Hi Antti-Jussi,

Your description is clear enough for us to understand. Here are some thoughts on your q

We could start from finding the probability P' of a number of the latter set NOT in the first set. Then the probability of it in the first set is 1-P'.

Below's a possible way to find P':

1. Let s_{1}and s_{2}represent the first and second item in the latter set. Either of s_{1}and s_{2}could be 1, 2, or 3 with probability of 1/3. For example, s_{1}= 1 is with probability 1/3 since s_{1}is randomly chosen from three numbers.

2. Now let's consider a specific case, say when s_{1}= 1 AND s_{1}is NOT in the first set. In this case, all items of the first set should be chosen from the other two numbers 2 and 3 (right?). Since there are totally three choices (i.e., 1, 2, 3) but only two (i.e., 2 ,3) are feasible, the probability for s_{1}= 1 NOT in the first set should be .

3. Step 2 considers one specific case (s_{1}= 1) out of three possible cases (i.e., s_{1}= 1, 2, or 3). So combining all the three cases, the probability for s_{1}NOT in the first set should be .

4. Given the probability of s_{1}NOT in the first set is , the probability that s_{1}exists in the first set is (right?)

5. Still remember the other item s_{2}? I think now you could similarly find that the probability of s_{2}in the first set is also (according to Steps 2-4).

5. So now we can go for the probability for both s_{1}AND s_{2}in the first set: .

So the above is how I think to solve the question. I try to make it specific enough for us to more easily spot where could go wrong. Let me know if more explanations could be useful and welcome any further discussion.