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Math Help - probability/expected value exercise

  1. #1
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    Question probability/expected value exercise

    how many
    Last edited by shirel; August 28th 2007 at 12:03 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shirel View Post
    how many times does one have to roll a die in order to have face 5 showing up with a 90 % certainty?
    if face 5 did not show up among the estimated number of tries, would that legitimate you to suspect the die be unfair?
    consider the same questions with an underlying 99 % certainty!

    calculations are necessary, but important is the discussion of the questions..
    please help me with all your knowledge and mathematical way of thinking

    thanks!
    In N trials the number of times, n that face 5 shows on a fair dice has a binomial distribution B(N, 1/6).

    So your question is asking for the smallest N such that:

     <br />
p(n=0) \le 0.1<br />

    so:

     <br />
p(n=0) = b(0;N,1/6)=\left(\frac{1}{6}\right)^N \le 0.1<br />

    so now find N.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by shirel View Post
    how many times does one have to roll a die in order to have face 5 showing up with a 90 % certainty?
    if face 5 did not show up among the estimated number of tries, would that legitimate you to suspect the die be unfair?
    consider the same questions with an underlying 99 % certainty!
    Whether you suspect that a coin is unfair if it fails a test that a fair coin
    passes 90% of the time depends on you, do you have some other reasom to
    suspect the coin? Otherwise you may suspect it and possibly do somemore tests.

    If it fails a 99% test you may have more reason to suspect the coin.

    RonL
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  4. #4
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    Hello, shirel!

    Another (slightly different) explanation . . .


    How many times does one have to roll a die in order
    to have face 5 showing up with a 90% certainty?

    We know that for one roll: . P(5) = \frac{1}{6},\;\;P(\text{not 5}) = \frac{5}{6}

    Consider n consecutive rolls without a 5.
    . . This probability is: . \left(\frac{5}{6}\right)^n

    And we want this probability to be less than 10%.

    . . Hence, we have: . \left(\frac{5}{6}\right)^n \:<\:0.1

    Take logs: . \log\!\left(\frac{5}{6}\right)^n \:<\:\log(0.1) \quad\Rightarrow\quad n\cdot\log\!\left(\frac{5}{6}\right) \;<\;\log(0.1)

    . . Divide by \log\!\left(\frac{5}{6}\right) which is negative: . n \;> \;\frac{\log(0.1)}{\log\!(\frac{5}{6})} \;\approx\;12.629


    Therefore, one must roll a die at least 13 times.

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