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.
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