Question

data compiled by the highway patrol department regarding the use of seat-belts by drivers in a certain area after the passage of a compulsory seat-belt law are shown in the accompanying table. (round your answers to three decimal places.) drivers percentage of drivers in group percent of group stopped for moving violation group i (using seat-belts) 65 0.3 group ii (not using seat-belts) 35 0.3 (a) if a driver in that area is stopped for a moving violation, what is the probability that he or she will have a seat-belt on? (b) if a driver in that area is stopped for a moving violation, what is the probability that he or she will not have a seat-belt on?

Answer 1

(a) The probability that a driver stopped for a moving violation has a seat-belt on is 0.65. (b) The probability that a driver stopped for a moving violation does not have a seat-belt on is 0.35.

Let's use the given data to calculate the probabilities:

(a) Probability that a driver with a seat-belt on is stopped for a moving violation:

`\[ P(\textSeat-belt ) = \frac{\text{Percentage of drivers with seat-belts stopped}}{\text{Percentage of drivers stopped}} \]\[ P(\text Violation) = (0.3 * 0.65)/(0.3) = 0.65 \]`

(b) Probability that a driver without a seat-belt is stopped for a moving violation:

`\[ P(\textNo Seat-belt ) = \frac{\text{Percentage of drivers without seat-belts stopped}}{\text{Percentage of drivers stopped}} \]\[ P(\textNo Seat-belt ) = (0.3 * 0.35)/(0.3) = 0.35 \]`

So, the probabilities are:

(a) The probability that a driver with a seat-belt on is stopped for a moving violation is 0.65.

(b) The probability that a driver without a seat-belt is stopped for a moving violation is 0.35.