Help on probability again please

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SUMMARY

The discussion focuses on solving a probability problem involving a geometric distribution where the probability of success in each trial is 0.13 and failure is 0.87. The probability distribution for the random variable Y, representing the number of trials until the first success, is expressed as Pr(Y=y) = 0.87^(y-1)*0.13. The participants compute Pr(Y<9) by summing the probabilities from Y=1 to Y=8. Additionally, they derive the conditional probability P(Y≤12|Y>8) using the formula P(Y<13 ∩ Y>8) / P(Y>8).

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Homework Statement


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.


Homework Equations


n/a


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.
 
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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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