Classical probability has two characteristics:
(1) The number of possible basic events in the experiment is limited;
(2) The possibility of each basic event is equal.
Here are some problem-solving strategies for classical probability.
1. Find the probability with mutually exclusive events or the opposite event.
In order to avoid complicated calculation, sometimes we can turn the requested events into simple and easy-to-find sum events in mutually exclusive events, or we can use opposite events to find them.
There are three white balls and three black balls in the schoolbag. If you choose two of them, what is the probability of having at most one black ball?
Analysis: To discuss or make use of opposing events by classification.
Solution 1: two balls are randomly taken out of the bag, and * * has 6ⅹ5÷2= 15 possible results. "If you choose two, there will be at most one black ball" is regarded as the union of two mutually exclusive events events "All White Ball" and "One Black Ball, One White Ball". "All-white ball" has 3ⅹ2÷2=3 possible results, and "one black and one white ball" has 3ⅹ3=9 possible results. Let event A be "there is at most one black ball". Event A contains 9+3= 12 basic events.
Therefore, event P(A) = = probability p (a) of 0.8.
Solution 2: The opposite of event A is: "Both are black balls (recorded as event B)", and the number of basic events contained in event B is 3ⅹ2÷2=3.
Therefore, the probability of event A is p (a) =1-p (b) =1-= 0.8.
Use formula
P(A)= number of basic events contained in event a/total number of basic events.
Example: 1 There are 10 pieces in a batch of products, of which 8 pieces are genuine and 2 pieces are defective.
(1) If you take one out of it, then put it back, take another one, then put it back, take another one, and find the probability that the products taken out three times in a row are genuine.
(2) If you take three pieces at a time, find the probability that all three pieces are genuine.
Analysis: (1) is a sample with release; (2) Non-substitution sampling
Solution: (1) is extracted for three times, and the results (x, y, z) are recorded in the order of extraction, so there are 10 possibilities for x, y and z, so all the results of the experiment are10 ⅹ10 ⅹ/.
Let event A be "all products taken out for three consecutive times are genuine", and according to the above calculation method, the basic event * * * is 8ⅹ8ⅹ8=5 12.
Therefore, the probability of event A is P(A)= =0.5 12.
(2) Method 1: It can be seen that the sampling is not put back three times, the order is different, and the basic events are different. If the results (x, y, z) are recorded in sequence, x has 10 possibilities, y has 9 possibilities and z has 8 possibilities, then all the results of the test are 10ⅹ9ⅹ8=720 possibilities.
Let event B be "all three items are authentic", and according to the above calculation method, there are 8ⅹ7ⅹ6=336 basic events * *. So the probability of event b P(B)= ≈0.467.
Method 2: It can be regarded as three times of sampling without putting it back. First, record the results (x, y, z) in the order of sampling, then there are 10 possibilities for x, 9 possibilities for y and 8 possibilities for z, but (x, y, z), (x, z, y), (y, x, z).
So the probability of event b P(B)= ≈0.467.
Comments: When calculating the number of basic events without sampling, it can be considered as orderly or disorderly, and the result is the same, but no matter which method is chosen, the observation angle must be consistent, otherwise there will be errors.
3. Find the probability through the intersection and union of sets.
Because all possible results of the experiment can be regarded as a set, that is, a complete set, and each event can be regarded as a subset of the complete set, the relationship between set and event probability is established through the correspondence between events and sets. Therefore, we can solve the probability problem simply and easily by using the operation and properties of sets.
Example 3 Randomly take an integer from 1∽ 100, and find: (1) the probability that it can be divisible by 6 and 8 at the same time; (2) The probability of being divisible by 6 or 8.
Analysis: (1) Take an integer at random, and there are 100 possible results. Numbers divisible by 6 and 8 are numbers divisible by 24, from 1≤24n≤ 100(n∈), 1 ≤ n ≤
(2) From 1≤6n 1≤ 100, we get 1 ≤ n1≤/kloc-6, from1≤ 8n2 ≤/kloc-. Numbers divisible by 8 have 12 possible results, and numbers divisible by 6 and 8 have 4 possible results, so the possible results of numbers divisible by 6 or 8 are 16+ 12-4=24. Write "divisible by 6 or 8" as event b, P(B)= =.
4. Establish a classical probability model
Classical probability has the characteristics of strong application, and many phenomena in life conform to the characteristics of classical probability after analysis. So we can build a model to solve it.
In order to investigate the number of some wild animals in a wildlife reserve, the investigators caught 1200 such animals a day, marked them and put them back. A week later, they captured 1000 such animals, including 1000 such animals. How to estimate the number of such animals in the nature reserve?
Analysis: First of all, this is a practical problem in life. It is impossible for us to count the number of such wild animals one by one, and it is completely unnecessary. Because it wastes the necessary human, material and financial resources, it is necessary to establish a mathematical model. According to the method of probability, this problem can be solved well.
Solution: Because every animal has the same possibility of being caught, and all animals are limited, a classical probability model can be established. Suppose there are x wild animals in the nature reserve, and the probability of each animal being caught is the same. So x/1200 =100/1000. According to this method, there are about 12000 such animals in the nature reserve.
Comment; It is through the application of mathematical knowledge that classical probability can be established and estimated. Practice has proved that the error estimated by this probability method is quite small, which saves manpower, material resources and financial resources.
5. Study probability with the idea of equation.
There are 36 students in a class. Now choose two to complete a task. It can also be assumed that every student is elected. If the probability that two people are of the same sex is 0.5, find the number of boys and girls in the class.
Idea: First, find out the total number of all basic events; If there are n boys and 36-n girls, find out the number of basic events of the same sex; List the equations that require solution; Check whether the value of n meets the meaning of the question.
Solution: Choose 2 people from 36 people and record the results (x, y) in the order of appearance. Since everyone is equally likely to be elected, there are 36 possibilities for X and 35 possibilities for Y, but (x, y) is the same as (y, x), so all the election results are 36ⅹ35÷2=630. According to the same calculation method, if two people are boys, there are n(n- 1)÷2 results; If two people are girls, there are (36-n)(35-n)÷2 kinds of results. Let event A be of the same sex, then the number of basic events contained in event A is n(n- 1)÷2+(36-n)(35-n)÷2. Judging from the meaning of the question:
p(A)=[n(n- 1)÷2+(36-n)(35-n)÷2]/630 = 1/2。
That is 2n-36n+ 15=0.
The solution is n= 15, or n=2 1.
After investigation, we know that all the students meet the requirements, so this class is 15 for boys and 2 1 for girls, or 2 1 for boys and 0/5 for girls.
6. Use a computer (or calculator) to estimate the probability of the event by simulating the experiment immediately.
With the popularization of computer, it has been widely used in many fields such as teaching and scientific research. We can solve the probability problem by computer simulation of random experiments.
Taking coin toss as an example, the method of generating random numbers by computer is given. Every software with statistical function has random function. Take Excel software as an example, open Excel software and perform the following steps:
1. Select cell A 1, type "=RANDBETWEEN(0, 1)" and press Enter, then the numbers in this cell will randomly generate 0 or 1.
2. Select the cell A 1, press Ctrl+C, and then select the cells of 0 and 1 to generate randomly. For example, if A2 presses Ctrl+V to A 100, then all the numbers in A2 100 are randomly generated zeros or 1, and we will soon get 65438. 、
3. Select the cell C 1, type in the frequency function "= frequency (A 1: A 100,0.5)", and press Enter, then the number in this cell ranges from a1to a100, and the ratio is. 5 the number of small numbers, that is, the frequency of 0, that is, the frequency in the opposite direction.
4. Select the cell D 1, type "=1-c1100", and then press Enter. The number in this cell is the frequency of 1 in these 100 tests, that is, the face-up frequency.
The coin toss experiment is simulated by computer. We call this method stochastic simulation.
The weather forecast says that the probability of rain every day in the next three days is 1/2. What are the chances that it will rain one of these three days?
Analysis: There are only a limited number of possible results in this experiment, and each result is equally possible. The probability of computer simulation rain is 2/5.
Solution: We solve this problem by designing a simulation experiment. Computers can generate random numbers with integer values between 0 and 9. We use 0, 1, 2, 3, 4 for rain, and 5, 6, 7, 8, 9 for no rain. This shows that the probability of rain is 1/2. Because it is three days, there are three random numbers. For example, 20 groups of random numbers are generated.
537 1 13 989 907 966 19 1 925 27 1 932 8 12
458 056 683 43 1 257 393 027 556 488 730
It is equivalent to doing 20 experiments. In this set of numbers, if there is just one number in 0, 1, 2, 3 and 4, it just rained all day. They are 537, 907, 925, 458, 056, 683, 257 and 488 respectively, that is to say, * * * has eight numbers. We get that the probability that it will rain one day in three days is about 8/20=2/5.
In short, many problems in life, such as touching the ball, housing allocation, birthdays, matching, winning lottery tickets, weather forecast, etc., can often be solved by classical probability.
References:
1 Wei Zongshu, A Course in Probability Theory and Mathematical Statistics, Beijing: Higher Education Press, 1999.
2 Compulsory Mathematics for Senior High Schools in Liu Shaoxue 3 Beijing: People's Education Press, 2005.