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Math Help - Discrete Probability Distribution

  1. #1
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    Question Discrete Probability Distribution

    from walpole 5.32

    From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
    a) all 4 will fire
    b) at most 2 will not fire ?

    I believe this is a hypergeometric distribution.
    pls correct me if i am wrong.

    a) since all 4 , i took P(x= 0) + .....P(x=4)
    4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
    and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?

    b) at most 2 will not fire means P (X<=2) ?
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  2. #2
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    Quote Originally Posted by hazel View Post
    from walpole 5.32

    From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
    a) all 4 will fire
    b) at most 2 will not fire ?

    I believe this is a hypergeometric distribution. Mr F says: Correct.
    pls correct me if i am wrong.

    a) since all 4 , i took P(x= 0) + .....P(x=4)
    4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
    and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?

    b) at most 2 will not fire means P (X<=2) ?
    You must define the random variable before doing anything else.

    Let X be the random variable 'number of defective missiles in sample'.
    X ~ Hypergeometric(N = 10, n = 4, D = 3).

    a) Calculate Pr(X = 0).

    b) Calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 - Pr(X = 3) - Pr(X = 4).
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  3. #3
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    Hello, Hazel!

    From a lot of 10 missiles, 4 are selected at random and fired.
    If the lot contains 3 defective missiles that will not fire,
    what is the probability that:

    a) all 4 will fire?

    Let: . \begin{array}{ccc} D &=& \text{de{f}ective} \\ G &=& \text{good} \end{array}

    There are: . {10\choose4} \:=\:210 possible samples.

    There are: . {7\choose4} \:=\:35 samples with 4 G's.

    Therefore: . P(\text{4 G}) \:=\:\frac{35}{210} \:=\:\frac{1}{6}



    b) at most 2 will not fire ?

    The opposite of "at most 2 D's" is "3 D's (and 1 G)".

    The number of samples with 3 D's and 1 G is: . {3\choose3}{7\choose1} \:=\:7

    Hence: . P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}

    Therefore: . P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Hazel!


    The opposite of "at most 2 D's" is "3 D's (and 1 G)".

    The number of samples with 3 D's and 1 G is: . {3\choose3}{7\choose1} \:=\:7

    Hence: . P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}

    Therefore: . P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}

    But for at most 2 will fire :
    means 1G, 3D and also 2G 2D right ?
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