If the events H 1 , H 2 , …, H n form a complete group, then to calculate the probability of an arbitrary event, you can use the total probability formula:
P (A) \u003d P (A / H 1) P (H 1) + P (A / H 2) P (H 2)
In accordance with which the probability of the occurrence of the event A can be represented as the sum of the products of the conditional probabilities of the event A under the condition of the occurrence of events H i by the unconditional probabilities of these events H i . These events H i are called hypotheses.
The Bayes formula follows from the total probability formula:
The probabilities P(H i) of the hypotheses H i are called a priori probabilities - the probabilities before the experiments.
The probabilities P(A/H i) are called a posteriori probabilities - the probabilities of the hypotheses H i refined as a result of the experiment.
Service assignment. The online calculator is designed to calculate the total probability with the design of the entire course of the solution in Word format (see examples of problem solving).
Example #1. The store receives light bulbs from two factories, with the share of the first factory being 25%. It is known that the share of defects in these factories is 5% and 10%, respectively, of all manufactured products. The seller randomly takes one light bulb. What is the probability that it will be defective?
Solution: Denote by A the event - "the light bulb will be defective." The following hypotheses about the origin of this light bulb are possible: H1- "The light bulb came from the first factory." H2- "The light bulb came from the second factory." Since the share of the first plant is 25%, the probabilities of these hypotheses are respectively 
; 
.
The conditional probability that a defective light bulb was produced by the first factory is 
, the second plant - p(A/H2)=
the desired probability that the seller took a defective light bulb, we find by the total probability formula
0.25 0.05+0.75 0.10=0.0125+0.075=0.0875
Answer: p(A)= 0,0875.
Example #2. The store received two batches of the same product of the same name, equal in quantity. It is known that 25% of the first batch and 40% of the second batch is the goods of the first grade. What is the probability that a randomly selected unit of a commodity will not be of the first grade?
Solution:
Denote by A the event - "the product will be of the first grade." The following hypotheses about the origin of this product are possible: H1- "goods from the first batch." H2- “goods from the second batch”. Since the share of the first party is 25%, then the probabilities of these hypotheses are equal, respectively ; 
.
The conditional probability that the item in the first batch is 
, from the second batch - 
the desired probability that a randomly selected unit of goods will be of the first grade
p(A) \u003d P (H 1) p (A / H 1) + P (H 2) (A / H 2) \u003d 0.25 0.5+0.4 0.5=0.125+0.2=0.325
Then, the probability that a randomly selected unit of goods will not be the first grade will be equal to: 1- 0.325 = 0.675
Answer:
.
Example #3. It is known that 5% of men and 1% of women are color blind. A randomly selected person was not color blind. What is the probability that this is a man (assume that men and women are equally divided).
Solution.
Event A - a randomly selected person was not colorblind.
Find the probability of this event occurring.
P(A) = P(A|H=male) + P(A|H=female) = 0.95*0.5 + 0.99*0.5 = 0.475 + 0.495 = 0.97
Then the probability that this is a man will be: p = P(A|H=male) / P(A) = 0.475/0.97 = 0.4897
Example #4. 4 students from the first year, from the second - 6, from the third - 5 take part in the sports Olympiad. The probabilities that a student from the first, second, third year will win the Olympiad are equal to 0.9, respectively; 0.7 and 0.8.
a) Find the probability of a randomly chosen participant winning.
b) Under the conditions of this problem, one student won the Olympiad. To which group does he most likely belong?
Solution.
Event A - win at random selected participant.
Here P(H1) = 4/(4+6+5) = 0.267, P(H2) = 6/(4+6+5) = 0.4, P(H3) = 5/(4+6+5) = 0.333,
P(A|H1) = 0.9, P(A|H2) = 0.7, P(A|H3) = 0.8
a) P(A) = P(H1)*P(A|H1) + P(H2)*P(A|H2) + P(H3)*P(A|H3) = 0.267*0.9 + 0.4*0.7 + 0.333*0.8 = 0.787
b) The solution can be obtained using this calculator.
p1 = P(H1)*P(A|H1)/P(A)
p2 = P(H2)*P(A|H2)/P(A)
p3 = P(H3)*P(A|H3)/P(A)
From p1, p2, p3 choose the maximum.
Example number 5. The company has three machines of the same type. One of them gives 20% of the total production, the second - 30%, the third - 50%. At the same time, the first machine produces 5% of rejects, the second 4%, the third - 2%. Find the probability that a randomly selected unusable product is produced by the first machine.
Probability event is the ratio of the number of elementary outcomes that favor a given event to the number of all equally possible outcomes of experience in which this event may occur. The probability of an event A is denoted by P(A) (here P is the first letter of the French word probabilite - probability). According to the definition
(1.2.1)
where is the number of elementary outcomes favoring event A; - the number of all equally possible elementary outcomes of experience, forming a complete group of events.
This definition of probability is called classical. It arose at the initial stage of the development of probability theory.
The probability of an event has the following properties:
1. The probability of a certain event is equal to one. Let's designate a certain event by the letter . For a certain event, therefore
(1.2.2)
2. The probability of an impossible event is zero. We denote the impossible event by the letter . For an impossible event, therefore
(1.2.3)
3. The probability of a random event is expressed as a positive number less than one. Since the inequalities , or are satisfied for a random event, then
(1.2.4)
4. The probability of any event satisfies the inequalities
(1.2.5)
This follows from relations (1.2.2) -(1.2.4).
Example 1 An urn contains 10 balls of the same size and weight, of which 4 are red and 6 are blue. One ball is drawn from the urn. What is the probability that the drawn ball is blue?
Solution. The event "the drawn ball turned out to be blue" will be denoted by the letter A. This trial has 10 equally possible elementary outcomes, of which 6 favor event A. In accordance with formula (1.2.1), we obtain 
Example 2 All natural numbers from 1 to 30 are written on identical cards and placed in an urn. After thoroughly mixing the cards, one card is removed from the urn. What is the probability that the number on the card drawn is a multiple of 5?
Solution. Denote by A the event "the number on the taken card is a multiple of 5". In this test, there are 30 equally possible elementary outcomes, of which 6 outcomes favor event A (numbers 5, 10, 15, 20, 25, 30). Hence, 
Example 3 Two dice are thrown, the sum of points on the upper faces is calculated. Find the probability of the event B, consisting in the fact that the top faces of the cubes will have a total of 9 points.
Solution. There are 6 2 = 36 equally possible elementary outcomes in this trial. Event B is favored by 4 outcomes: (3;6), (4;5), (5;4), (6;3), so 
Example 4. A natural number not exceeding 10 is chosen at random. What is the probability that this number is prime?
Solution. Denote by the letter C the event "the chosen number is prime". In this case, n = 10, m = 4 (primes 2, 3, 5, 7). Therefore, the desired probability 
Example 5 Two symmetrical coins are tossed. What is the probability that both coins have digits on the top sides?
Solution. Let's denote by the letter D the event "there was a number on the top side of each coin". There are 4 equally possible elementary outcomes in this test: (G, G), (G, C), (C, G), (C, C). (The notation (G, C) means that on the first coin there is a coat of arms, on the second - a number). Event D is favored by one elementary outcome (C, C). Since m = 1, n = 4, then 
Example 6 What is the probability that the digits in a randomly chosen two-digit number are the same?
Solution. Two-digit numbers are numbers from 10 to 99; there are 90 such numbers in total. 9 numbers have the same digits (these are the numbers 11, 22, 33, 44, 55, 66, 77, 88, 99). Since in this case m = 9, n = 90, then 
,
where A is the "number with the same digits" event.
Example 7 From the letters of the word differential one letter is chosen at random. What is the probability that this letter will be: a) a vowel b) a consonant c) a letter h?
Solution. There are 12 letters in the word differential, of which 5 are vowels and 7 are consonants. Letters h this word does not. Let's denote the events: A - "vowel", B - "consonant", C - "letter h". The number of favorable elementary outcomes: - for event A, - for event B, - for event C. Since n \u003d 12, then
, And .
Example 8 Two dice are tossed, the number of points on the top face of each dice is noted. Find the probability that both dice have the same number of points.
Solution. Let us denote this event by the letter A. Event A is favored by 6 elementary outcomes: (1;]), (2;2), (3;3), (4;4), (5;5), (6;6). In total there are equally possible elementary outcomes that form a complete group of events, in this case n=6 2 =36. So the desired probability 
Example 9 The book has 300 pages. What is the probability that a randomly opened page will have a sequence number that is a multiple of 5?
Solution. It follows from the conditions of the problem that there will be n = 300 of all equally possible elementary outcomes that form a complete group of events. Of these, m = 60 favor the occurrence of the specified event. Indeed, a number that is a multiple of 5 has the form 5k, where k is a natural number, and , whence 
. Hence, 
, where A - the "page" event has a sequence number that is a multiple of 5".
Example 10. Two dice are thrown, the sum of points on the upper faces is calculated. What is more likely to get a total of 7 or 8?
Solution. Let's designate the events: A - "7 points fell out", B - "8 points fell out". Event A is favored by 6 elementary outcomes: (1; 6), (2; 5), (3; 4), (4; 3), (5; 2), (6; 1), and event B - by 5 outcomes: (2; 6), (3; 5), (4; 4), (5; 3), (6; 2). There are n = 6 2 = 36 of all equally possible elementary outcomes. Hence, And .
So, P(A)>P(B), that is, getting a total of 7 points is a more likely event than getting a total of 8 points.
Tasks
1. A natural number not exceeding 30 is chosen at random. What is the probability that this number is a multiple of 3?
2. In the urn a red and b blue balls of the same size and weight. What is the probability that a randomly drawn ball from this urn is blue?
3. A number not exceeding 30 is chosen at random. What is the probability that this number is a divisor of zo?
4. In the urn A blue and b red balls of the same size and weight. One ball is drawn from this urn and set aside. This ball is red. Then another ball is drawn from the urn. Find the probability that the second ball is also red.
5. A natural number not exceeding 50 is chosen at random. What is the probability that this number is prime?
6. Three dice are thrown, the sum of points on the upper faces is calculated. What is more likely - to get a total of 9 or 10 points?
7. Three dice are tossed, the sum of the dropped points is calculated. What is more likely to get a total of 11 (event A) or 12 points (event B)?
Answers
1. 1/3. 2 . b/(a+b). 3 . 0,2. 4 . (b-1)/(a+b-1). 5 .0,3.6 . p 1 \u003d 25/216 - the probability of getting 9 points in total; p 2 \u003d 27/216 - the probability of getting 10 points in total; p2 > p1 7 . P(A) = 27/216, P(B) = 25/216, P(A) > P(B).
Questions
1. What is called the probability of an event?
2. What is the probability of a certain event?
3. What is the probability of an impossible event?
4. What are the limits of the probability of a random event?
5. What are the limits of the probability of any event?
6. What definition of probability is called classical?
Let's not think about the lofty for a long time - let's start right away with a definition.
The Bernoulli scheme is when n independent experiments of the same type are performed, in each of which an event of interest to us A may appear, and the probability of this event is known P (A) \u003d p. It is required to determine the probability that event A will occur exactly k times during n trials.
The tasks that are solved according to the Bernoulli scheme are extremely diverse: from simple ones (such as “find the probability that the shooter hits 1 time out of 10”) to very severe ones (for example, tasks for percentages or playing cards). In reality, this scheme is often used to solve problems related to product quality control and the reliability of various mechanisms, all the characteristics of which must be known before starting work.
Let's go back to the definition. Since we are talking about independent trials, and in each trial the probability of the event A is the same, only two outcomes are possible:
- A is the occurrence of event A with probability p;
- "not A" - event A did not appear, which happens with probability q = 1 − p.
The most important condition without which the Bernoulli scheme loses its meaning is constancy. No matter how many experiments we conduct, we are interested in the same event A that occurs with the same probability p.
Incidentally, not all problems in probability theory can be reduced to constant conditions. Any competent tutor in higher mathematics will tell you about this. Even something as simple as picking colored balls out of a box is not an experiment with constant conditions. They took out another ball - the ratio of colors in the box changed. Therefore, the probabilities have also changed.
If the conditions are constant, one can accurately determine the probability that event A will occur exactly k times out of n possible. We formulate this fact in the form of a theorem:
Bernoulli's theorem. Let the probability of occurrence of event A in each experiment be constant and equal to p. Then the probability that in n independent trials event A will appear exactly k times is calculated by the formula:
where C n k is the number of combinations, q = 1 − p.
This formula is called the Bernoulli formula. It is interesting to note that the problems below are completely solved without using this formula. For example, you can apply probability addition formulas. However, the amount of computation will be simply unrealistic.
Task. The probability of producing a defective product on the machine is 0.2. Determine the probability that in a batch of ten parts produced on a given machine exactly k will be without defects. Solve the problem for k = 0, 1, 10.
By assumption, we are interested in the event A of the release of products without defects, which happens each time with a probability p = 1 − 0.2 = 0.8. We need to determine the probability that this event will occur k times. Event A is opposed to the event “not A”, i.e. production of a defective product.
Thus, we have: n = 10; p=0.8; q = 0.2.
So, we find the probability that all parts in the batch are defective (k = 0), that only one part is defective (k = 1), and that there are no defective parts at all (k = 10):
Task. The coin is tossed 6 times. The loss of the coat of arms and tails is equally probable. Find the probability that:
- the coat of arms will drop three times;
- the coat of arms will drop once;
- the coat of arms will appear at least twice.
So, we are interested in the event A when the coat of arms falls. The probability of this event is p = 0.5. The event A is opposed to the event “not A”, when tails come up, which happens with the probability q = 1 − 0.5 = 0.5. It is necessary to determine the probability that the coat of arms will fall out k times.
Thus, we have: n = 6; p = 0.5; q = 0.5.
Let us determine the probability that the coat of arms fell out three times, i.e. k = 3:
Now let's determine the probability that the coat of arms fell out only once, i.e. k = 1:
It remains to determine with what probability the coat of arms will fall out at least twice. The main snag is in the phrase “no less”. It turns out that any k will suit us, except for 0 and 1, i.e. you need to find the value of the sum X \u003d P 6 (2) + P 6 (3) + ... + P 6 (6).
Note that this sum is also equal to (1 − P 6 (0) − P 6 (1)), i.e. out of all possible options, it is enough to “cut out” those when the coat of arms fell out 1 time (k = 1) or did not fall out at all (k = 0). Since P 6 (1) we already know, it remains to find P 6 (0):
Task. The probability that a TV has hidden defects is 0.2. The warehouse received 20 TVs. Which event is more likely: that there are two TVs with hidden defects in this batch or three?
Event of interest A is the presence of a latent defect. Total TVs n = 20, the probability of a hidden defect p = 0.2. Accordingly, the probability of getting a TV set without a hidden defect is q = 1 − 0.2 = 0.8.
We get the starting conditions for the Bernoulli scheme: n = 20; p = 0.2; q = 0.8.
Let's find the probability of getting two "defective" TVs (k = 2) and three (k = 3):
\[\begin(array)(l)(P_(20))\left(2 \right) = C_(20)^2(p^2)(q^(18)) = \frac((20}{{2!18!}} \cdot {0,2^2} \cdot {0,8^{18}} \approx 0,137\\{P_{20}}\left(3 \right) = C_{20}^3{p^3}{q^{17}} = \frac{{20!}}{{3!17!}} \cdot {0,2^3} \cdot {0,8^{17}} \approx 0,41\end{array}\]!}
Obviously, P 20 (3) > P 20 (2), i.e. the probability of getting three TVs with hidden defects is more likely to get only two such TVs. Moreover, the difference is not weak.
A small note about factorials. Many people experience a vague feeling of discomfort when they see the entry "0!" (read "zero factorial"). So, 0! = 1 by definition.
P. S. And the biggest probability in the last task is to get four TVs with hidden defects. Do the math and see for yourself.
Whether we like it or not, our life is full of all kinds of accidents, both pleasant and not very. Therefore, each of us would do well to know how to find the probability of an event. This will help you make the right decisions under any circumstances that are associated with uncertainty. For example, such knowledge will be very useful when choosing investment options, evaluating the possibility of winning a stock or lottery, determining the reality of achieving personal goals, etc., etc.
Probability Formula
In principle, the study of this topic does not take too much time. In order to get an answer to the question: "How to find the probability of a phenomenon?", you need to understand the key concepts and remember the basic principles on which the calculation is based. So, according to statistics, the events under study are denoted by A1, A2,..., An. Each of them has both favorable outcomes (m) and the total number of elementary outcomes. For example, we are interested in how to find the probability that an even number of points will be on the top face of the cube. Then A is roll m - rolling 2, 4, or 6 (three favorable choices), and n is all six possible choices.
The calculation formula itself is as follows:
With one outcome, everything is extremely easy. But how to find the probability if the events go one after the other? Consider this example: one card is shown from a deck of cards (36 pieces), then it is hidden back into the deck, and after shuffling, the next one is pulled out. How to find the probability that at least in one case the Queen of Spades was drawn? There is the following rule: if a complex event is considered, which can be divided into several incompatible simple events, then you can first calculate the result for each of them, and then add them together. In our case, it will look like this: 1/36 + 1/36 = 1/18. But what about when several occur at the same time? Then we multiply the results! For example, the probability that when two coins are tossed at the same time, two tails will fall out will be equal to: ½ * ½ = 0.25.
Now let's take an even more complex example. Suppose we enter a book lottery in which ten out of thirty tickets are winning. It is required to determine:
- The probability that both will win.
- At least one of them will bring a prize.
- Both will be losers.
So let's consider the first case. It can be broken down into two events: the first ticket will be lucky, and the second one will also be lucky. Let's take into account that the events are dependent, since after each pulling out the total number of options decreases. We get:
10 / 30 * 9 / 29 = 0,1034.
In the second case, you need to determine the probability of a losing ticket and take into account that it can be both the first in a row and the second one: 10 / 30 * 20 / 29 + 20 / 29 * 10 / 30 = 0.4598.
Finally, the third case, when even one book cannot be obtained from the lottery: 20 / 30 * 19 / 29 = 0.4368.
"Accidents are not accidental"... It sounds like a philosopher said, but in fact, studying accidents is the lot of the great science of mathematics. In mathematics, chance is the theory of probability. Formulas and examples of tasks, as well as the main definitions of this science will be presented in the article.
What is Probability Theory?
Probability theory is one of the mathematical disciplines that studies random events.
To make it a little clearer, let's give a small example: if you toss a coin up, it can fall heads or tails. As long as the coin is in the air, both of these possibilities are possible. That is, the probability of possible consequences correlates 1:1. If one is drawn from a deck with 36 cards, then the probability will be indicated as 1:36. It would seem that there is nothing to explore and predict, especially with the help of mathematical formulas. Nevertheless, if you repeat a certain action many times, then you can identify a certain pattern and, on its basis, predict the outcome of events in other conditions.
To summarize all of the above, the theory of probability in the classical sense studies the possibility of the occurrence of one of the possible events in a numerical sense.
From the pages of history
The theory of probability, formulas and examples of the first tasks appeared in the distant Middle Ages, when attempts to predict the outcome of card games first arose.
Initially, the theory of probability had nothing to do with mathematics. It was justified by empirical facts or properties of an event that could be reproduced in practice. The first works in this area as a mathematical discipline appeared in the 17th century. The founders were Blaise Pascal and Pierre Fermat. For a long time they studied gambling and saw certain patterns, which they decided to tell the public about.
The same technique was invented by Christian Huygens, although he was not familiar with the results of the research of Pascal and Fermat. The concept of "probability theory", formulas and examples, which are considered the first in the history of the discipline, were introduced by him.
Of no small importance are the works of Jacob Bernoulli, Laplace's and Poisson's theorems. They made probability theory more like a mathematical discipline. Probability theory, formulas and examples of basic tasks got their present form thanks to Kolmogorov's axioms. As a result of all the changes, the theory of probability has become one of the mathematical branches.
Basic concepts of probability theory. Events
The main concept of this discipline is "event". Events are of three types:
- Reliable. Those that will happen anyway (the coin will fall).
- Impossible. Events that will not happen in any scenario (the coin will remain hanging in the air).
- Random. The ones that will or won't happen. They can be influenced by various factors that are very difficult to predict. If we talk about a coin, then random factors that can affect the result: the physical characteristics of the coin, its shape, initial position, throw force, etc.
All events in the examples are denoted by capital Latin letters, with the exception of R, which has a different role. For example:
- A = "students came to the lecture."
- Ā = "students didn't come to the lecture".
In practical tasks, events are usually recorded in words.
One of the most important characteristics of events is their equal possibility. That is, if you toss a coin, all variants of the initial fall are possible until it falls. But events are also not equally probable. This happens when someone deliberately influences the outcome. For example, "marked" playing cards or dice, in which the center of gravity is shifted.
Events are also compatible and incompatible. Compatible events do not exclude the occurrence of each other. For example:
- A = "the student came to the lecture."
- B = "the student came to the lecture."
These events are independent of each other, and the appearance of one of them does not affect the appearance of the other. Incompatible events are defined by the fact that the occurrence of one precludes the occurrence of the other. If we talk about the same coin, then the loss of "tails" makes it impossible for the appearance of "heads" in the same experiment.
Actions on events
Events can be multiplied and added, respectively, logical connectives "AND" and "OR" are introduced in the discipline.
The amount is determined by the fact that either event A, or B, or both can occur at the same time. In the case when they are incompatible, the last option is impossible, either A or B will drop out.
The multiplication of events consists in the appearance of A and B at the same time.
Now you can give a few examples to better remember the basics, probability theory and formulas. Examples of problem solving below.
Exercise 1: The firm is bidding for contracts for three types of work. Possible events that may occur:
- A = "the firm will receive the first contract."
- A 1 = "the firm will not receive the first contract."
- B = "the firm will receive a second contract."
- B 1 = "the firm will not receive a second contract"
- C = "the firm will receive a third contract."
- C 1 = "the firm will not receive a third contract."
Let's try to express the following situations using actions on events:
- K = "the firm will receive all contracts."
In mathematical form, the equation will look like this: K = ABC.
- M = "the firm will not receive a single contract."
M \u003d A 1 B 1 C 1.
We complicate the task: H = "the firm will receive one contract." Since it is not known which contract the firm will receive (the first, second or third), it is necessary to record the entire range of possible events:
H \u003d A 1 BC 1 υ AB 1 C 1 υ A 1 B 1 C.
And 1 BC 1 is a series of events where the firm does not receive the first and third contract, but receives the second one. Other possible events are also recorded by the corresponding method. The symbol υ in the discipline denotes a bunch of "OR". If we translate the above example into human language, then the company will receive either the third contract, or the second, or the first. Similarly, you can write other conditions in the discipline "Probability Theory". The formulas and examples of solving problems presented above will help you do it yourself.
Actually, the probability
Perhaps, in this mathematical discipline, the probability of an event is a central concept. There are 3 definitions of probability:
- classical;
- statistical;
- geometric.
Each has its place in the study of probabilities. Probability theory, formulas and examples (Grade 9) mostly use the classic definition, which sounds like this:
- The probability of situation A is equal to the ratio of the number of outcomes that favor its occurrence to the number of all possible outcomes.
The formula looks like this: P (A) \u003d m / n.
And, actually, an event. If the opposite of A occurs, it can be written as Ā or A 1 .
m is the number of possible favorable cases.
n - all events that can happen.
For example, A \u003d "pull out a heart suit card." There are 36 cards in a standard deck, 9 of them are of hearts. Accordingly, the formula for solving the problem will look like:
P(A)=9/36=0.25.
As a result, the probability that a heart-suited card will be drawn from the deck will be 0.25.
to higher mathematics
Now it has become a little known what the theory of probability is, formulas and examples of solving tasks that come across in the school curriculum. However, the theory of probability is also found in higher mathematics, which is taught in universities. Most often, they operate with geometric and statistical definitions of the theory and complex formulas.
The theory of probability is very interesting. Formulas and examples (higher mathematics) are better to start learning from a small one - from a statistical (or frequency) definition of probability.
The statistical approach does not contradict the classical one, but slightly expands it. If in the first case it was necessary to determine with what degree of probability an event will occur, then in this method it is necessary to indicate how often it will occur. Here a new concept of “relative frequency” is introduced, which can be denoted by W n (A). The formula is no different from the classic:
If the classical formula is calculated for forecasting, then the statistical one is calculated according to the results of the experiment. Take, for example, a small task.
The department of technological control checks products for quality. Among 100 products, 3 were found to be of poor quality. How to find the frequency probability of a quality product?
A = "the appearance of a quality product."
W n (A)=97/100=0.97
Thus, the frequency of a quality product is 0.97. Where did you get 97 from? Of the 100 products that were checked, 3 turned out to be of poor quality. We subtract 3 from 100, we get 97, this is the quantity of a quality product.
A bit about combinatorics
Another method of probability theory is called combinatorics. Its basic principle is that if a certain choice A can be made in m different ways, and a choice B in n different ways, then the choice of A and B can be made by multiplying.
For example, there are 5 roads from city A to city B. There are 4 routes from city B to city C. How many ways are there to get from city A to city C?
It's simple: 5x4 = 20, that is, there are twenty different ways to get from point A to point C.
Let's make the task harder. How many ways are there to play cards in solitaire? In a deck of 36 cards, this is the starting point. To find out the number of ways, you need to “subtract” one card from the starting point and multiply.
That is, 36x35x34x33x32…x2x1= the result does not fit on the calculator screen, so it can simply be denoted as 36!. Sign "!" next to the number indicates that the entire series of numbers is multiplied among themselves.
In combinatorics, there are such concepts as permutation, placement and combination. Each of them has its own formula.
An ordered set of set elements is called a layout. Placements can be repetitive, meaning one element can be used multiple times. And without repetition, when the elements are not repeated. n is all elements, m is the elements that participate in the placement. The formula for placement without repetitions will look like:
A n m =n!/(n-m)!
Connections of n elements that differ only in the order of placement are called permutations. In mathematics, this looks like: P n = n!
Combinations of n elements by m are such compounds in which it is important which elements they were and what is their total number. The formula will look like:
A n m =n!/m!(n-m)!
Bernoulli formula
In the theory of probability, as well as in every discipline, there are works of outstanding researchers in their field who have taken it to a new level. One of these works is the Bernoulli formula, which allows you to determine the probability of a certain event occurring under independent conditions. This suggests that the appearance of A in an experiment does not depend on the appearance or non-occurrence of the same event in previous or subsequent tests.
Bernoulli equation:
P n (m) = C n m ×p m ×q n-m .
The probability (p) of the occurrence of the event (A) is unchanged for each trial. The probability that the situation will happen exactly m times in n number of experiments will be calculated by the formula that is presented above. Accordingly, the question arises of how to find out the number q.
If event A occurs p number of times, accordingly, it may not occur. A unit is a number that is used to designate all outcomes of a situation in a discipline. Therefore, q is a number that indicates the possibility of the event not occurring.
Now you know the Bernoulli formula (probability theory). Examples of problem solving (the first level) will be considered below.
Task 2: A store visitor will make a purchase with a probability of 0.2. 6 visitors entered the store independently. What is the probability that a visitor will make a purchase?
Solution: Since it is not known how many visitors should make a purchase, one or all six, it is necessary to calculate all possible probabilities using the Bernoulli formula.
A = "the visitor will make a purchase."
In this case: p = 0.2 (as indicated in the task). Accordingly, q=1-0.2 = 0.8.
n = 6 (because there are 6 customers in the store). The number m will change from 0 (no customer will make a purchase) to 6 (all store visitors will purchase something). As a result, we get the solution:
P 6 (0) \u003d C 0 6 × p 0 × q 6 \u003d q 6 \u003d (0.8) 6 \u003d 0.2621.
None of the buyers will make a purchase with a probability of 0.2621.
How else is the Bernoulli formula (probability theory) used? Examples of problem solving (second level) below.
After the above example, questions arise about where C and p have gone. With respect to p, a number to the power of 0 will be equal to one. As for C, it can be found by the formula:
C n m = n! /m!(n-m)!
Since in the first example m = 0, respectively, C=1, which in principle does not affect the result. Using the new formula, let's try to find out what is the probability of buying goods by two visitors.
P 6 (2) = C 6 2 ×p 2 ×q 4 = (6×5×4×3×2×1) / (2×1×4×3×2×1) × (0.2) 2 × (0.8) 4 = 15 × 0.04 × 0.4096 = 0.246.
The theory of probability is not so complicated. The Bernoulli formula, examples of which are presented above, is a direct proof of this.
Poisson formula
The Poisson equation is used to calculate unlikely random situations.
Basic formula:
P n (m)=λ m /m! × e (-λ) .
In this case, λ = n x p. Here is such a simple Poisson formula (probability theory). Examples of problem solving will be considered below.
Task 3 A: The factory produced 100,000 parts. The appearance of a defective part = 0.0001. What is the probability that there will be 5 defective parts in a batch?
As you can see, marriage is an unlikely event, and therefore the Poisson formula (probability theory) is used for calculation. Examples of solving problems of this kind are no different from other tasks of the discipline, we substitute the necessary data into the above formula:
A = "a randomly selected part will be defective."
p = 0.0001 (according to the assignment condition).
n = 100000 (number of parts).
m = 5 (defective parts). We substitute the data in the formula and get:
R 100000 (5) = 10 5 / 5! X e -10 = 0.0375.
Just like the Bernoulli formula (probability theory), examples of solutions using which are written above, the Poisson equation has an unknown e. In essence, it can be found by the formula:
e -λ = lim n ->∞ (1-λ/n) n .
However, there are special tables that contain almost all the values of e.
De Moivre-Laplace theorem
If in the Bernoulli scheme the number of trials is sufficiently large, and the probability of the occurrence of event A in all schemes is the same, then the probability of the occurrence of event A a certain number of times in a series of trials can be found by the Laplace formula:
Р n (m)= 1/√npq x ϕ(X m).
Xm = m-np/√npq.
To better remember the Laplace formula (probability theory), examples of tasks to help below.
First we find X m , we substitute the data (they are all indicated above) into the formula and get 0.025. Using tables, we find the number ϕ (0.025), the value of which is 0.3988. Now you can substitute all the data in the formula:
P 800 (267) \u003d 1 / √ (800 x 1/3 x 2/3) x 0.3988 \u003d 3/40 x 0.3988 \u003d 0.03.
So the probability that the flyer will hit exactly 267 times is 0.03.
Bayes formula
The Bayes formula (probability theory), examples of solving tasks with the help of which will be given below, is an equation that describes the probability of an event, based on the circumstances that could be associated with it. The main formula is as follows:
P (A|B) = P (B|A) x P (A) / P (B).
A and B are definite events.
P(A|B) - conditional probability, that is, event A can occur, provided that event B is true.
Р (В|А) - conditional probability of event В.
So, the final part of the short course "Theory of Probability" is the Bayes formula, examples of solving problems with which are below.
Task 5: Phones from three companies were brought to the warehouse. At the same time, part of the phones that are manufactured at the first plant is 25%, at the second - 60%, at the third - 15%. It is also known that the average percentage of defective products at the first factory is 2%, at the second - 4%, and at the third - 1%. It is necessary to find the probability that a randomly selected phone will be defective.
A = "randomly taken phone."
B 1 - the phone that the first factory made. Accordingly, introductory B 2 and B 3 will appear (for the second and third factories).
As a result, we get:
P (B 1) \u003d 25% / 100% \u003d 0.25; P (B 2) \u003d 0.6; P (B 3) \u003d 0.15 - so we found the probability of each option.
Now you need to find the conditional probabilities of the desired event, that is, the probability of defective products in firms:
P (A / B 1) \u003d 2% / 100% \u003d 0.02;
P (A / B 2) \u003d 0.04;
P (A / B 3) \u003d 0.01.
Now we substitute the data into the Bayes formula and get:
P (A) \u003d 0.25 x 0.2 + 0.6 x 0.4 + 0.15 x 0.01 \u003d 0.0305.
The article presents the theory of probability, formulas and examples of problem solving, but this is only the tip of the iceberg of a vast discipline. And after all that has been written, it will be logical to ask the question of whether the theory of probability is needed in life. It is difficult for a simple person to answer, it is better to ask someone who has hit the jackpot more than once with her help.
