SOLUTION: Your boss gave you key ring with 6 different keys. However, you forgot which key is the right key to open the door. You keep track of each key that doesn't work so that you don't t

Question 1078156: Your boss gave you key ring with 6 different keys. However, you forgot which key is the right key to open the door. You keep track of each key that doesn't work so that you don't try using that key again. What is the probability that you will be able to open the door with the third key?
Answer by ikleyn(52905) (Show Source): You can put this solution on YOUR website!
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Let {1, 2, 3, . . . , 6} be the initial (starting) permutation of the keys numbered as shown, 
with the "right" key in the first position marked as "1".


Then the full space of events is the set of all permutations of 6 objects, which counts 6! = 1*2*3*4*5*6 = 720 events (permutations).


The "lucky"/fortunate/favorable events are those where "1" is in the third position.
The other keys can be in ANY other positions.


With this model, you have 5! = (6-1)! = 120 permutations where "1" is in the third position.


Hence, the probability under the question is  = .

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