Probability problems are calculated problems and universal formulas are derived from particulars?

1. It is equivalent to a five-element set to select a non-empty subset as the result of machine selection, and it is obvious that there are 2 5=32 subsets, and 31 are removed from the empty set, so the probability is 1 31.

One problem is that there is no upper limit, because if it is repeated, it can be repeated indefinitely, and for the time being, it will be up to 5.

2. Disordered and repeatable numbers. Set A, B......The number of occurrences is xa, xb, ......xe。

Find xa+xb+......xe=1,2,3,4,5 non-negative integer solutions.

In the partition method, the result is 1, that is, 1 letter corresponds to c(5,4)=5 kinds, and 2 corresponds to c(6,4).

The total is c(5,4)+c(6,4)+c(7,4)+c(8,4)+c(9,4)=251

3. The order cannot be repeated, and it is good to follow the number of arrangements.

a(5，1)+a(5，2)+a(5，3)+a(5，4)+a(5，5)=325

Of course, according to the order of non-repeatability, there can be no AAAE or EAAAA, then the order of repeatability also gives a result.

It's simple, 1 bit has 5 1 choices, 2 bit has 5 2 choice ......

The corresponding is 5 1 + 5 2 + 5 3 + 5 4 + 5 5 = 3905

Ma Yun and Li Xianli" five characters are represented by ABCDE, then there are: ABCD+ABED=EDCAD1, two four-digit numbers are added to get five digits, and the number of 10,000 digits must be 1, that is, E=1: ABCD+AB1D=1DCAD.

2.D+D=D in a single digit is weird. In this case, it will only appear when d=0:

abc0+ab10=10ca0。3. Look at the thousands, a+a=10, then a=5 must be, get: 5bc0+5b10=10c50.

4.Look at the ten digits, c+1=5, c=4, and the formula is: 5b40+5b10=10450.

5. Finally, look at the hundreds, b+b=4, b=2. The final result is: 5240 + 5410 = 10450, that is:

When learning mathematics as a subject, in fact, for different types of problems, in fact, this is very difficult for us, and many times we don't know where to start, especially for most of the girls, they are very difficult to learn mathematics, some people will have such doubts, that is, how to calculate the probability problem in mathematics? What kind of skills are the fastest for this problem, in my personal opinion, I think we should start with the simplest numbers, and then we should draw a chart for him, let's understand it in detail. 1 Starting from simple numbers, I believe that everyone has encountered such a problem when they are solving math problems, that is to say, when we get a probability problem, there will generally be corresponding data in it, and at this time, we will read the problem first, and then observe the relationship between each group of data in it.

2 Drawing a chartFor this kind of probability problem, in fact, I think some are very simple, but some of them do have a certain degree of difficulty, so when we don't know where to start, we should change the chart, when a complex problem comes to us in our minds, draw this chart for him, we can clearly see the various relationships in it, which can quickly answer the problem. Therefore, we should pay more attention to this aspect of the problem in the usual life, for everyone, to understand this aspect of the problem is a certain benefit to us, and now learning mathematics is indeed of great help to us, because mathematics is mainly to exercise our logical thinking ability, if the logical thinking ability is stronger, then they solve the problem, the yield is quite high. And it can also improve the individual's reflexes, which are of great help to a person's intellectual development.

Yes, the sum of p(ai) is the sum of p(a1) p(a2) p(a3)p(aiaj), p(a1a2) p(a1a3) p(a2a3). It should be noted that the sum of the high split I is less than jp(aiajak) is naturally p(a1a2a3) when the qi faction n = 3.

Draw a picture of the number of years and miss a picture, you should know where you are rotten.

OK. AUB means that at least one of A, B occurs, then his negative is none of them, i.e., neither A nor B. So: p(aub)=1-p(non-a, non-b).

Why can't p(a b) be equal to 1-p(non-a, not b)? Yes.

This problem belongs to the confidence interval problem in statistics, the mean variance is unknown, using the t-statistic, the lower and upper limits of the 95% confidence interval are and , according to the actual problem, we can know that the number of grapes eaten between 5 and 10 is normal, more than 10 (11 and above) or less than 5 (4 and below) is abnormal, and the probability is less than 5%.

A formula for the upper and lower limits of confidence.

The mean plus or minus (the points of the t-distribution with degrees of freedom of 7) is multiplied by (sample standard deviation) divided by (square root of sample size 8).

If you need a formula, you can see it below**.

Rounding is the answer above.

Solution: Eat x a day.

e(x)=sum of 8 days 8=60 8=

The range is 7 to 8 pieces.

d(x)= ∑(x-e(x))²/8=

When the quantity is a, p< (a-e(x)) d(x)) >

a> i.e. more than 12 are abnormal.

The same goes for a-e(x)<

a《Less than 3 is abnormal.

Again, for the following reasons:

1) The probability of finding 2 yellow balls at the same time: C10 takes 1/2 = 1 45

The probability that 2 of them are yellow balls one by one: 2 out of 10 times 1 out of 9 = 1 45

The probability of finding two at the same time.

p1=1/c(2,10)=1/45

The probability of taking it out and putting it back again.

p2=c(1,2)*c(1,2)/c(1,10)*c(1,10)=4/100=1/25

It is clear that the probabilities of the two methods are different.