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Find P(X = 3) porbability

  1. Jul 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that 0.03% of plastic containers manufactured by a certain process have small
    holes that render them unfit for use. Let X represent the number of containers in a
    random sample of 10,000 that have this defect. Find P(X = 3).

    2. Relevant equations
    binomial?


    3. The attempt at a solution

    .97^9997
    *
    .03^3
    *
    10000 choose 3

    i bet this actually does equal the answer maybe, but my calculator of course spits out 0 since .97^9997 is really really tiny...and doesnt take into account that 10000 choose 3 is humungo

    help?
    the answer is .224, but i cant match it
     
  2. jcsd
  3. Jul 7, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: probability

    It's not that hard to work it out "by hand".

    [tex]\left(\begin{array}{c}10000 \\ 3\end{array}\right)= \frac{10000!}{3!9997!}[/tex]
    [tex]= \frac{10000(9999)(9998)}{6}=\frac{10000}{2}\frac{9999}{3}(9998)[/tex]
    [tex]= 166616670000[/tex]

    Now, .033=0.000027 and (this is the hard one) .979997 is about .6 x 10-133. Since 166616670000 is about 1.7 x 1011 that product will be on the order of 1011- 5- 133= 10-127.

    However, is the proportion of of "unfit" cups .03% or 3%? You say ".03%" but you use 3%. If the correct value is .03%, then you want
    [tex]\left(\begin{array}{c}10000 \\ 3\end{array}\right)(.0003)^3)(.9997)^{9997}[/tex]
    Now that would be about .224.
     
    Last edited: Jul 7, 2009
  4. Jul 8, 2009 #3
    Re: probability

    i was sitting in class today and wondered if i can use poisson distribution on this?

    o and yes that is my mistake that problem is copy and pasted, so yes it is actually 0.03% PERCENT...i didnt notice

    edit: ah yes i figured it out, [tex]\lambda[/tex] = .0003*10000 = 3

    P(X=3)

    [tex]\frac{e^{-3}*3^{3}}{3!}[/tex] = .224

    thanks alot though
     
    Last edited: Jul 8, 2009
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