Help on probability again please

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The discussion focuses on solving a probability problem related to a geometric distribution, where the probability of success in each trial is 0.13 and failure is 0.87. For part (c), the probability distribution of the random variable Y, representing the number of trials until the first success, is correctly identified as Pr(Y=y) = 0.87^(y-1)*0.13. In part (d), the cumulative probability Pr(Y<9) is calculated by summing from y=1 to y=8. Part (e) involves finding the conditional probability P(Y≤12|Y>8), which is expressed using the intersection of events and requires summing probabilities for the specified ranges. The thread emphasizes the need for MATLAB to compute and visualize these probabilities effectively.
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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) = \sum(0.87^(y-1)*0.13

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.

P(Y&lt;13|Y&gt;8) = \frac {P(Y&lt;13 \cap Y&gt;8)}{P(Y&gt;8)} = \frac {P(8&gt;Y&gt;13)}{P(Y&gt;8)}

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