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Math Help - binomial probability

  1. #1
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    binomial probability

    HI ,
    I would like to have some help with this problems:

    A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.

    1-Find the probability that no part is scrapped in the next 10 experiments.
    My solution:
    b(0;10,0.1)

    b(x;n;p) is the binomial distribution.
    where x =number os success
    n=number of trial
    p=probability of success

    2- find the probability that at most two parts will be scrapped in the next operations.
    My solution: B(2;10,0.10)
    B(x;n,p) is the cumulative probabilities

    3- Find the expected number of scrapped parts in the next 25 operations.
    NO solution

    4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
    No solutions

    5- Find the probability that the third part scrapped is the 20th part cast.
    No solution

    To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.

    Thank you
    B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    HI ,
    I would like to have some help with this problems:

    A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.

    1-Find the probability that no part is scrapped in the next 10 experiments.
    My solution:
    b(0;10,0.1)
    b(x;n;p) is the binomial distribution.
    where x =number os success
    n=number of trial
    p=probability of success
    b(0;10,0.1) = 10!/(0! (10-0)!) (0.1)^0 (0.9)^10 = 0.9^10 ~= 0.349

    2- find the probability that at most two parts will be scrapped in the next operations.
    My solution: B(2;10,0.10)
    B(x;n,p) is the cumulative probabilities
    B(2;10,0.1) = b(0;10,0.1) + b(1;10,0.1) + b(2;10,0.1)

    ................= 0.9^10 + 10 0.1 0.9^9 + 45 0.1^2 0.9^8 ~= 0.930

    3- Find the expected number of scrapped parts in the next 25 operations.
    NO solution

    4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
    No solutions

    5- Find the probability that the third part scrapped is the 20th part cast.
    No solution

    To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.

    Thank you
    B
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    HI ,
    I would like to have some help with this problems:

    A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.

    1-Find the probability that no part is scrapped in the next 10 experiments.
    My solution:
    b(0;10,0.1)
    b(x;n;p) is the binomial distribution.
    where x =number os success
    n=number of trial
    p=probability of success

    2- find the probability that at most two parts will be scrapped in the next operations.
    My solution: B(2;10,0.10)
    B(x;n,p) is the cumulative probabilities

    3- Find the expected number of scrapped parts in the next 25 operations.
    NO solution
    The expected number of scrapped parts in 25 trials is:

    E(n) = Sum n*pr(n) n=0, .., 25 = Sum n*b(n;25,0.1) n=0,..,25

    Now we can do this sum, or we can use the fact that we know the expected
    number of failures in N Bernoulli trials with a probability of failur on a single
    trial of p is N*p, so here it is 25*0.1 = 2.5.

    4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
    No solutions
    The expected cost of scrapped parts in 25 trials is:

    E(n) = Sum (n^2)*pr(n) n=0, .., 25 = Sum (n^2)*b(n;25,0.1) n=0,..,25

    but I don't know a short cut to calculate this.

    5- Find the probability that the third part scrapped is the 20th part cast.
    No solution

    To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.

    Thank you
    B
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    HI ,
    I would like to have some help with this problems:

    A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.


    5- Find the probability that the third part scrapped is the 20th part cast.
    No solution


    This probabability is equal to the probability of exactly two scrapped in the
    first 19 times the probability that the 20th is scrapped:

    P = b(2;19,0.1)*0.1 = [19*18/2 * 0.1^2*0.9^17 ]* 0.1 ~= 0.0285

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    The expected cost of scrapped parts in 25 trials is:

    E(n) = Sum (n^2)*pr(n) n=0, .., 25 = Sum (n^2)*b(n;25,0.1) n=0,..,25

    but I don't know a short cut to calculate this.
    On second thoughts I do know a short cut for this:

    Var(n) = E((n - m)^2) = E(n^2) - m^2 =

    but the variance of the number of success of a Binomial random variable in N trials is

    Var(n)=N*p*(1-p)

    and the mean m=Np, so:

    N*p*(1-p) = E(n^2) - N^2p^2,

    or:

    E(n^2) = N*p*(1-p) + N^2*p^2 = 8.5

    RonL
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