# Help on probability again please

1. May 25, 2009

### zhfs

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) = $$\sum(0.87^(y-1)*0.13$$

and got no idea for part (e), need help please.

2. May 25, 2009

### 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.

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

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