So 1 1 the probability that an apple box appears in A's hand = 1 1 the probability that an apple box appears in B's hand;
There are many traps in the topic, but it can be seen that there are only three results. Let's assume that 1 1 Apple's box appears in A's hand as A 1.
A doesn't have A0.
1 1 The apple box appears in B's hand as B 1.
B doesn't have B0.
Neither A nor B has it, which is A0B0.
Then the results are: A 1B0, A0B 1, A0B0.
This is a combination problem.
Because A, B, 1 1 apple boxes are all random. According to the law of permutation and combination, you can wait for the same law as the probability of the lottery sequence mentioned above.
The subjects of all three are 1/3.
Is it simple? Just don't be confused by the topic.
The one on the fifth floor is nonsense, which violates the sequential lottery method. Sorry, I forgot the name of that law.