Question

7. A car dealership has 8 red, 9 silver, and 5 black cars on the lot. Seven cars are randomly chosen to be displayed(in no particular order) in front of the dealership. Find the probability that (a) all are silver (b) 5 are red and 2 are black (c) 4 are red and the rest are silver

Answer 1

a) 0.0002 = 0.02% probability that all are silver.

b) 0.0033 = 0.33% probability that 5 are red and 2 are black

c) 0.0345 = 3.45% probability that 4 are red and the rest are silver

Explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question, the order in which the cars were selected is not important, so we use the combinations formula.

Combinations formula:

`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)!)`

(a) all are silver

Desired outcomes:

7 silver cars, from a set of 9. So

`D = C_(9,7) = (9!)/(7!(9-7)!) = 36`

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

`T = C_(22,7) = (22!)/(7!(22-7)!) = 170544`

Probability:

`p = (D)/(T) = (36)/(170544) = 0.0002`

0.0002 = 0.02% probability that all are silver.

(b) 5 are red and 2 are black

Desired outcomes:

5 red, from a set of 8

2 black, from a set of 5

`D = C_(8,5)*C_(5,2) = (8!)/(5!(8-5)!)*(5!)/(2!(5-2)!) = 56*10 = 560`

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

`T = C_(22,7) = (22!)/(7!(22-7)!) = 170544`

Probability:

`p = (D)/(T) = (560)/(170544) = 0.0033`

0.0033 = 0.33% probability that 5 are red and 2 are black

c) 4 are red and the rest are silver

Desired outcomes:

4 red, from a set of 8

3 silver, from a set of 9

`D = C_(8,4)*C_(9,3) = (8!)/(4!(8-4)!)*(9!)/(3!(9-3)!) = 70*84 = 5880`

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

`T = C_(22,7) = (22!)/(7!(22-7)!) = 170544`

Probability:

`p = (D)/(T) = (5880)/(170544) = 0.0345`

0.0345 = 3.45% probability that 4 are red and the rest are silver