Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Attempting to find an equation, not sure how to present it.

  1. Sep 29, 2008 #1
    Hey, all! I haven't been here for a long time, but I've got a problem I'm working on, and I'm not quite sure how to present my equation.

    My problem is, for any given random event P with 1/D chance of occurring, what is the probability L, given T attempts, of event P occurring at least once?

    The probability of P occurring when T is one is obvious, just 1/D, but past that, it seems to get rather complicated.

    For D = 2, 3, and 4, the chances are charted below:

    Code (Text):

    D=2
    T | L
    1 | 1/2
    2 | 3/4
    3 | 7/8
    4 | 15/16
    5 | 31/32

    D = 3
    T | L
    1 | 1/3
    2 | 5/9
    3 | 19/27
    4 | 65/81
    5 | 211/243

    D = 4
    T | L
    1 | 1/4
    2 | 7/16
    3 | 37/64
    4 | 175/256
    5 | 781/1024
     
    Now, I've figured an equation which works for solving for any L in terms of T, but the problem is that it requires the numerator of the previous L, and I don't want to have to get to T = 5 by solving 4, 3, 2, and 1 each time. I'd like an equation that solves for L given only T and D. What's it called if my equation requires a previous result to determine the next result?

    I think I put the equation down correctly, please tell me it makes sense.


    [tex]L(T) = \frac{N_T}{D_T} = \frac{(D-1)*(N_{T-1})+D^{T-1}}{D^{T}}[/tex]
    Did I explain everything all right?
     
    Last edited: Sep 29, 2008
  2. jcsd
  3. Sep 29, 2008 #2

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Probability P occurs (1 try): 1/D
    Probability P does not occur (1 try): 1 - 1/D
    Probability P does not occur (T tries): (1 - 1/D)^T
    Probability P occurs at least once (T tries): 1 - (1 - 1/D)^T
     
  4. Oct 1, 2008 #3
    Thank you, but now I have another problem.

    Let's say we had five marbles, 2 green and 3 red.

    What are the chances, if you draw two marbles without replacing any, of getting at least one green?

    That equation wouldn't work here, because N/D changes/
     
  5. Oct 1, 2008 #4

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Try working out some small examples. The formula should make itself obvious.

    You'll probably use the factorial in the formula.
     
  6. Oct 1, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Use the same logic as CRGreathouse showed you for the first problem: the probability of getting "at least one green" is 1- the probability of getting no green- i.e. the probability of getting two red.

    Initially there are 2 green and 3 red: 5 altogether so the probability of getting RED is __

    Assuming red is drawn (which you must in order to get "2 red") there are 2 green and 2 red left: 4 altogether, so the probability of getting RED is ____.

    The probability of getting both red is the product of those two.

    The probability of getting "at least one green" is 1 minus that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Attempting to find an equation, not sure how to present it.
Loading...