The drunk man and the six keys

[iplraf]

A drunken man has 6 keys, one of which opens the door to his house.
He tries the keys at random, one by one, and independently.
What's the probability of he opens the door in the last try if the wrong keys are eliminated?
A. 1
B. 5/6
C. 1/6
D. 1/2
E. 2/3

[parallel_chase]

5/6 * 4/5 * 3/4 * 2/3 * 1/2 = 1/6


whats the oA?

[iplraf]

Sure!
That's it!

[Stuart@KaplanGMAT]

A drunken man has 6 keys, one of which opens the door to his house.
He tries the keys at random, one by one, and independently.
What's the probability of he opens the door in the last try if the wrong keys are eliminated?
A. 1
B. 5/6
C. 1/6
D. 1/2
E. 2/3

We really don't need calculations to figure this one out.

There are 6 keys total. The probability that the first 5 keys are wrong is the same as the probability that the last key is correct, which is simply 1/6.

[parallel_chase]

Very true Stuart, i figured that out once I did all the calculations.

Thanks

[Ian Stewart]

What's the probability of he opens the door in the last try

I'm sure the intended answer is 1/6, and Stuart's solution above is perfect, but I find the wording of the question ambiguous. Isn't it absolutely certain that the last key he tries is the right one? After all, once he finds the right key, won't he just go inside? If the first key he tries is the right key, then his first try will also be his last try, after all. Or are we supposed to assume he's so sauced that he's going to keep trying other keys even after he opens the door? But still sober enough to go through the keys systematically, eliminating wrong keys?

Anyway, I'm curious about the source of the question- it doesn't read like a real GMAT question!