Question

20 points Return to questionItem 4Item 4 20 points Police records in the town of Saratoga show that 13 percent of the drivers stopped for speeding have invalid licenses. If 14 drivers are stopped for speeding, find the probabilities that (a) None will have an invalid license. (Round your answer to 4 decimal places.) Probability .13 .13 Incorrect (b) Exactly one will have an invalid license. (Round your answer to 4 decimal places.) Probability Not attempted (c) At least 2 will have invalid licenses. (Round your answer to 4 decimal places.) Probability Not attempted

Answer 1

a) 0.1423

b) 0.2977

c) 0.56

Explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that
`p = 0.13`

14 drivers are stopped

This means that
`n = 14`

(a) None will have an invalid license.

This is
`P(X = 0)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(14,0).(0.13)^(0).(0.87)^(14) = 0.1423`

(b) Exactly one will have an invalid license.

This is
`P(X = 1)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 1) = C_(14,1).(0.13)^(1).(0.87)^(13) = 0.2977`

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

`P(X < 2) + P(X \geq 2) = 1`

We want
`P(X \geq 2)`

So

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44`

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56`