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tomo

  • 2 years ago

A computer has a list of n possible passwords. If only one is accepted, what is the probability of guessing the password right on the first try?

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  1. blurbendy
    • 2 years ago
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    1 / n

  2. tomo
    • 2 years ago
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    thank you. that makes sense. 1/total. However, I am now trying to find the P(correct on 3rd try). Is that P(wrong) * P(wrong) * P(right). if so is that (n-1)/n * (n-1)/(n-1) * 1/(n-2)?

  3. tomo
    • 2 years ago
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    on the second term i mean n-2/n-1

  4. KingGeorge
    • 2 years ago
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    Assuming we can't repeat passwords, that looks correct.

  5. tomo
    • 2 years ago
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    ok, so extrapolating to (if i <n, on the ith try). I think that it should be something like ((n-1)^(i-1) * 1)/ (n*(n-1)...(n-i)). I can't remember what the formula for the sum of products is though.

  6. tomo
    • 2 years ago
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    actually on the last term of the denominator i think it would be n-(i-1)

  7. KingGeorge
    • 2 years ago
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    I've got to go now, but you certainly could extrapolate like that. To write it exactly, you could do a lot of canceling, or you could write it in product notation using this symbol. \[\prod\]

  8. tomo
    • 2 years ago
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    thank you very much KingGeorge.

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