A car rental agency has 9 midsized and 15 compact cars on its lot, from which 6 will be selected. Assuming that each car is equally likely to be selected and the cars are selected at random, determine the probability that the cars selected consist of all midsized cars.

Question

bahrom7893 · Answer

The probability that the first car selected is midsized:
(9/(9+15)) = 9/24
2nd car is midsized:
8/23, and so on, so:
(9/24)*(8/23)*(7/22)*(6/21)*(5/20)*(4/19) did I get this right?

anonymous · Answer

nope

bahrom7893 · Answer

okay hmmm hold on i must have not considered something.. ughh i hate probability

anonymous · Answer

could this work. see attachment

bahrom7893 · Answer

chemwile that's what I got.

anonymous · Answer

Alright, so the selection of cars will go like this. (9/24)(8/23)(7/22)... and so on and so forth, until all six slots are filled. Thus, (9!/3!)/(24!/18!) gives us the probability of one configuration. If we add the probabilities of all configurations, there being six, it will end up as 6(9!/3!)/(24!/18!) or 0.37%.

I think.

anonymous · Answer

.0006 was the answer.

bahrom7893 · Answer

yes katlyns, u said i was wrong.. I WAS RIGHT! WOOT!

bahrom7893 · Answer

so was chem.

anonymous · Answer

Darn, lol, yes they were right.

anonymous · Answer

Great Team!

anonymous · Answer

i didn't see that on either one of yalls. i typed all your fractions in and got a huge fraction and decimal

anonymous · Answer

but thank you!

bahrom7893 · Answer

katlyns, let's go for another one