Help on probability again please

[zhfs]

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.

 

[Random Variable]

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.

$$P(Y<13|Y>8) = \frac {P(Y<13 \cap Y>8)}{P(Y>8)} = \frac {P(8>Y>13)}{P(Y>8)}$$

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

 

[Architeuthis Dux]

For part (c), you are correct. The probability distribution for Y is given by Pr(Y=y) = 0.87^(y-1)*0.13, where y represents the number of trials up to and including the first success.

For part (d), you need to sum up the probabilities for y=1 to y=8, since we want to find the probability that Y is less than 9. So the expression would be Pr(Y<9) = \sum_{y=1}^8 (0.87^(y-1)*0.13).

For part (e), we want to find the probability that Y is less than or equal to 12, given that Y is greater than 8. This can be written as Pr(Y<=12 | Y>8). Using conditional probability, we can rewrite this as Pr(Y<=12 and Y>8)/Pr(Y>8). The expression for Pr(Y<=12 and Y>8) can be found by summing up the probabilities for y=9 to y=12, and Pr(Y>8) can be found by summing up the probabilities for y=9 to infinity. So the final expression would be Pr(Y<=12 | Y>8) = (\sum_{y=9}^{12} (0.87^(y-1)*0.13)) / (\sum_{y=9}^{\infty} (0.87^(y-1)*0.13)). You can use MATLAB to compute this value.