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Thread: Solving for # of attempts using known indepentant chance and cumulative chance

  1. #1

    Solving for # of attempts using known indepentant chance and cumulative chance

    In a program I am working on I am using the formula z = 1 - ((1 - x) ^ y), where 'z' is the cumulative probability of an event occurring over 'y' attempts with a 'x' percent chance of the event occurring on each individual attempt.
    This works fine except I need the formula rearranged so it solves for 'y' attempt using a known 'z' and 'x'.

    Appreciate any help, thanks!

  2. #2
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    Quote Originally Posted by TunnelCat View Post
    In a program I am working on I am using the formula z = 1 - ((1 - x) ^ y), where 'z' is the cumulative probability of an event occurring over 'y' attempts with a 'x' percent chance of the event occurring on each individual attempt.
    This works fine except I need the formula rearranged so it solves for 'y' attempt using a known 'z' and 'x'.

    Appreciate any help, thanks!
    z = 1 - ((1 - x) ^ y)

    1 - z = (1 - x)^y

    Now take "ln" of both sides and solve for "y". If (x) > 1, you need to be careful!!
    Last edited by Subhotosh Khan; 01-27-2018 at 02:20 PM.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    I don't think x can be greater than 1 given that it is the probability of success on each trial.

  4. #4
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    Quote Originally Posted by j-astron View Post
    I don't think x can be greater than 1 given that it is the probability of success on each trial.
    The number 5% can be used as 5 or 0.05 - depending on the context of the problem (e.g. rule of 72 in interest rate calculation). I do not quite know how 'x' is expressed here.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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