Conditional Probability Problem
Problem: Balls are drawn (one at a time) from an urn containing M balls numbered from 1 to M. The ball is kept if it is numbered 1, and returned to the urn otherwise. What is the probability that the second ball drawn is numbered 2?
Here is what I have figured. If the ball numbered 1 is chosen first:
P(2) = 1/M * 1/(M-1)
If the ball numbered 1 is not chosen first:
P(2) = (M-1)/M * 1/M
My question is: Are these added together to get the overall probability? Meaning, is the final answer:
[1/M * 1/(M-1)] + [(M-1)/M * 1/M] ? Thanks!