Results 1 to 8 of 8

Math Help - binomcdf, binompdf

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    46

    binomcdf, binompdf

    Ok, these problems should be easy, as I can use the calculator function to solve, but deriving the numbers has proved to be difficult.

    Find probability that at least 2 computers are defective:

    23 computers, probability of defective is .093

    What is the probability of a multiple choice test, 3 possible answers, 10 questions that you will get a 80%

    Couldn't figure out how to type them in righ.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by skyslimit View Post
    Ok, these problems should be easy, as I can use the calculator function to solve, but deriving the numbers has proved to be difficult.

    Find probability that at least 2 computers are defective:

    23 computers, probability of defective is .093

    What is the probability of a multiple choice test, 3 possible answers, 10 questions that you will get a 80%

    Couldn't figure out how to type them in righ.
    Let X be the random variable number of defective computers.
    X ~ Binomial(n = 23, p = 0.093)
    Calculate \Pr(X \geq 2) = 1 - \Pr(X \leq 1).


    Let Y be the random variable number of correctly answered questions.
    X ~ Binomial(n = 10, p = 1/3)
    Calculate \Pr(X = 8).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    The first one should be hypergeometric since you can't pick the same computer twice.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by matheagle View Post
    The first one should be hypergeometric since you can't pick the same computer twice.
    The wording of the first one is ambiguous and misleading I think.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    Quote Originally Posted by matheagle View Post
    The first one should be hypergeometric since you can't pick the same computer twice.
    I disagree. It should be binomial. If it's hypergeometric, what's the parameter value for the number of successes in the population?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    The bottom line is, you need to know if you're sampling with or without replacement. Usually when you sample items looking to see if they are defective you do not replace them. Like buying milk or eggs. If you walk out with 4 containers of eggs, they were not replaced. The other problem here is 23 times .093 is not an integer. I had expected it to be an integer, meaning that were were 2 or 3 defectives computers here.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    Even if 23 times 0.093 is an integer, I still don't see why, in your first post, you jump to the conclusion that it's hypergeometric.

    First, the question is clearly about the whole population, ie, it's a census, not a sampling problem.

    Second, with hypergeometric problems, you are given the number of defective parts in the population, not the probability that each sample item is defective.

    Third, it's not just about sampling with or without replacement, it's about whether the probability of picking up defective parts is constant from sample to sample. If the probability is constant, it's a binomial problem, even if you're sampling without replacement. If it's not, because the number of defective parts are fixed in the population, then it's hypergeometric.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I did not realize we were looking at all of the computers. I though we were looking at a subset, like there were 23 total machines and 21 good and 2 bad and how can we pick 2 bad and some good.... But it seems we weren't.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum