(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

There are two locks on the door and the keys are among the six different ones you carry. In a hurry, you dropped one somewhere. What is the probability that you can still open the door? What is the probability that the first two keys you try will open the door?

3. The attempt at a solution

So for the first part, I supposed that the six keys have numbers on them, 1 through six. Thus, my complete set of keys in {1 2 3 4 5 6}. If I lose one, I can construct the following subsets:

{1 2 3 4 5}

{1 2 3 4 6}

{1 2 3 5 6}

{1 2 4 5 6}

{1 3 4 5 6}

{2 3 4 5 6}

This is all very arbitrary so let us suppose that keys 1 and 2 are required to open the door. Thus, we can see from inspection that I can still open the door 4 times out of the six total possibilities. Thus, the probability is 4/6 = 2/3.

For the second part, I said that the remaining five keys can permute among themselves in 5! ways. We multiply this value by six since we have 6 different possibilities for the key that we lose. We fix the two required keys, and the remaining 3 keys in our subset can permute in 3! ways. Thus, the probability is [itex]\frac{(2)(3!)}{(6)(5!)}[/itex]. But this seems very low to me. Did I do something wrong?

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# Homework Help: Counting Probability

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