Solve mathematical probability problems. 20. Solving mathematical probability problems.

This is the easiest calculation problem in the second year of high school, and you can't do it.

Solution: 1) f(x) is the increasing function that the derivative of f(x) (-2x 2+2ax+4) (x 2+2) 2>=0 holds constant over the interval [-1,1].

That is, -2x 2+2ax+4>=0 holds constant over the interval [-1,1].

Then f(-1)>=0 f(1)>=0, i.e., -1<=a<=1

2) F(x)=1 x is organized into quadratic functions, in the form of x 2-ax-2=0, and the two roots are x1 and x2

then x1+x2=a x1x2=-2 thus has.

x1-x2i^2=(x1+x2)^2-4x1x2=a^2+8

So m 2+TM+1>=|x1-x2|That is, m 2 + TM + 1> = (a 2 + 8).

It is required that m 2+tm + 1> = (a 2+8) is constant for any a belongs to [-1,1] and t belongs to [-1,1].

Then it is required that the minimum value of the left f(t) = mt+m 2+1 of the above equation must be "" = the maximum value of f(a) on the right.

f(t) is a primary function, so let's talk about it.

When m>0, the minimum value f(t)=f(-1)=m2-m+1>=3 gives m>=2

When m>0, the minimum value f(t)=f(-1)=m2-m+1>=3 gives m>=2

When m=0, this is obviously not true.

So the range of m is m>=2 or m<=-2

x|=x (x≥0)-x (x<0)

1-1|x|dx=0-1|x|dx+01|x|dx

0-1(-x)DX+01XDX, so C. should be selected

4.Let f(x)=x2 (0 x<1)2-x (1 x 2), then 02f(x)dx equals ( ).

d.does not exist.

Answer] C Analysis] 02f(x)dx=01x2dx+12(2-x)dx

Take f1(x)=13x3, f2(x)=2x-12x2, then f1(x)=x2, f2(x)=2-x

02f(x)dx=f1(1)-f1(0)+f2(2)-f2(1)

13-0+2×2-12×22-2×1-12×12=56.Therefore, C. should be chosen)

Solution: Place this line between the point o and the point 1 on the number axis, and let the two points taken be x and y respectively, then the event "arbitrarily take 2 points to find the small 1 2", that is.

Event "x-y|<1 2 "i.e. event".

It is equivalent to enclosing 13 points in a circle, fixing one point, and then selecting 6 points, and if there are fixed points and two points on the left and right, they will be taken out. The probability of taking out more than 2 points (including 2 points) is (c(3,2)c(10,4)+c(3,3)c(10,3)) c(13,6) (3 10 9 8 7 4!). ＋10x9x8/3！

The chocolate is numbered 1-16, and the total method is 16*15, both times are obtained milk chocolate in total 6*5, and both times are obtained pure chocolate in 10*9, then the probability of obtaining milk chocolate is p=(6*5) (16**15) both times, and the probability of obtaining pure chocolate is p=(10*9) (16*15) twice.

Of the 10 performers, 2 can only sing, 3 can sing and dance, and 5 can dance.

There are two types of programs where one person sings a solo and four people dance.

1. Soloists can only sing but not dance.

There are c(2,1)*c(8,4)=140 options.

2. Soloists can sing and dance.

There is c(3,1)*c(5,4)+c(3,2)*c(5,2)+c(3,3)*c(5,2).

There are a total of 195 options.

Good luck.

1. (1) The probability of taking out the white ball is 1-1 4=3 4

2) The red ball is 18*1 3=6

2. Closing S1 and S3, S2 and S3 at the same time can make the bulb glow, but S2 and S3 cannot, so the luminous rate of the bulb is 2 3

Either the red ball is taken out, or the white ball is taken out, and the probability of the two is 1 The probability of taking out the red ball is 1 4, then the probability of taking out the white ball is naturally 1-1 4 = 3 4, in fact, this probability shows that the probability of the white ball is three times the probability of the red ball.

In other words, there are actually three times as many white balls as there are red balls. There are 18 white balls, 18 red balls 3=6 As for the second question, there are three options, s1s2, s1s3, s2s3s1s3 closed, s2s3 closed bright, s1s2 closed and not bright, so the probability of light is 2 3

1.Except that the red ball is the white ball, so it's 1-1 4=3 4

2.If 18 white balls account for 3 4, then there is a total of 18 (3 4) = 24 and white balls are 24-18 = 6

3.It could be (s1 s2) (s1 s3) (s2 s3), two of which can make the bulb glow, and the chance is 2 3