probability--gmatprep

[ildude02]

Is there a way to use combination technique to solve this problem?

[elanmorin5]

First post for me, so take it easy...

I thought of it in terms of probabilities:

We have to pick a different car each time, and it is equally likely that any of the three will be picked.
So, in the first selection, we can think of the probability of picking a car as "1"
In the second selection, the probability of picking a different car is 2/3 (since two of the cars will be different from the first one).
In the third selection, only one car will be untried, so the probability of choosing it is 1/3.
Thus, in order to find the probability for the events to all occur together, we multiply the probabilities of the individual events:
1*2/3*1/3 = 2/9, the answer

[lightbulb]

Let the cars be A, B and C.
Total number of ways in which he could have taken the ride:

1st time: Any of three cars, so 3 ways
2st time: Any of three cars, so 3 ways
3rd time: Any of three cars, so 3 ways

Exhaustive cases: 3*3*3 = 27.

Favorable cases:

1st time 3 ways (either A, B or C)
2nd time 2 ways (because he can't ride the car he rode the first time)
3rd time 1 way (because he can't ride the cars in which he rode the first and second time)

Total favorable ways: 3*2*1 = 6

Pr = 6/27 = 2/9