Question

a car rental agency at a local airport has available 5 fords, 7 chevrolets, 4 dodges, 3 hondas, and 4 toyotas. if the agency randomly selects 9 of these cars to chauffeur delegates from the airport to the downtown convention center, find the probability that 2 fords, 3 chevrolets, 1 dodge, 1 honda, and 2 toyotas are used.

Answer 1

Explanation:

Sup? Pay attention closely (or not XD)

Short answer: 0.0308 or 3.08%

We use the combination formula C = n!/(r! * (n-r)!) for each brand. It basically tells us the number of ways to select a certain number of cars:

1) Fords: C(5, 2) = 5! / (2! * (5 - 2)!) = 10 ways to pick 2 Fords

2) Chevrolets: C(7, 3) = 7! / (3! * (7 - 3)!) = 35 was to pick 3 Chevys

3) Dodges: C(4, 1) = 4! / (1! * (4 - 1)!) = 4 ways to pick 1 Dodge

4) Hondas: C(3, 1) = 3! / (1! * (3 - 1)!) = 3 ways to pick 1 Honda

5) Toyotas: C(4, 2) = 4! / (2! * (4 - 2)!) = 6 ways to pick 2 Toyotas

Then you multiply all them results. Peep this:

10*35*4*3*6 = 25,200 ways to pick your target cars from each brand in total (sheesh)

Now how many ways would it be to pick 9 out of the 23 cars??

Use your best friend combination problem to solve this!

C(23, 9) = 23! / (9! * (23 - 9)!) = 817,190 (sheesh)

Now divide! 25,200/817,190 = 0.0308 or 3.08%

Thanks for listening to my TED talk. Hope this helps.