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Mathematical probability problems of liberal arts in college entrance examination
If the positive integer is not a multiple of 3

Then his square must be a multiple of 3+1

[(3k+ 1)^2

=

9k^2

+

6K

+

1]

modern

three

=

1

[(3k+2)^2

=

9k^2

+

12K

+

4]

modern

three

=

1

T

=

a^2

+

b^2

+

c^2

+

d^2

T

modern

three

=

There are two possibilities.

A: All four items are multiples of 3, and the probability is PA.

B: 1 is a multiple of 3, others are not multiples of 3 (3k+ 1), and the probability is PB.

therefore

P

=

PA+PB

=(2/6)^4

+

c(4 1)*

(2/6)^ 1

*

C (3,3)

*

(4/6)^3

= 1/8 1

+

4* 1/3*8/27

=

33/8 1

1 building SB

The second floor is wrong. This is the sample that was put back. We should consider the ball that should be taken out.

In the case of 1, 1, 1, the overall situation is 6 4.

3rd floor, answer: Yes.