Project 32: Geometcdf

Suppose you have a 20% chance of winning a particular game of chance.  You plan to play until you win.  The distribution of games you will play is defined by the geometric distribution with p = .2.

The tree diagram for the first several possibilities is shown below.

If X = the number of games played 

P(X = 1) = .2

P(X = 2) = .8*.2 = .16

P(X = 3) = .8*.8*.2 = .128

P(X = 4) = .8^3*.2 = .1024

X can possibly be any number from 1 to infinity but the probabilities continually decrease.  The sum of all the probabilities for X still must equal 1.

If we wanted the probability that X is between 2 and 4 inclusively, we would add the probabilities:

.16 + .128 + .1024

to get .3904

Project 32: The variables 'p', 'low', 'high' have been initialized.  There is also a working method called geometpdf.

p represents the probability of success on any trial (.2 in the example shown above)

low represents the lower bound for X (inclusively)

high represents the upper bound for X (inclusively)

geometpdf(p, k) returns the probability of getting the first success on trial k (when every trial has 'p' probability of success)

example: geometpdf(.2, 3) = .1024

Task: Appropriately initialize the value of 'geometcdf' that represents the probability that the number of trials needed is between low and high inclusively.

Example: if p = .2, low = 2, and high = 4, then geometcdf should be .3904

**If your code works for 5 test cases, you can enter your e-mail address

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0