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Math Help - Help with stats concepts. Help. Statistics overwhelm me.

  1. #1
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    Help with stats concepts. Help. Statistics overwhelm me.

    1. A corporation has four investment possibilities: A, B, and C with respective gains of $10, $20, and $50 million and investment D with a loss of $30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.


    2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car. b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.
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  2. #2
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    Quote Originally Posted by bombo31
    1. A corporation has four investment possibilities: A, B, and C with respective gains of $10, $20, and $50 million and investment D with a loss of $30 million. If one investment will be made and the probabilities of choosing A, B, C, or D are .1, .4, .2, and .3, respectively, find the expected gain for the corporation.
    My intution telle me you need to find,
    (.1)10+(.4)20+(.2)50-.3(30)=1+8+10-9=10\mbox{ million}
    Cuz since 10% probability means you are expected to get 10% of A which means 10% of 10 million, similarly the other 3. Then you need to add them up to see what they give, while remember to subtract D
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    Quote Originally Posted by bambi31
    2.. A marketing research survey shows that approximately 80% of the car owners surveyed indicated that their next car purchase will be either a compact or economy car. Assume the 80% figure is correct and five prospective buyers are interviewed. a. Find the probability that all five indicate that their next car purchase will be either a compact or economy car.
    The probability of the first person saying yes is .8
    The probability of two people is .8\times .8
    The probability of three .8\times .8\times.8
    The that all five is (.8)^5\approx .33

    Quote Originally Posted by bambi31
    b. Find the probability that at most one indicates that her next purchase will be either a compact or economy car.
    Find the probability that NONE will says yes. That is,
    (.2)^5 because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
    1-(.2)^5=.99968
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  4. #4
    MHF Contributor Quick's Avatar
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    Question 2


    Find the probability that NONE will says yes. That is,
    (.2)^5 because given that probability is .8 for saying yes then the probability of saying no is 1-.8=.2 since there are five people you rasie to the fifth az in the first question. But you need at most 1 which is the same as saying what is the probability of this (no one saying yes) will not happen which is,
    1-(.2)^5=.99968
    The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
      \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}  + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16%
    Last edited by Quick; June 3rd 2006 at 08:04 PM.
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  5. #5
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    Quote Originally Posted by Quick
    The question was not what are the chances that at least one will say yes but AT MOST one will say an yes. Heres what I get
      \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}  + \frac{4}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}=\frac{5}{3125} = 0.0016 = 0.16%
    You want:

    P((n_y=0) \lor (n_y=1))=P(n_y=0)+P(n_y=1)

    Now:

    P(n_y=0)=0.2^5\approx 0.00032,

    and:

    P(n_y=1)={5 \choose 1}\ 0.2^4\ 0.8=5\ 0.2^4\ 0.8\approx 0.0064

    Therefore:

    P((n_y=0) \lor (n_y=1))\approx 0.00672,

    or \approx 0.67 \%

    RonL
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  6. #6
    MHF Contributor Quick's Avatar
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    Why

    Well, it appears I forgot to multiply by five.
    Last edited by Quick; June 4th 2006 at 04:55 AM.
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    Quote Originally Posted by Quick
    Well, it appears I forgot to multiply by five.
    Because you only considered one case, namely,
    YNNNN

    But there are 4 more:
    NYNNN
    NNYNN
    NNNYN
    NNNNY

    This is why you need to multiply by five.
    ----
    Note, in may probability exacmple you need to use binomial coeificcients because of this reason.
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  8. #8
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    Thank you

    Sorry about the double posts. I did not realize, and thanks for your help.
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