Listening to tracks on a cd player

[toysoy]

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 second selection will not be his favourite track?
b)What is the expected waiting time before Louis hears his favourite track?
c)If Louis has two favourite tracks, what is the expected waiting time before he hears both tracks?

Repeat part b assuming the CD player does not repeat selections.

 

[tkhunny]

a) This is a very odd question. Is #2 his favorite? 100% Favorites don't come randomly, as a general rule. Perhaps we are to assume he has a favorite and he has forgotten which it is? One favorite and 12 tunes? 1/12?

b) No memory. That's a bad random process for a CD. Anyway, the chance of a tune being picked is 1/12 at each stage.

Pr(first) = 1/12
Pr(second) = (11/12)*(1/12)
Pr(third) = [(11/12)^2]*(1/12)
Pr(fourth) = [(11/12)^3]*(1/12)
...

There's the entire distribution. What is its mean? Do you recognize this distribution?

Let's see how far you get with this before you demonstrate your efforts on the third part.

 

[toysoy]

Ok I actually think I figured out how to do the first parts but can you verify my answers?

a) q^x*p, q = 11/12, p = 1/12, x = 2
=(11/12)^2*(1/12)
= .840277778

b) E(x) = q/p , q = 11/12, p = 1/12
=(11/12)/(1/12)
=11

c) E(x) = q/p , q = 10/12, p = 2/12
=(10/12)/(2/12)
=5

 

[tkhunny]

a) I think I get it. He has a favorite track. It could be selected at any time.

Pr(favorite is selected on first pick) = 1/12
Pr(favorite is selected on second pick) = 1/12
Pr(favorite is selected on third pick) = 1/12

Remember. the process has no memory, so previous selections have nothing to do with subsequent selections.

(11/12)^2 * (1/12) is the probability that something else will be selected twice and favorite will them be selected - as in, the favorite will be selected third. I do not believe this answers the question.

b) You didn't answer my questions on this one. I still think I had this one right in the first place. I seriously doubt the mean of the distribution is 11.