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Math Help - Probability Distribution

  1. #1
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    Probability Distribution

    I have no idea if this is right....
    for a i get .021%
    for b i get 3
    for c i get 24.38%
    please help asap!
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  2. #2
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    Let me remind the binomial probability formula, an event m out of n times exactly with probability p is,
    \left( \begin{array}{c} n\\m \end{array} \right) p^m (1-p)^{n-m}.

    Over here
    \left\{ \begin{array}{c} n=20\\m=3\\p=.15 \end{array}
    Thus,
    \left( \begin{array}{c} 20\\3 \end{array} \right) (.15)^3(.85)^{17}\approx .2428
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  3. #3
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    For the second you need to find m such as,
    \left( \begin{array}{c} 20\\m \end{array} \right) (.15)^m(.85)^{20-m} is maximized.
    Basic trail and error, you get,
    m=3
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  4. #4
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    For the third to first.
    You need to find, (m=3,4,5,....,20) because "at least"
    \left( \begin{array}{c} 20\\3 \end{array} \right)(.15)^3(.85)^{17}+\left( \begin{array}{c} 20\\4 \end{array} \right)(.15)^4(.85)^{16}+... +\left( \begin{array}{c} 20\\20 \end{array} \right)(.15)^{20}(.85)^0
    You can add them up but that is the n00bish way of doing this problem. It is easier to find,
    \left( \begin{array}{c}20 \\0\end{array} \right)(.15)^0(.85)^{20}+\left( \begin{array}{c}20 \\1 \end{array} \right)(.15)^1(.85)^{19}+ \left( \begin{array}{c} 20\\2 \end{array} \right) (.15)^2 (.85)^{18}\approx .4049
    Now from one subtract that to get,
    \approx .5951
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