2. The probability is the same (according to the meaning of the question, don't put it back)
Consider grasping it in the order of a b c d e.
The probability of the first catch is 3/5.
B the probability of the second catch: P 1: A There are 2 places left (3/5)*(2/4).
P2: If A doesn't win, there are still 3 places (2/5)*(3/4).
P=P 1+P2=3/5
The probability of catching C for the third time: P 1: 1 position in A and B (3/5) * (2/4) * (1/3) =1/0.
P2: The number of unsuccessful places in A is (3/5)*(2/4)*(2/3)= 1/5.
P3: Miss A comes second, B (2/5)*(3/4)*(2/3)= 1/5.
P4: A missed B missed 3 digits (2/5) * (1/4) * (3/3) =110.
P=P 1+P2+P3+P4=3/5
Similarly, D E is more troublesome.
In short, the probability is 3/5.