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Math Help - Why isn't this example a negative binomial

  1. #1
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    Why isn't this example a negative binomial

    A metal detector misses 1% of the time
    What is the probability that the first person missed is among the fist 50 persons scanned ?

    I saw this as we want 1 success in 50 trials so p=1 we apply p (1-p)^ {49}

    After some thought I think I spotted my mistake. The negative binomial is x trials until a success so applying it to this situation means that the 50th person is missed.

    The solution is P(X < or equal to 50) = 1 - 0.99^{50}
    Can some one help me clearly distinguish between then negative binomial and this situation but explaining the logic behind the solution ?
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  2. #2
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    "1 success in 50 trials" is no good. There can be 50 failures in 50 trials and the first still will be int eh first 50.

    Your second try is also no good, not because you don't have it right, but because you don't understand it.

    Who cares what the distribution is. Think it through.

    Pr(1st failure = 1) = 0.01
    Pr(1st failure = 2) = 0.01 * 0.99
    Pr(1st failure = 3) = 0.01 * 0.99^2
    Pr(1st failure = 4) = 0.01 * 0.99^3
    Pr(1st failure = 5) = 0.01 * 0.99^4
    Pr(1st failure = 6) = 0.01 * 0.99^5
    ...
    Pr(1st failure = 50) = 0.01 * 0.99^49

    Can you add those up?

    0.01 * (1 + 0.99 + 0.99^2 + ... + 0.99^49)

    I hope so.

    If, after solving the problem, you recognize the distribution - more power to you. If you really want to play with properties, find the distribution and play with them. If you really want to solve the problem, think it through - don't go shopping for distribution names.
    Last edited by TKHunny; July 20th 2010 at 05:31 AM.
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  3. #3
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    Hello, Hitman6267!

    A metal detector misses 1% of the time.
    What is the probability that the first person missed is among the fist 50 persons scanned?

    We have: . \begin{array}{ccc}P(\text{Miss}) &=& 0.01 \\ P(\text{Hit}) &=& 0.99 \end{array}

    The opposite of "there is at least one Miss among the first 50 scans"
    . . is "the first 50 scans are all Hits."

    Hence: . P(\text{first 50 are Hits}) \:=\:(0.99)^{50}

    Therefore: . P(\text{a Miss among first 50}) \;=\;1-(0.99)^{50}

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  4. #4
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    understood, thank you.
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