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Counting Probability

  1. Apr 14, 2012 #1
    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?
     
  2. jcsd
  3. Apr 14, 2012 #2
    Can you open the door or not? I think that the first question can be solved by the binomial distribution.
    The basic probability is 2/6 i.e. the probability of not being able to open the door. I get the result 26% for not being able, so 64% would be my answer.

    The 2nd question is very basic i'd say. First you have to multiply the 64% with the probability of having the first key correct and then the 2nd. i get the result 5%
     
  4. Apr 14, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Why do you say that 2/6 = 26%?

    For the second question, let C = {can open the door} and F = {first two keys work}. We have P{F} = P{F|C}P{C}, with P{C} = 2/3. Given C we have 5 keys among which are the two good ones. To get P{F|C}, note that the first key drawn must be one of the two good ones, then the next key must be the remaining good one. Binomial coefficients C(n,m) do not apply in this problem because the order in which the events occur (not just their number) is important in this question.

    RGV
     
  5. Apr 14, 2012 #4
    I think the probability is 1/60.
     
  6. Apr 15, 2012 #5
    I used the binomial distribution i didn't say that 2/6 = 26%.
     
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