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by suhir.kumar » Sun Apr 15, 2012 11:01 am
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
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by Neo Anderson » Sun Apr 15, 2012 11:09 am
for this person to choose wrong key in the 1st try P = 5/6 (5 wrong keys 1 right key)
to choose wrong key in the 2nd try P = 4/5 (4 wrong keys 1 right key)
to choose wrong key in the 3rd try P = 3/4 (3 wrong keys 1 right key)
to choose wrong key in the 4th try P = 2/3 (2 wrong keys 1 right key)
to choose wrong key in the 5th try P = 1/2 (1 wrong key 1 right key)
to choose right key in the last try P = 1 (only 1 right key left)

multiplying all of these = 5/6*4/5*3/4*2/3*1/2*1 = 1/6
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by Shalabh's Quants » Sun Apr 15, 2012 11:14 am
suhir.kumar wrote: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
Prob. of selecting 1st wrong key out of 5 wrong keys = 5/6;
Prob. of selecting 2nd wrong key out of 4 wrong keys = 4/5;
Prob. of selecting 3rd wrong key out of 3 wrong keys = 3/4;
Prob. of selecting 4th wrong key out of 2 wrong keys = 2/3;
Prob. of selecting 5th wrong key out of 2 wrong keys = 1/2;

Prob.of wrong trials before opening the door = 5/6*4/5*3/4*2/3*1/2 = 1/6
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by shubham_k » Sun Apr 15, 2012 11:16 am
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.
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by GMATGuruNY » Mon Apr 16, 2012 3:49 am
suhir.kumar wrote: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
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by GMATGuruNY » Mon Apr 16, 2012 3:51 am
suhir.kumar wrote: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
Almost no math is needed here.

The probability that he chooses the correct key on ANY GIVEN ATTEMPT is equal to probability that he chooses the correct key on THE FIRST ATTEMPT: 1/6.

If the problem asked for the probability that he selects the correct key on his second attempt, the answer would be the same: 1/6.
If the problem asked for the probability that he selects the correct key on his third attempt, the answer would be the same: 1/6.
And so on.

The correct answer is C.

For more on this concept, check my posts here:

https://www.beatthegmat.com/probablity-ques-t60161.html
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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