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Math Help - I dont know how to find these answers for these problems...probability

  1. #1
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    I dont know how to find these answers for these problems...probability

    2. A process follows the binomial distribution with n = 7 and p = .4. Find
    a. P(x = 3)
    b. P(x > 5)
    c. P(equal than or less than)2 (I dont have that keyboard symbol sorry)
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  2. #2
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    Quote Originally Posted by jwells1999 View Post
    2. A process follows the binomial distribution with n = 7 and p = .4. Find
    a. P(x = 3)
    b. P(x > 5)
    c. P(equal than or less than)2 (I dont have that keyboard symbol sorry)
    If you are working on these types of questions, you must have been taught that for a binomial distribution with n trials and probability of success p in a single trial:

    Pr(X = x) = {^nC_x} \, p^x (1 - p)^{n-x}

    a. Apply the above formula.

    b. Pr(X = 6) + Pr(X = 7). Apply the above formula on each.

    c. Pr(X = 0) + Pr(X = 1) + Pr(X = 2). Apply the above formula on each.

    And please do not double post.
    Last edited by mr fantastic; February 7th 2008 at 12:43 PM.
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    First of all I just wanted to say sorry for the double post, I didnt know where to post my question.

    As with the problem...

    I was taught a different formula and I think thats what is getting me confused.
    I was taught f(x)=n!/x!(n-x)!*p^x(1-p)^n-x

    For a and c I got the following result
    a)F(x)=5040/144*.064(.6)^4
    F(x)=35*.064*.1296
    F(x)=35*.0082944
    F(x)=.290304 or 29.03%


    c)F(x)=(7!) / (2!(7-2)! * (.4^2)(.6)^5
    F(x)=[5040/240]*.16(.07776)
    F(x)=21*.0124416
    F(x)=.2612736 or 26.13%


    but I cant figure out b. Should I be solving whereas x=6 since the problem is x>5??

    If so I would get

    F(x)=(7!) / (6!)(7-6)! *.4^6(1-.4)^7-6
    F(x)=(7)*.004096(.6)
    F(x)=7*.0024576 or .25%

    but Im not sure if thats the right way to do it.

    Any help would be much appreciated
    Thanks
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  4. #4
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    Quote Originally Posted by jwells1999 View Post
    First of all I just wanted to say sorry for the double post, I didnt know where to post my question.

    As with the problem...

    I was taught a different formula and I think thats what is getting me confused.
    I was taught f(x)=n!/x!(n-x)!*p^x(1-p)^n-x

    Mr F says: \frac{n!}{x! (n - x)!} is the same as {^nC_x}. In fact, it's the basic formula for {^nC_x}

    For a and c I got the following result
    a)F(x)=5040/144*.064(.6)^4
    F(x)=35*.064*.1296
    F(x)=35*.0082944
    F(x)=.290304 or 29.03%


    c)F(x)=(7!) / (2!(7-2)! * (.4^2)(.6)^5
    F(x)=[5040/240]*.16(.07776)
    F(x)=21*.0124416
    F(x)=.2612736 or 26.13%


    but I cant figure out b. Should I be solving whereas x=6 since the problem is x>5?? Mr F says: read what I posted for this question.

    If so I would get

    F(x)=(7!) / (6!)(7-6)! *.4^6(1-.4)^7-6
    F(x)=(7)*.004096(.6)
    F(x)=7*.0024576 or .25%

    but Im not sure if thats the right way to do it.

    Any help would be much appreciated
    Thanks
    ..
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