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
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probability--gmatprep
Source: Beat The GMAT — Problem Solving |
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
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
