It should be116.

Because the first ball will definitely go into one of the cups.

All that remains is the possibility that the other two balls will enter the same cup.

For example, ABCD cup. 123 ball

The first ball will definitely fall to A.B.C.D

If it is a cup.

Then the second cup will have a chance of 1/4 in the A cup, and so will the third ball.

So 1/4 *1/4 =116,5, all the time four,1,there are four kinds of three balls in the same one.

Put one of the three balls at random, and one of them is also four kinds.

Two balls and one cup+one ball and one cup are 12 kinds.

There are 20 kinds. Four of these 20 kinds are possible.

4/20 = 1/5

. . . . . . . . .

It may be wrong to graduate from primary school, but it is definitely not 1/64.

There is a positive solution on the fourth floor and the fifth floor, 1, and the above answer is correct: 1/64. 0, one tenth.

Because 1 cup has three balls, there are four possibilities; There are 24 possibilities to put a ball in each of the three cups; 2 balls 1 cup, 1 ball 1 cup. There are 12 possibilities. There are 40 possibilities, so the answer is one in ten. 0, 1/4 *1/4 is equal to a quarter, 0, 3 balls are different, there are 64 ways to play * * * (4*4*4), and four results meet the question, that is, all three balls are placed in1.

Three goals are consistent and four results are consistent. What matters is the way it is published. All put a cup, 4 kinds; Put three cups separately, because the balls are the same, and the results of putting them in cups 1, 2, 3 are the same as those of putting them in cups 3, 2, 1, so there are four kinds (3 out of 4 cups, C 43); Put two in one cup, and put 1 in one cup. Because the balls are all the same, there is no need to choose the ball. Just choose two cups from four cups to sort, A 4...,0,