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Math Help - Binomial Probability Problem

  1. #1
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    Binomial Probability Problem

    hey guys i got a final exam on wednesday and i am working on some practice questions
    jus need some help in doing this 1 dont really care about the answer
    i jus want the steps in solving this

    every now and again even a good diamond cutter has a problem and the diamond breaks for one particular cutter the rate of breakage is 0.1
    i. give reasons why the binomial probability model is suited for this problem
    if the cutter works on 75 stones
    a. what is the probability that he breaks exactly 3 stones
    b. what is the probability that he breaks 2 or more stones
    c. find the mean of this distribution
    d. find the stand dev of this distribution
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  2. #2
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    Quote Originally Posted by math321 View Post
    hey guys i got a final exam on wednesday and i am working on some practice questions
    jus need some help in doing this 1 dont really care about the answer
    i jus want the steps in solving this

    every now and again even a good diamond cutter has a problem and the diamond breaks for one particular cutter the rate of breakage is 0.1
    i. give reasons why the binomial probability model is suited for this problem
    if the cutter works on 75 stones
    a. what is the probability that he breaks exactly 3 stones
    b. what is the probability that he breaks 2 or more stones
    c. find the mean of this distribution
    d. find the stand dev of this distribution
    The answers to c and d can be found here
    Binomial distribution - Wikipedia, the free encyclopedia

    The reason is that these are Bernoulli trials there are exactly two possible outcomes the diamond breaks or it doesn't.

    P(X=x)=\binom{75}{x}(0.1)^x(0.9)^{75-x}

    Hint for b use the compliment of the event.
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  3. #3
    Senior Member Sambit's Avatar
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    i. Binomial probability model is applicable whenever the outcome of a random experiment (here cutting of a diamond) can be classified into 2 categories: success (here the breaking of diamond) and failure (not breaking). So it is applicable here.

    P.S.: Don't get confused with the practical meaning of "success" and "failure". You are to find out probability of breaking, so we take "probability of breaking" as success.

    a. If X is the number of diamonds that break amongst the 75 diamonds, then X\sim Bin(n=75,p=0.1).

    Then P(X=x) = {{n}\choose{x}}p^x(1-p)^{n-x}

    So the probability that he breaks exactly 3 stones = P(X=3)

    b. the probability that he breaks 2 or more stones = P(X\geq 2) = 1-P(X<2) = P(X=0) + P(X=1)

    c. mean = np

    d. standard deviation = \sqrt{np(1-p)}
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    so when is not applicable
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  5. #5
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    Quote Originally Posted by math321 View Post
    so when is not applicable
    Quote Originally Posted by math321 View Post
    so when is not applicable
    Essentially, the trials must be independent.
    If the outcomes are not independent it is not applicable.
    For example, if we deal five cards the outcomes are not independent.
    On the other hand, if we draw one card at a time returning the card to the deck each time then the outcomes are independent.
    Last edited by Plato; May 9th 2011 at 03:17 PM.
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    cud u check to see if p=.001 or .1
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  7. #7
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    Quote Originally Posted by math321 View Post
    cud u check to see if p=.001 or .1
    Ask whoever gave you these practice questions for the solutions or, at least, the answers. (It is not our job to do the work of your teacher).
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  8. #8
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    Quote Originally Posted by math321 View Post
    cud u check to see if p=.001 or .1
    The way you have written the question it says 0.1.
    However, that is 10%. That is high failure rate.
    All you can do is read the textbook or your notes.
    Perhaps there is something about the wording that implies 0.001.
    That is 0.1% which is more like it.
    BUT as written it is 0.1.
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  9. #9
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    yep the question says .1% so i made a typo
    working on it right now thanks though
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  10. #10
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    got this when i solved
    P (X = 3) = \binom{75}{3} . 0.001^3 ((1-0.001)^(75-3))
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  11. #11
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    for my final answer i got .000062832
    cud u verify that for me
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  12. #12
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    Quote Originally Posted by math321 View Post
    for my final answer i got .000062832
    cud u verify that for me
    Yes P(X=3)=.000062832 if p=.001
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