A high school math probability problem, a high school math probability problem

Buy 5 bags of food, and the possible situation is a total of 3 to the 5th power. To win, 3 out of 5 bags must have different cards, i.e. C35. Among these 3 bags, it doesn't matter which card is the first bag, that is, C13, the second bag must be the same as the first bag, so C12, the third bag is C11

Therefore p=c35*c13*c12*c11 3 5=

It is equivalent to throwing the marked trihedral dice five times, and the principle of multiplication is known 3 5 kinds. If the situation does not conform to the situation, only the surface or the surface or the surface will appear, and the principle of multiplication knows that there are 2 5 kinds each, and the five occurrences of 1 or 2 or 3 are repeated once, so subtract 3So p=[3 5-(3*2 5-3)] 3 5=50 81

3 kinds of cards are numbered 123, and each bag of food can hold one of the 3 kinds of cards, and the total number is 3 5. Considering not winning, it may be that 5 bags only put 12 or 13 or 23 cards, so there are 2 5 + 2 5 + 2 5 = 96 Here there are 5 bags of 1 or 2 or 3 repeats, so the probability p=[3 5-(3*2 5-3)] 3 5=50 81 Hope to adopt!

Black Black & White + Black & White + White Black & White + White & White.

White, black, white: After taking the white ball, change to a black ball, the black ball will have one more, and the white ball will be one less, so the probability of taking the black ball again is 5 6, and the probability of taking the white ball again is 1 6

The probability of opening the door for the third time is.

The first door cannot be opened, which is 3 5 * the second door cannot be opened, 3 4 * the third can open the door, 2 3 = 3, 5 * 3 4 * 2 3 = 3 10

The frequency of students from 80 to 100 is 390, and the probability of drawing 39 students within 390 is 1 10

The probability of throwing the heads and tails of a silver coin is 1 in 2, so the person who answers (1) and answers (2) is the same, and the odd and even student numbers also account for half, so there are four corresponding situations: odd (1), odd (2), even (1), even (2). So the number of people in each case is 500, the answer of odd (1) is definitely "yes", here are 500 people, and the answer of even (1) is definitely "no", so 10 out of 1000 people who answer question (2) answer "yes", so 20 out of 2000 people should cheat.

That's how I think about it.

The person who answered the first question accounted for 1 2

and the odd number of students accounts for 1 2

So out of these 510 people, 500 people can be divided into the people who answer question (1) and then there are 10 people left, and these 10 people are the people who answer question (2) in those 1000, so out of the total of 2000 people, there should be 10*2 20 people who cheat.

There is a question in the math book that seems to be the same, so let's see for yourself.

The answer you gave is wrong, first of all, there are two situations: the first situation is the probability that the first six times to win two products, and the seventh time must be a defective product; The second case is that the first seven times are all **, then the remaining three must all be defective, and the sum of these two probabilities is the correct answer. p＝p1+p2＝2/15

One in fifty-six, seven must be defective molecules: two of the three products are selected by six and two are arranged again; The full permutation of the denominator is seven, i.e., the factorial of seven.

The problem can be transformed into finding the probability of drawing three ** in a row.

p = c7 3 divided by c10 3 = 7 24

The answer is not wrong, it is you who think wrong. In this situation, you should doubt yourself, not the answer.

The idea is as follows: in the second replacement of the bulb, the bulb that needs to be replaced is divided into two situations:1. The initial installation life is 2 years.

The probability of this situation should be P1 (1-P2), first of all, the bulb must meet the life of more than 1 year to ensure the occurrence of P2. Therefore, it should be satisfied for more than one year, and then (1-p2), that is, the bulb can work for two years. The two must be multiplied, because they are conditional.

1-p1）^2。You know this, so I won't talk about it.

You are right in your thinking that the probability of a lifespan of more than 1 year and less than 2 years should be p1-p2The answer of the person in front of me was that he didn't know how to pretend to understand, and he was superstitious about the answer. The answers to the college entrance examination questions of that year were wrong.

However, the proposition group has found that there is a problem with the answer before grading. When organizing the grading, it is actually processed according to P1-P2. I didn't expect a lot of information to blindly copy the answers, so that the correct answers were lost.

I accidentally saw your question, which is kind of clarification.

1) There are 6*6=36 choices for (a,b) pairs.

f(x) is on (-1, on which is the increasing function ==> and the opening of the parabola f(x) is upward, and the axis of symmetry x=2b a is located on the left side of (-1, i.e., a>0 and 2b a -1==a>0 and b -a 2==a,b) desirable(1,-2),(1,-1),(2,-2),(2,-1),(3,-2),(4,-2); There are 6 types.

Therefore, the probability of the null suspicion is 6 36=1 6

2) According to the problem, (a, b) is selected as a feasible domain problem, which is solved by the area method.

With a as the x-axis and b as the y-axis, then a,b satisfies the enclosed area as a straight triangle with an area of 32.

f(x) is on (1, on which is the increase function ==> the opening of the parabolic blind line f(x) is upward, and the symmetry axis x=2b a is located on the left side of (1), i.e., a>0 and 2b a 1, and the area enclosed is 32 3

Therefore, the probability is (32 3) 32 1 3

The ridge may be incorrect, whether f(x) should be ax 2-4bx+1

Solution: (1) Let a: "z-3i" be a real number.

That is, b=3, and a has no restriction.

p(a)=1/6

2)|z-3|3 is equivalent to (a-3) 2+b 2<=9, i.e., a 2-6*a+b 2<=0

b:“|z-3|≤3”a1:a=1,b=1;

a2:a=1,b=2;

a3:a=2,b=1;

a4:a=2,b=2;

a5:a=3,b=1;

a6:a=3,b=2;

a7:a=3,b=3;

a8:a=4,b=1;

a9:a=4,b=2;

a10:a=5,b=1;

a11:a=5,b=2;

p(b)=p(a1)+…p(a11)=11/36

(1) When z-3i=a+bi-3i is a real number, there is bi-3i=0, b=3, p(3)=1 6

2) By |a+bi-3|3, (a-3) + b 9, when b takes 1, a can take 1,2,3,4,5, when b takes 2, a can take 1,2,3,4,5, when b takes 3, a can take 3, there are 11 eligible ways to take it.

By the principle of addition: p(|z-3|≤3）=11/36.

(1) Adopt the assignment method, a general method for many topics!

Let a: "z-3i" be a real number.

That is, b=3, and a has no restriction.

Therefore p(a) = 1 6

2)|z-3|3 is equivalent to (a-3) 2+b 2<=9, i.e., a 2-6*a+b 2<=0

b:“|z-3|≤3”a1:a=1,b=1;

a2:a=1,b=2;

a3:a=2,b=1;

a4:a=2,b=2;

a5:a=3,b=1;

a6:a=3,b=2;

a7:a=3,b=3;

a8:a=4,b=1;

a9:a=4,b=2;

a10:a=5,b=1;

a11:a=5,b=2;

p(b)=p(a1)+…p(a11)=11/36

It should be: p=(1 6)*(1 5) on the line, because the meaning of the title is that there are male A and female B among the 3 people selected, and it doesn't matter who the third person is, and the selection is not divided into men and women, so the denominator should be 6, so it is necessary to select A among the 6 people first, and B among the remaining 5 people.

Choose 3 out of 6 people, a total of c(6,3)=20 choices, A and B both choose c(1,1)*c(1,1)*c(4,1)=4 choices, so the probability is 4 20=1 5.

Male A and Female B are selected at the same time, that is, Male A and Female B have been selected by the confirmation, and one more person is selected from the remaining 4 people, that is, there are 4 possibilities, and there are 20 possibilities among the 6 people, so the probability is 4 20, p=(1 4)*(1 2)*(1 4) is completely wrong, and the multiplication of probability is not suitable here.

There are 20 kinds of candidates from six candidates, and I use C to calculate, and I have four kinds of male and female A at the same time, so the probability is one in five.

It doesn't matter who the third person is.

As long as the initial p = (1 4) * (1 2) is sufficient.

The whole is 6 people, no distinction between men and women... Pig...

Choose 3 out of 6 for a total of c63=20

Male A and female B are selected at the same time, and a c41=4 is selected from the remaining 4 people

4 divided by 20 = 1 5