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Binomial Probability Distribution

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    The rejection rate of a certain journal is 45%. If the journal accepts articles at random, what is the minimum number of articles someone has to submit to have a probability of more than 0.75 of getting at least one article accepted?


    2. Relevant equations
    I'm almost sure this is a binomial distribution question where you take p and n to kook up the P(X) in the binomial probabilities table. Only thing is, I don't know what is n.


    3. The attempt at a solution
    p=1-0.45=0.55

    P(1) = 1-P(X<=0)
    >0.75 = 1-P(X<=0)
    P(X<=0) < 0.25

    But then what? Is my potential n the minimum nr of articles or 1?

    {{Also, this is my first post, would someone please tell me where to get the scientific notation for the formulas to put in the posts? Pls and tx! }}}
     
  2. jcsd
  3. Feb 17, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi luv2learn! Welcome to PF! :smile:

    (have an leq: ≤ :wink:)

    No, it's not binomial …

    you're right (if I'm reading you properly: your notation is a bit weird :confused:) that the question is the same as what is 1 - Qn,

    where Qn is the probability that all n articles are rejected.

    ok, rejections are independent, so what is Qn ? :smile:
     
  4. Feb 17, 2010 #3
    Ok, so apparently I've got this whole question wrong, LOL

    So the probability that n artiles are rejected is Qn = 0.45 x n
     
  5. Feb 17, 2010 #4

    tiny-tim

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    erm :redface: … with n = 3, that's greater than 1 ! :biggrin:

    Try again! :smile:
     
  6. Feb 17, 2010 #5
    (I'm really losing it, been at it for 10hrs.)

    Qn=0.45n ;
    Rejecting 1 is: Q1=0.451; Which implies accepting n-1, which is = 1-0.451 = 0.55
    Q2=0.452; accept n-2 = 1-0.452 = 0.798; etc.

    So if x = minimum nr of articles to be submitted, then I'm actually trying to find
    Accept n-x = 1-0.45x > 0.75 ???
     
  7. Feb 17, 2010 #6

    tiny-tim

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    Now you're confusing me :confused:

    you're looking for n such that 0.45n < 0.25 :wink:

    (either use logs or just trial-and-error! :biggrin:)
     
  8. Feb 17, 2010 #7
    Yeah, tx. I got the same thing but in a very long (and confusing) way.
    In the end n > 1.74 i.e. n = 2

    Tx a lot. But is there a simple way of seeing if its a binomial distribution or not? I thought I know but clearly I don't. Or can the same answer be reached if I use binomial distribution probability rules?
     
  9. Feb 17, 2010 #8

    tiny-tim

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    (How did you get 1.74? :confused:)

    You're misunderstanding which bit of the binomial is which.

    For (p + q)n, the figure for k successes is pkqn-k nCk

    in this case, technically, you did use the binomial theorem, but with k = n and therefore nCk = 1. :wink:
     
  10. Feb 17, 2010 #9
    0.45n>0.25
    log (0.45n>log (0.25)
    nlog(0.45)>log(0.25)
    n=log0.25/log0.45
    n=1.736
     
  11. Feb 17, 2010 #10

    tiny-tim

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    oh yes, that's fine. :smile:
     
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