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Homework Help: Binomial Probability Distribution

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    The College Board reports that 2% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities. Consider a random sample of 25 students who have recently taken the test.

    a.) What is the probability that exactly 1 received a special accommodation?
    b.) What is the probability that at least 1 received a special accommodation?


    2. Relevant equations

    [tex]\begin{pmatrix}
    n\\
    x\\
    \end{pmatrix}p^x(1-p)^{(n-x)}[/tex]

    3. The attempt at a solution

    a.) [tex]\begin{pmatrix}
    25\\
    1\\
    \end{pmatrix}.02^x(1-.02)^{(25-1)} = .3079[/tex]

    30% seems kind of high. If 40,000 students out of 2,000,000 were {success} and chose 25 of the 2,000,000, there's a 30% chance I'd get 1 of the 40,000? That, just seems too high for some reason.

    b.) [tex]
    \sum_{x=1}^{25}
    \begin{pmatrix}
    25\\
    x\\
    \end{pmatrix}.02^x(.98)^{(25-x)}[/tex]

    Does this look right? How can I calculate this? I don't believe I have to simplify it, per instructor's orders...but I'm not sure if I have it right to begin with.
     
  2. jcsd
  3. Feb 21, 2010 #2
    As far as A is concerned, it looks OK to me... wait for other opinions :\

    as far as B is concerned, you can find the probability that zero out of the 25 students get the special accommodation and calculate
    1-(zero out of 25)= at least one

    they are the complements of each other.

    (none)+(at least 1)= 1
     
  4. Feb 21, 2010 #3
    That is exactly what I thought. however, how can you calculate that 0 of 25 are {success}? It would come out as this:

    [tex]1-[{
    \begin{pmatrix}
    25\\
    0\\
    \end{pmatrix}.02^0(1-.02)^{(25-0)}}]
    [/tex]

    Do I ignore the 25C0 at the beginning? If I do, I get:

    [tex]1-[{(1-.02)^{(25)}}] = 1 - .603 = .397
    [/tex]
     
  5. Feb 21, 2010 #4
    25 choose 0 is 1. Intuitively there is only one way to choose 0 things (namely by choosing nothing). But 0! is defined to be 1. Also your part a) is correct. It's only a sample after all.
     
  6. Feb 21, 2010 #5
    c(25,0)=1
     
  7. Feb 21, 2010 #6
    Ooh. Right. 0!=1. Somehow, that slipped my mind.

    Thanks a lot.
     
  8. Feb 21, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    25C0 = 1. There's only one way to select nothing from 25.

    Your answer to part a is correct. It may seem high, but it's not. The effect of what seems like small probabilities can build up very quickly.
     
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