The sophomore probability question is very simple!!!

50% of the engines are ready to fly, two of the four engines are running, and one of the twin engines is enough.

So the four-engine aircraft cannot fly in the case of a three-engine failure and a four-engine failure, the total probability is.

c（1，4）*a*(1-a)*(1-a)(1-a)+(1-a)^4=4a(1-a)^3 -(1-a)^4

1-a)^3(5a-1)

c(2,4) is the arrangement of 2 out of 4).

A twin-engine aircraft cannot fly when both engines fail at the same time, and the total probability is (1-a) 2 If a four-engine aircraft is safer than a twin-engine aircraft, the probability of not being able to fly should be less than that of a twin-engine aircraft.

i.e. (1-a) 3 (5a-1) < (1-a) 21-a) (5a-1) < 1

5a^2-6a+2>0 5(a^2-6/5a+9/25-9/25+2/5)>0

5(a-2/5)^2+1/5>0

So when a is a real number, the inequality holds, so 0When < a<1, all four engines are safer than twin engines.

The safety probability of a twin-engine aircraft is m=2a(1-a)+a*a, and the safety probability of a four-engine aircraft is n=(3*4) 2*a*a(1-a)(1-a) +4*a*a*a*(1-a)+a*a

In order to make a four-engine aircraft safer than a twin-engine aircraft, it is necessary to meet n>m, i.e., n-m>0 After collation, it can be obtained: (a-1)(a-1)(3a-2)>0 So only 3a-2>0

So 1>a>2 3

The square of c42*(1-a)*the square of a=c21*(1-a)*a

Just solve the equation.

I'm a sophomore now, and I'm not familiar with high school.

Touch 1 ball at a time, and stop touching the ball when both colors of the ball are touched. Description x>=2If the first three balls are all red balls, the fourth time is the yellow ball, which stops at this time, and x=4. Hence 2=(1) x 2 3 4

p is set to a is set to b is set to c

then p(a) (i.e., the first two are red and yellow) = [c(1,2) c(1,5)]*c(1,3) c(1,4)]+c(1,3) c(1,5)]*c(1,2) c(1,4)]=6 20+6 20=12 20=3 5

p(b) (i.e. the first three are red and the third is yellow or the first three are yellow and the third is red) =

c(2,3)/c(2,5)]*c(1,2)/c(1,3)]+c(2,2)/c(2,5)]*c(1,3)/c(1,3)]=3/10*2/3+1/10=6/20

p(c) (i.e. the first four three are red and the fourth is yellow) =

c(3,3)/c(3,5)]*c(1,2)/c(1,2)]=2/20

Validation: 12+6+2 is exactly =20

2) Distribution Columns:

Variance doesn't count. Let's substitute the formula.

The first question: your problem is that the first time you ask for the selection of 3 people are boys, so the girls do not participate in the speech directly, directly exclude, you directly choose 3 from 6 people, so there are 6 cases, according to the requirements only from 4 boys, only 4 cases.

The second question: find the origin of 1 girl among the 3 people selected, and 2 boys and 1 girl, choose 2 from the 4 boys with c42 = 6 (in fact, c42 was not originally written like this, it should be 4 in the lower right corner of c, 2 in the upper right corner of c, we just can't play that form, so we wrote c42, which means that 2 members are selected from 4 (for example: this member)), and 1 c21 = 3 out of 3 girls, so the probability is c42 * c21 20 = 12 20 = 3 5.

Question 3: Find the probability that there is at least 1 girl in the selected 3 people, this question should be divided into 2 situations, and then the probability of each situation is added;

1): There is a case of 1 girl, that is, the case of the second question is 3 5;

2): There are 2 girls c22 and 1 boy c41, the probability is c22*c41 20=4 20=1 5.

So the final probability is 3 5 + 1 5 = 4 5.

You can use a treemap, but you should number it as a1 a2 a3 a4 for boys and b1 b2 for girls.

2c 4 is 2 out of 4 people, and choose a1 a2 is very the same as a2 a1, it is best to choose people without queuing, and there is no hurry to go back to school after calculation.

There is also a way to draw a table.

In the case of 1 2 3 4 for boys and 5 6 for girls.

6. Note that 11 is repetitive, to cross out 12 21 is the same, to cross out one, after writing in the put.

Note that 111 121 can only appear once in a person This method is better than a treemap.

Enough for liberal arts students.

1.In line with the hypergeometric distribution, high 2 you will learn, the most basic problem. Combined methods. (I can't hit a combination of symbols).

2.Same. 3。Add the probability of one girl to the probability of two girls.

To do this mathematically, let x be the number of beers you need to drink before you can select a coin. Then the distribution of x is listed as:

Then the mathematical expectation of x is e(x)=, i.e., the average bottle of wine is randomly selected.

So if you want to drink more, choose one bottle at a time, and give up if you want to drink less!

Give up if you want to drink less, and drink more bottle by bottle. Intuition is this. It feels like you can't start with probability. Ask the great god to help.

1 - [(364 365) to the power of 200].

The probability that everyone will have a birthday on October 1 is 1 365, because there are 365 days in a year, and no one is not on October 1, that is, 1-1 365 = 364 365.

At least one person has a birthday, and the opposite is that no one has a birthday on October 1. Then his probability is (364 365) to the power of 200.

And at least one person's birthday is October 1, which is 1-[(364 365) to the 200th power].

(1) The probability of A getting the black ball is 3 5, and the probability of B getting the black ball is 1 2 (2) The probability of two people getting the same color ball is divided into both taking the black ball and both going to the white ball, and the probability of both getting the black ball is 3 5 x 1 2=3 10, and the probability of both going to the white ball is 2 5 x 1 4=1 10, that is, the probability of A winning is 4 10 = 2 5, then the probability of B winning is 1-2 5 = 3 5

If you can choose any three ways to have c(6,3)=20, you may wish to set x(1,1),x(2,1),x(3,2),x(4,2), x(5,3), x(6,4) represent six routes, the first number is the number of roads, and the second number is the amount of information passed.

1) The information is not smooth, there are x(1,1),x(2,1),x(3,2)+x(1,1),x(2,1),x(4,2)+x(1,1),x(2,1),x(5,3)+x(1,1),x(3,2),x(4,2)+x(2,1),x(3,2),x(4,2)=5

So the probability of unobstructed is 1-5 20=3 4

The method is to determine an x(1,1) first, then consider the last two, after searching, the first one is replaced by x(2,1), then consider the last two, and so on.

2）ex＝4×2/20＋5×3/20＋6×5/20＋7×5/20＋8×3/20+9×2/20＝13/2

Of course, you can also look at the diagram and write it as a three-dimensional vector. For example, (1,1,2), as long as you look at it, don't miss it.

1) The information is not smooth, there are x(1,1),x(2,1),x(3,2)+x(1,1),x(2,1),x(4,2)+x(1,1),x(2,1),x(5,3)+x(1,1),x(3,2),x(4,2)+x(2,1),x(3,2),x(4,2)=5

So the probability of unobstructed is 1-5 20=3 4

There are c(6,3) 20 ways to choose any three lines, and there are 5 kinds of situations where the information is not smooth, so the probability of information flow is 1 5 20 3 4

x 4 5 6 7 8 probability 2 20 3 20 5 20 5 20 5 20ex 4 2 20 5 3 20 6 5 20 7 5 20 8 5 20

Upstairs is wrong p(x=8)=3 20, and p(x=9)=2 20 event probability is missing.

Therefore, ex 4 2 20 5 3 20 6 5 20 7 5 20 8 3 20 + 9 2 20 13 2