Three balls are drawn from an urn containing M M3 balls numb

Three balls are drawn from an urn containing M (M>3) 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 third ball drawn is numbered 3?

This is what I have so far and I am not sure where to go or if it is right:

P= 1/m of drawing a 1 on the first pull.

P= 1/(m-1) of drawing the second ball.

P=(m-1)/m of not drawing a 1 on the first pull .

P= (m-1)/(m-1) of not drawing a 3 on the last pull .

Solution

see there will be two cases.

as the number of ball will reduced if ball with number 1 is picked in either 1st time or second time, other wise the number of balls in the urn will remain same. and there is only one ball with each number.

so case 1: if the ball with number 1 is picked in either 1st or 2nd chance

as 1 is picked so total no: of balls in the urn will be m-1

and the probability of being 3 ball with number 3 will be 1/(m-1)

case 2: when ball with number 1 is not picked in 1st and 2nd chance;

as ball with number 1 is not picked then the number of balls in the urn will be m

so the probability of ball being of number 3 is 1/m