Question

It has been discovered that 5% of drivers stopped at a traffic stop had traces of alcohol on their person, and 10% of those stopped do not use seat belts. Furthermore, it has been discovered that the two offenses are unrelated to one another. Calculate the probability that exactly three of the five drivers stopped at random have committed either of the two offenses.

Answer 1

Let A denote the event "traces of alcohol", B denote the event "seat belts".

Then

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=P(A)+P(B)-P(A)P(B)$$

$$=0.05+0.10-0.05(0.10)=0.145$$

Let $X=$ the number of drivers committed either of the two offenses:$X\sim Bin(n, p).$

Given $n=5, p=0.145, q=0.855.$

$$P(X=3)=\dbinom{5}{3}(0.145)^3(0.855)^{5-3}$$

$$=0.02228621091$$