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Help on probability again please

  1. May 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Trials of a biology experiment are performed, and each trial is independent of the other
    trials. Each experiment trial has a 0.13 probability of a success, and 0.87 probability of
    failure.
    (c) Let Y denote the random variable representing the number of trials up to and including
    the first success. Find an expression for the probability distribution of Y ,
    then use MATLAB to compute and plot this function for the first 50 possible values.
    (d) Let Y be defined as in (c). Find an expression for and use MATLAB to compute
    Pr(Y < 9).
    (e) Let Y be defined again as in (c). Find a mathematical expression for and compute
    (using MATLAB) the probability that Y is less than or equal to 12 given that Y is
    greater than 8.


    2. Relevant equations
    n/a


    3. The attempt at a solution

    is part(c) just
    Pr(Y=y) = 0.87^(y-1)*0.13

    part(d)
    Pr(Y<9) = [tex]\sum(0.87^(y-1)*0.13[/tex]

    and got no idea for part (e), need help please.
     
  2. jcsd
  3. May 25, 2009 #2
    It's a geometric distribution, so yes, your answer to (c) is correct.

    For part (d) you have to sum from y=1 to y=8.

    [tex] P(Y<13|Y>8) = \frac {P(Y<13 \cap Y>8)}{P(Y>8)} = \frac {P(8>Y>13)}{P(Y>8)} [/tex]

    [tex]=\frac { \sum^{12}_{y=9} 0.87^{y-1}(0.13)} { \sum^{\infty}_{y=9} 0.87^{y-1}(0.13)} [/tex]
     
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