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"A computer system uses passwords [. . .]": Probabilities=?

  1. Jan 23, 2015 #1

    s3a

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    1. The problem statement, all variables and given/known data
    PROBLEM(S):
    A computer system uses passwords that contain exactly eight characters, and each character is one of 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Let Ω denote the set of all possible passwords, and let A and B denote the events that consist of passwords with only letters or only integers, respectively. Suppose that all passwords in Ω are equally likely.

    Determine the probability of each of the following:
    (a) A
    (b) B
    (c) A password contains at least 1 integer.
    (d) A password contains exactly 2 integers.

    ANSWERS:
    (a) 0.2448
    (b) 4.58E−7
    (c) 0.7551
    (d) 0.254

    2. Relevant equations
    Multiplication rule.

    3. The attempt at a solution
    Could someone please show me how to do (a), so that I can work on the rest on my own? For (a), I tried P(A) = 1/(52^8), but that's wrong.
     
  2. jcsd
  3. Jan 23, 2015 #2

    Mark44

    Staff: Mentor

    For each character chosen, you have 52 possible characters out of a total of 62.
     
  4. Jan 23, 2015 #3

    s3a

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    Oh, I see!

    So, (a) is (52/62)^8!

    Thanks!
     
  5. Jan 23, 2015 #4

    s3a

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    I also get (b) to be roughly 4.58E-7, but could you please help me with parts (c) and (d) now?
     
  6. Jan 23, 2015 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Assuming that "!" is not a factorial, yes! (And, of course, that fraction can be reduced- 52/62= 26/31.)
     
  7. Jan 23, 2015 #6

    s3a

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    For (c), i get [1/10 * 10/62]^8, but that's wrong.
     
  8. Jan 23, 2015 #7

    s3a

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    The "!" was not for a factorial; it was just because of my excitement. :)

    Could you help me with part (c)?
     
  9. Jan 23, 2015 #8

    Dick

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    The set of all passwords containing at least one integer is equal to the set of ALL passwords minus the set of passwords consisting of only letters. So?
     
  10. Jan 23, 2015 #9

    s3a

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    I get 62^8 - 52^8 ~= 1.65E14 passwords containing at least one integer.
     
  11. Jan 23, 2015 #10

    Dick

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    Homework Helper

    Now translate that into a probability.
     
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