Challenging Statistics-Probability Question?

A student needs to know details of a class assignment that is due the next day and decides to call fellow class members for this information. She believes that for any particular call the probability of obtaining the necessary information is 0.40. She decides to continue calling class members until the information is obtained. Let the random variable X denote the number of calls needed to obtain the information.


a. Find the probability function of X.


b. Find the cumulative probability function of X.


c. Find the probability that at least three calls are required.





thanks in advance.

Challenging Statistics-Probability Question?
Let X be the number of calls needed to get the required information. X has the geometric distribution with success probability p = 0.40.





Let X have the geometric distribution. X is the number of trials until the first success.





X ~ Geom(p) where p is the success probability





the probability mass function is:





f(x) = P(X = x) = p * (1 - p) ^ (x - 1) for x = 1, 2, 3, 4, 5, ....


f(x) = 0





the mass function is found by looking for x - 1 failures followed by 1 success.





the cumulative probability function is





P( X ≤ x) = ∑ P(X = t)





the limits on the sum are t = 1 to t = x





this can reduce down to:





p * ∑(1 - p) ^ n for n = 0 to x - 1





P(X ≤ 3) = 0.4 + 0.4 * 0.6 + 0.4 * 0.6 ^2 = 0.784