How can i solve this tricky question in statistics? (details inside, tricky for me)

I'm stuck on this question and i don't know how to solve it.

in a box there are 3 balls, numbered 1,2 and 3. we randomly pull 2 balls out of it, and we write the lower number on the higher number, and then retrieve them to the box.. so for instance if (2,3) were pulled, we retrieve (2,2). we repeat the process until all of the balls in the box are marked 1. if we get the same number twice, we just put them back to the box without doing anything.

a)if after the first pull, there is at least one ball with the number 2 on it, what are the odds that the proccess will end after only one additional ball pulling?

b)what are the odds that the proccess will be over after exactly two pulls?

thank you very much for your help

1 Answer

a. #1/3#; b. #4/9#

Explanation:

a

On the first draw, there are 3 possibilities:

1, 2; 1, 3; 2, 3

Each draw has an equal probability of occurring, i.e. #1/3#

Question a is asking for the situation where, after the first draw, there is at least one ball marked 2. That occurs in the event of 1,3 (the initial 2 is still in the box, giving 1, 1, 2), and 2, 3 (the 3 is overwritten with a 2, giving 1, 2, 2).

We're asked for the probability of all the balls having 1 written on them in one more draw. Of the two scenarios, only the one with 1, 1, 2 has a chance of occurring in one more draw (1, 2, 2 is impossible to reach 1, 1, 1 in one draw).

Within scenario 1, 1, 2, we can have draws 1, 1; 1, 2; and 1, 2 (one of those 1's is the former 3 and so we end up with two ways to get 1, 2). And so there is a #2/3# probability of having all three balls having 1 written on them if we get to this scenario.

The total probability, then, of having all three balls with a 1 written on them, given that we have at least 1 ball with a 2 on it after a single draw is:

#1/2(2/3)=1/3#

b

What are all the ways we can have the balls have 1s on them within 2 draws?

One is the scenario we did in a:

#1, 2, 3 => 1, 1, 2 => 1, 1, 1#

The only other scenario I see is for this to occur:

#1, 2, 3 => 1, 1, 3 => 1, 1, 1#

For each of these two scenarios, we have a probability of the first draw leading to having two 1s of #1/3#, and then a #2/3# probability of having the second draw giving us all the balls with 1s on them. That's:

#2xx(1/3)(2/3)=4/9#