There are 6 more draws that would satisfy the conditions for part A:
(1, 3, 2), (1, 4, 2), (1, 4, 3), (1, 5, 2), (1, 5, 3), (1, 5, 4)
Not sure if I did part a correctly. Part B, not even sure how to do it.
Problem:
A bin contain balls numbered 1 through 5. Balls are removed, one at a time, until a ball with a similar value tan the previous number is drawn.
a. Find the probability that you stop after three draws (the third ball has a smaller value)
I first figured that there were 14 possible formations that would meet the requirements. They are:
(2,3,1), (2,4,1), (2,4,3), (2,5,1), (2,5,3), (2,5,4), (3,4,1), (3,4,2), (3,5,1), (3,5,2), (3,5,4), (4,5,1), (4,5,2), (4,5,3)
I then figured out the total possible formations 5P3 = 60
14/60 = .2333 is the probability that I would stop after three draws. Was I correct in completing this problem?
b. Let X represent the value of the ball that is the first ball with a smaller value. Determine the probability mass function on X.
Here's a hint to help with part B...
The ways that 1 can be the first smaller number are
2,1
2,3,1......2,4,1.....2,5,1
2,3,4,1...2,3,5,1...2,3,4,5,1...2,3,5,4,1
2,4,3,1...2,4,5,1...2,4,3,5,1...2,4,5,3,1
2,5,3,1...2,5,4,1...2,5,3,4,1...2,5,4,3,1
Starting with 2, there are 8 valid sequences
The same type of procedure will count the others.