How do the three brothers draw lots?

First of all, I want to know how many tickets you have that are not tickets.

Suppose there are x tickets, and there are two in them.

Secondly, I don't know if the ticket you drew should be put back.

Assuming that the winning ticket is not put back, after all, no one will have to put it back.

As can be seen from the meaning of the question, B has drawn the ticket.

Then we just need to discuss whether A can get the ticket.

For a, the probability that he will get the ticket is 2/x.

Knowing that Party B has drawn the tickets, the probability that both parties have drawn the tickets at the same time = (2/x) * [1/(x-1)] = 2/[x * (x-1)]

For C, the probability of getting the ticket is that A can't get the ticket.

(x-2)/x * 1/(x-2)= 1/x

It can be concluded that the 1 you need to explain is wrong.

Because [1-(1/x)] ≠ 2/[x * (x-1)]