Hypergeometric distribution?

1)Louis inserts a 12-track CD into a CD player and presses the random play button. This CD player's random function chooses each track independently of any previously played tracks.


a)What is the probability that the CD player will select Louis' favourite track first?


b)what is the probability that the second selection will not be his favourite track?


c)wha is the expected waiting time before Louis hears his favourite track?


d)sketch a graph of the probability distribution for the waiting times.


e)explain how having a different number of tracks on the CD would affect the graph in part d)


f) If Louis has two favourite tracks, what is the expected waiting time before he hears both tracks?

Hypergeometric distribution?
First, this problem does not use the hypergeometric distribution, it uses the geometric distribution. The hypergeometric distribution is used for sampling without replacement. Since the player's random function can choose any track from the CD at any time we are sampling with replacement and thus we will look at the geometric distibution. (other possible distributions for sampling with replacement are the binomial, Poisson and others.)





Let X be the number of trials until Louis hears his favorite song played. X has the geometric distribution with success probability p = 1/12.





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 otherwise





the mean of the geometric is 1/p


the variance of the geometric is (1-p)/p²





a) X ~ Geom(1/12)


P(X = 1) = 1/12





b) X ~ Geom(1/12)


P(X = 2) = (1/12) * (1 - 1/12) = 11/144





c) E(X) = 1/(1/12) = 12





d) this is left for you to do as graphing in this forum is not possible.





e) the graph will always be positively skewed. However, as the number of tracks on the disc increases the size of the "hump" will decrease and the skewness will decrease.





f) X ~ Geom(2/12 = 1/6)


E(X) = 1/(1/6) = 6