

12% of all drivers do not have a valid driver’s license

This means that $A = 0.12$

So

$A = a + (A \cap B)$

$0.12 = a + 0.04$

$a = 0.08$

The probability that a randomly selected driver either fails to have a valid license or fails to have insurance is about

$P = a + b + (A \cap B) = 0.08 + 0.02 + 0.04 = 0.14$

So the correct answer is:

e) 0.14