Ask Math Masters to Solve Several Math Problems in Senior High School

(1). All counterpoint, x = 0；; There are only 1 cases.

(2) Two dislocations may be as follows: 1 2 or 1 3 or 1 4 or 2 3 or 2 4 or 3 4 dislocations.

X = 2, 4, 6 (only 6 cases * * * in total, x = 2: 3 respectively; X = 4: 2 species; X = 6: 1 species)

(3). There are many possibilities for three mismatches, but you only need to calculate the possible value of x and find the place different from the above.

1 para, X = 4；; ; 2 para position, X = 6；; ; 3 para position, X = 6；; ; 4 counterpoint, X = 4. (There are only eight cases in total, that is, x = 4: 4; X = 6: 4 species)

(4). All wrong, completely mixed. Maximum case of x, 1.4 exchange, 2.3 exchange, X = 8. (There are only 9 kinds of * * *, which are X = 4: 1 species 21.43; X = 6: 4 kinds of 2 4 1 3 or 2 3 4 1 or 3 1 4 2 or 465 438+0 23; X = 8: 4 kinds of 3 4 1 2 or 3 4 2 1 or 4 3 1 2 or 4 3 2 1)

So all the values are: {0,2,4,6,8}

The whole arrangement is 24.

p(X = 0)= 1/24；

P(X=2) = 3/24 = 1/8 (only two mismatches can be used and must be connected, 1 2 or 2/3 or 3/4 are interchanged).

P(X=4) = (2 + 4 + 1) / 24 = 7 / 24

p(X = 6)=( 1+4+4)/24 = 9/24 = 3/8

P(X=8) = 4 / 24 = 1 / 6

p(3 & lt； X & lt7)= P(X = 4)+P(X = 6)= 7/24+9/24 = 16/24 = 2/3

After four consecutive releases, they all appeared independently, so P = (2/3) 4 = 14/8 1.

I also suggest listing all the possibilities. There are only 24 situations. You can write them out according to the rules, such as:

1 2 3 4

1 2 4 3