What is the probability. The meaning of the word "probability. From the pages of history

Probability is a very easy topic if you focus on the meaning of the problems, and not on the formulas. But how to solve probability problems. First, what is a probability? This is the chance that some event will happen. If we say that the probability of some event is 50%, what does this mean? That it will either happen or not happen is one of two things. Thus, calculating the probability value is very simple - you need to take the number of options suitable for us and divide by the number of all possible options. For example, the chance of getting tails on a coin toss is ½. How do we get ½? In total, we have two possible options (heads and tails), of which one suits us (tails), so we get the probability ½.

As we have already seen, the probability can be expressed both in percentage and in ordinary numbers. Important: on the exam you will need to write down the answer in numbers, not in percentages. It is assumed that the probability varies from 0 (never happens) to 1 (absolutely will happen). You can also say that always

Probability of Suitable Events + Probability of Unsuitable Events = 1

Now we understand exactly how to calculate the probability of a single event, and even such tasks are in the FIPI bank, but it is clear that this does not end there. To make life more fun, in probability problems there are usually at least two events, and you need to calculate the probability taking into account each of them.

We calculate the probability of each event separately, then put signs between the fractions:

1. If you need the first AND second event, then multiply.

2. If you need the first OR second event, then add up.

Problems and solutions to problems on probability

Task 1. Among the natural numbers from 23 to 37, one number is randomly chosen. Find the probability that it is not divisible by 5.

Decision:

Probability is the ratio of favorable options to their total number.

There are 15 numbers in this interval. Of these, only 3 are divisible by 5, so 12 is not divisible.

Probability then:

Answer: 0.8.

Task 2. Two students in the class are randomly selected to serve in the dining room. What is the probability that two boys will be on duty if there are 7 boys and 8 girls in the class?

Decision: Probability is the ratio of favorable options to their total number. In the class of 7 boys, these are favorable options. And only 15 students.

Probability that the first duty boy:

Probability that the second duty boy:

Since both must be boys, we multiply the probabilities:

Answer: 0.2.

Task 3. There are 12 seats on board the aircraft next to the emergency exits and 18 seats behind the partitions separating the cabins. The rest of the seats are inconvenient for a tall passenger. Passenger V. is tall. Find the probability that at check-in, with a random choice of seat, passenger B. will get a comfortable seat if there are 300 seats on the plane.

Decision: Passenger B. is comfortable with 30 seats (12 + 18 = 30), and there are 300 seats in the plane. Therefore, the probability that passenger B will get a comfortable seat is 30/300, i.e. 0.1.

Task 4. There are only 25 tickets in the collection of tickets in mathematics, 10 of them contain a question on inequalities.

Find the probability that a student will not get a question on inequalities in a ticket randomly chosen at the exam.

Decision: Of the 25 tickets, 15 do not contain a question on inequalities, so the probability that in a randomly selected ticket a student will not get a question on inequalities is 15/25, i.e. 0.6.

Task 5. There are only 35 tickets in the collection of chemistry tickets, 7 of them contain a question on acids.

Find the probability that a student will not get a question on acids in a ticket randomly selected in the exam.

Decision: Out of 35 tickets, 28 do not contain a question on acids, so the probability that a student does not get a question on acids in a ticket randomly selected at the exam is 28/35, i.e. 0.8.

Task 6. On average, out of 500 garden pumps sold, 2 leak. Find the probability that one randomly selected pump does not leak.

Decision: If out of 500 pumps 2 are leaking, then 498 are not leaking. Therefore, the probability of choosing a good pump is 498/500, i.e. 0.996.

Task 7. The probability that a new vacuum cleaner will be repaired within a year is 0.065. In a certain city, out of 1000 vacuum cleaners sold during the year, 70 pieces arrived at the warranty workshop.

How different is the frequency of the “warranty repair” event from its probability in this city?

Decision: The frequency of the "warranty repair" event is 70/1000, i.e. 0.07. It differs from the predicted probability by 0.005 (0.07 - 0.065 = 0.005).

Task 8. 50 athletes participate in the gymnastics championship: 18 from Russia, 14 from Ukraine, the rest from Belarus. The order in which the gymnasts perform is determined by lot.

Find the probability that the athlete who performs first will be from Belarus.

Decision: There are 50 participants in the championship, and 18 athletes from Belarus (50 - 18 - 14 = 18).

The probability that an athlete from Belarus will be the first to perform is 18 out of 50, i.e. 18/50, or 0.36.

Task 9. The scientific conference is held in 5 days. A total of 80 reports are planned - the first three days, 12 reports each, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by a draw.

What is the probability that Professor M.'s report will be scheduled for the last day of the conference?

Decision: During the first three days, 36 reports will be read (12 ∙ 3 = 36), 44 reports are planned for the last two days. Therefore, 22 reports are scheduled for the last day (44: 2 = 22). This means that the probability that the report of Professor M. will be scheduled for the last day of the conference is 22/80, i.e. 0.275.

Task 10.

Before the start of the first round of the chess championship, the participants are randomly divided into game pairs by drawing lots. In total, 26 chess players participate in the championship, including 14 participants from Russia, including Yegor Kosov.

Find the probability that in the first round Yegor Kosov will play with any chess player from Russia?

Decision: In the first round, Egor Kosov can play with 25 chess players (26 - 1 = 25), of which 13 are from Russia. This means that the probability that in the first round Egor Kosov will play with any chess player from Russia is 13/25, or 0.52.

Task 11.

16 teams participate in the World Championship. By drawing lots, they must be divided into four groups of four teams each. Mixed in the box are cards with group numbers: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4.

Team captains draw one card each. What is the probability that the Russian team will be in the second group?

Decision: The probability that the Russian team will be in the second group is equal to the ratio of the number of cards with number 2 to the total number of cards, i.e. 4/16, or 0.25.

Task 12. There are 5 people in a group of tourists. With the help of lots, they choose two people who must go to the village for food. Tourist A. would like to go to the store, but he submits to the lot. What is the probability that A will go to the store?

Decision: Choose two tourists out of five. Therefore, the probability of being chosen is 2/5, i.e. 0.4.

Task 13. There are 30 people in a group of tourists. They are thrown by helicopter in several steps into a remote area, 6 people per flight. The order in which the helicopter transports tourists is random. Find the probability that tourist P. will take the first helicopter flight.

Decision: There are 6 seats on the first flight, a total of 30 seats. Then the probability that the tourist will fly on the first helicopter flight is 6/30, or 0.2.

Task 14. What is the probability that a randomly chosen natural number from 10 to 19 is divisible by 3?

Decision: There are ten natural numbers from 10 to 19, of which three numbers are divisible by 3: 12, 15 and 18. Therefore, the required probability is 3/10, i.e. 0.3.

Probability of multiple events

Task 1. Before the start of a volleyball match, the team captains draw fair lots to determine which team will start the ball game. The "Starter" team takes turns playing with the "Rotor", "Motor" and "Strator" teams. Find the probability that the "Starter" will only start the second game.

Decision:

The following option will suit us: Stator does not start the first game, starts the second game, does not start the third game. The probability of such a development of events is equal to the product of the probabilities of each of these events. The probability of each of them is 0.5, therefore: 0.5 0.5 0.5 = 0.125.

Task 2. To advance to the next round of the competition, a football team needs to score at least 4 points in two games. If a team wins, it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Decision:

Question type: combination of events.

The probability of the origin of any of these 3 options is equal to the sum of the probabilities of each of the options: 0.08 + 0.08 + 0.16 = 0.32.

Task 3. There are 21 students in the class. Among them are two friends: Anya and Nina. The class is randomly divided into 7 groups of 3 people each. Find the probability that Anya and Nina are in the same group.

Decision:

Question Type: Reduce Groups.

The probability of Anya falling into one of the groups is 1. The probability of Nina falling into the same group is 2 out of 20 (2 remaining places in the group, and 20 people left). 2/20 = 1/10 = 0.1.

Task 4. Petya had 4 ruble coins and 2 two ruble coins in his pocket. Petya, without looking, shifted some 3 coins into another pocket. Find the probability that both two-ruble coins are in the same pocket.

Decision:

Method number 1

Task type: group reduction.

Imagine that six coins are divided into two groups of three coins. The probability that the first one-ruble coin will fall into one of the pockets (groups) = 1.

The probability that two two-ruble coins will fall into the same pocket = the number of remaining places in this pocket / the number of remaining places in both pockets = 2/5 = 0.4.

Method number 2

Question type: combination of events.

The task is performed in several ways:

If Petya transferred three out of four ruble coins to another pocket (he did not transfer two-ruble coins), or if he transferred both two-ruble coins and one ruble coin to another pocket in one of three ways: 1, 2, 2; 2, 1, 2; 2, 2, 1. You can depict this in a diagram (Peter puts it in pocket 2, so we will calculate the probabilities in the “pocket 2” column):

Task 5. Petya had 2 coins of 5 rubles and 4 coins of 10 rubles in his pocket. Petya, without looking, shifted some 3 coins into another pocket. Find the probability that five-ruble coins are now in different pockets.

Decision:

Task type: group reduction.

Method number 1

Imagine that six coins are divided into two groups of three coins. The probability that the first two-ruble coin will fall into one of the pockets (groups) = 1. The probability that the second coin will fall into another pocket = the number of remaining places in the other / the number of remaining places in both pockets = 3/5 = 0.6.

Method number 2

Question type: combination of events.

The task is performed by several options:

In order for five-ruble coins to end up in different pockets, Petya must take one five-ruble and two ten-ruble coins from his pocket. This can be done in three ways: 5, 10, 10; 10, 5, 10 or 10, 10, 5. You can depict this in a diagram (Peter puts it in pocket 2, so we will calculate the probabilities in the “pocket 2” column):

The probability of the occurrence of any of these 4 options is equal to the sum of the probabilities of each of the options:

Task 6. In a random experiment, a symmetrical coin is tossed three times. Find the probability that it comes up heads exactly twice.

Decision: Question type: finding the desired and the actual \ combining events We are satisfied with three options:

Eagle - tails - eagle;

Eagle - eagle - tails;

Tails - eagle - eagle;

The probability of each case is 1/2, and of each option is 1/8 (1/2 ∙ 1/2 ∙ 1/2 = 1/8)

We will be satisfied with either the first, or the second, or the third option. Therefore, we add their probabilities and get 3/8 (1/8 + 1/8 + 1/8 = 3/8), i.e. 0.375.

Task 7. If grandmaster A. plays white, then he wins grandmaster B. with a probability of 0.5. If A. plays black, then A. beats B. with a probability of 0.34. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.

Decision:

Question type: combination of events.

In any case, A. will play both white and black, so we will be satisfied with the option when grandmaster A. wins playing white (probability 0.5) and also playing black (probability 0.34). Therefore, it is necessary to multiply the probabilities of these two events: 0.5 ∙ 0.34 = 0.17.

Task 8. The probability that the battery is defective is 0.02. The customer in the store selects a random package containing two of these batteries. Find the probability that both batteries are good.

Decision:

Question type: combination of events.

The probability that the battery is good is 0.98. The buyer needs both the first and second batteries to be in good order: 0.98 0.98 = 0.9604.

Task 9. Groups perform at the rock festival - one from each of the declared countries. The order of performance is determined by lot. What is the probability that a band from the USA will perform after a band from Canada and after a band from China? Round the result to the nearest hundredth.

Decision:

Question type: combination of events.

The total number of groups performing at the festival does not matter to answer the question. No matter how many there are, for these countries there are 6 ways of mutual arrangement among the speakers (KIT - China, CAN = Canada):

…USA, CAN, CHINA…

…USA, CHINA, CAN…

… KIT, USA, CAN…

... CAN, USA, CHINA ...

... KAN, KIT, USA ...

…KIT, CAN, USA…

The US is behind China and Canada in the last two cases. Therefore, the probability that the groups will be randomly distributed in this way is equal to:

Complementary Probability

Task 1.

The automatic line makes batteries. The probability that a finished battery is defective is 0.02. Before packaging, each battery goes through a control system. The probability that the system will reject a bad battery is 0.97. The probability that the system will mistakenly reject a good battery is 0.05.

Find the probability that a randomly selected battery will be rejected.

Decision:

There are 2 options that suit us:

Option A: the battery is rejected, it is defective;

Option B: the battery is rejected, it is working.

Probability of option A: 0.02 ∙ 0.97 = 0.0194;

Probability of option B: 0.05 ∙ 0.98 = 0.049;

Either the first or the second option will suit us: 0.0194 + 0.049 = 0.0684.

Task 2. Two factories produce the same glass for car headlights. The first factory produces 60% of these glasses, the second - 40%. The first factory produces 3% of defective glasses, and the second - 5%. Find the probability that a glass accidentally bought in a store will be defective.

Decision:

The probability that the glass was bought at the first factory and it is defective: 0.6 0.03 = 0.018.

The probability that the glass was bought at the second factory and it is defective: 0.4 0.05 = 0.02.

The probability that a glass accidentally bought in a store will be defective is 0.018 + 0.02 = 0.038.

Task 3. At the ceramic tableware factory, 10% of the produced plates are defective. During product quality control, 80% of defective plates are detected. The rest of the plates are for sale. Find the probability that a plate randomly selected at the time of purchase has no defects. Round the result to thousandths.

Decision:

Suppose we have x plates initially (after all, we are constantly dealing with percentages, so nothing prevents us from operating with specific values).

Then 0.1x are defective plates, and 0.9x are normal ones, which will go to the store immediately. Of the defective ones, 80%, that is, 0.08x, is removed, and 0.02x remains, which will also go to the store. Thus, the total number of plates on the shelves in the store will be: 0.9x + 0.02x = 0.92x. Of these, 0.9x will be normal. Accordingly, according to the formula, the probability will be 0.9x / 0.92x ≈ 0.978.

Task 4. According to customer reviews, Igor Igorevich assessed the reliability of two online stores. The probability that the desired product will be delivered from store A is 0.91. The probability that this product will be delivered from store B is 0.89. Igor Igorevich ordered the goods at once in both stores. Assuming that online stores operate independently of each other, find the probability that none of the stores will deliver the goods.

Decision. The probability that the first store will not deliver the goods is 1 − 0.91 = 0.09. The probability that the second store will not deliver the goods is 1 − 0.89 = 0.11. The probability of the occurrence of these two events at the same time is equal to the product of the probabilities of each of them: 0.09 0.11 = 0.0099.

Task 5. When manufacturing bearings with a diameter of 70 mm, the probability that the diameter will differ from the specified one by less than 0.01 mm is 0.961. Find the probability that a random bearing will have a diameter less than 69.99 mm or greater than 70.01 mm.

Decision: We are given the probability of an event in which the diameter will be between 69.99 mm and 70.01 mm, and it is equal to 0.961. We can find the probability of all other options using the principle of complementary probability: 1 − 0.961 = 0.039.

Task 6. The probability that a student correctly solves more than 9 problems on a history test is 0.68. The probability that more than 8 problems will be solved correctly is 0.78. Find the probability that exactly 9 problems will be solved correctly.

Decision: The probability that T. correctly solves more than 8 problems includes the probability of solving exactly 9 problems. At the same time, events in which O. solves more than 9 problems do not suit us. Therefore, subtracting the probability of solving more than 8 tasks from the probability of solving more than 9 tasks, we will find the probability of solving only 9 tasks: 0.78 - 0.68 = 0.1.

Task 7. A bus runs daily from the district center to the village. The probability that there will be less than 21 passengers on the bus on Monday is 0.88. The probability that there will be less than 12 passengers is 0.66. Find the probability that the number of passengers will be between 12 and 20.

Decision. The probability that a bus will have fewer than 21 passengers includes the probability that it will have between 12 and 20 passengers. At the same time, events in which there will be less than 12 passengers are not suitable for us. Therefore, subtracting from the first probability (less than 21) the second probability (less than 12), we will find the probability that there will be from 12 to 20 passengers: 0.88 - 0.66 = 0.22.

Task 8. There are two types of weather in Fairyland: good and excellent, and the weather, having settled in the morning, remains unchanged all day. It is known that with a probability of 0.9 the weather tomorrow will be the same as today. On April 10, the weather in Fairyland is good. Find the probability that on April 13th there will be great weather in Magicland.

Decision:

The task is performed by several options ("X" - good weather, "O" - great weather):

The probability of the origin of any of these 4 options is equal to the sum of the probabilities of each of the options: 0.081 + 0.081 + 0.081 + 0.001 = 0.244.

Task 9. There are two types of weather in Fairyland: good and excellent, and the weather, having settled in the morning, remains unchanged all day. It is known that with a probability of 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in Fairyland is fine. Find the probability that there will be great weather in Magicland on July 6th.

Decision:

The task is performed by several options ("X" - good weather, "O" - great weather):

The probability of the origin of any of these 4 - x options is equal to the sum of the probabilities of each of the options: 0.128 + 0.128 + 0.128 + 0.008 = 0.392.

Do you want to know what are the mathematical chances of your bet being successful? Then we have two good news for you. First: to calculate the patency, you do not need to carry out complex calculations and spend a lot of time. It is enough to use simple formulas, which will take a couple of minutes to work with. Second, after reading this article, you will easily be able to calculate the probability of passing any of your trades.

To correctly determine the patency, you need to take three steps:

Calculate the percentage of the probability of the outcome of an event according to the bookmaker's office;
Calculate the probability from statistical data yourself;
Find out the value of a bet given both probabilities.

Let us consider in detail each of the steps, using not only formulas, but also examples.

The first step is to find out with what probability the bookmaker evaluates the chances of a particular outcome. After all, it is clear that bookmakers do not bet odds just like that. For this we use the following formula:

PB=(1/K)*100%,

where P B is the probability of the outcome according to the bookmaker's office;

K - bookmaker odds for the outcome.

Let's say the odds are 4 for the victory of the London Arsenal in a duel against Bayern. This means that the probability of its victory by the BC is regarded as (1/4) * 100% = 25%. Or Djokovic is playing against South. The multiplier for Novak's victory is 1.2, his chances are equal to (1/1.2)*100%=83%.

This is how the bookmaker itself evaluates the chances of success for each player and team. Having completed the first step, we move on to the second.

Calculation of the probability of an event by the player

The second point of our plan is our own assessment of the probability of the event. Since we cannot mathematically take into account such parameters as motivation, game tone, we will use a simplified model and use only the statistics of previous meetings. To calculate the statistical probability of an outcome, we use the formula:

PAnd\u003d (UM / M) * 100%,

wherePAnd- the probability of the event according to the player;

UM - the number of successful matches in which such an event took place;

M is the total number of matches.

To make it clearer, let's give examples. Andy Murray and Rafael Nadal have played 14 matches. In 6 of them, total under 21 games were recorded, in 8 - total over. It is necessary to find out the probability that the next match will be played for a total over: (8/14)*100=57%. Valencia played 74 matches at the Mestalla against Atlético, in which they scored 29 victories. Probability of Valencia winning: (29/74)*100%=39%.

And we all know this only thanks to the statistics of previous games! Naturally, such a probability cannot be calculated for some new team or player, so this betting strategy is only suitable for matches in which opponents meet not for the first time. Now we know how to determine the betting and own probabilities of outcomes, and we have all the knowledge to go to the last step.

Determining the value of a bet

The value (valuability) of the bet and the passability are directly related: the higher the valuation, the higher the chance of a pass. The value is calculated as follows:

V=PAnd*K-100%,

where V is the value;

P I - the probability of an outcome according to the better;

K - bookmaker odds for the outcome.

Let's say we want to bet on Milan to win the match against Roma and we calculated that the probability of the Red-Blacks winning is 45%. The bookmaker offers us a coefficient of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V \u003d 45% * 2.5-100% \u003d 12.5%. Great, we have a valuable bet with good chances of passing.

Let's take another case. Maria Sharapova plays against Petra Kvitova. We want to make a deal for Maria to win, which, according to our calculations, has a 60% probability. Bookmakers offer a multiplier of 1.5 for this outcome. Determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be refrained from.

Bet Pass Probability: Conclusion

When calculating the passability of a bet, we used a simple model that is based only on statistics. When calculating the probability, it is desirable to take into account many different factors that are individual in each sport. It happens that it is not statistical factors that have more influence. Without it, everything would be simple and predictable. By choosing your niche, you will eventually learn to take into account all these nuances and give a more accurate assessment of your own probability of events, including many other influences. The main thing is to love what you do, gradually move forward and improve your skills step by step. Good luck and success in the exciting world of betting!

General statement of the problem: the probabilities of some events are known, but the probabilities of other events that are associated with these events need to be calculated. In these problems, there is a need for such operations on probabilities as addition and multiplication of probabilities.

For example, two shots were fired while hunting. Event A- hitting a duck from the first shot, event B- hit from the second shot. Then the sum of events A and B- hit from the first or second shot or from two shots.

Tasks of a different type. Several events are given, for example, a coin is tossed three times. It is required to find the probability that either all three times the coat of arms will fall out, or that the coat of arms will fall out at least once. This is a multiplication problem.

Addition of probabilities of incompatible events

Probability addition is used when it is necessary to calculate the probability of a combination or a logical sum of random events.

Sum of events A and B designate A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B- an event that occurs if and only if an event occurs during the observation A or event B, or at the same time A and B.

If events A and B are mutually inconsistent and their probabilities are given, the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.

The theorem of addition of probabilities. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:

For example, two shots were fired while hunting. Event BUT– hitting a duck from the first shot, event AT– hit from the second shot, event ( BUT+ AT) - hit from the first or second shot or from two shots. So if two events BUT and AT are incompatible events, then BUT+ AT- the occurrence of at least one of these events or two events.

Example 1 A box contains 30 balls of the same size: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball is taken without looking.

Decision. Let's assume that the event BUT– “the red ball is taken”, and the event AT- "The blue ball is taken." Then the event is “a colored (not white) ball is taken”. Find the probability of an event BUT:

and events AT:

Events BUT and AT- mutually incompatible, since if one ball is taken, then balls of different colors cannot be taken. Therefore, we use the addition of probabilities:

The theorem of addition of probabilities for several incompatible events. If the events make up the complete set of events, then the sum of their probabilities is equal to 1:

The sum of the probabilities of opposite events is also equal to 1:

Opposite events form a complete set of events, and the probability of a complete set of events is 1.

The probabilities of opposite events are usually denoted in small letters. p and q. In particular,

from which the following formulas for the probability of opposite events follow:

Example 2 The target in the dash is divided into 3 zones. The probability that a certain shooter will shoot at a target in the first zone is 0.15, in the second zone - 0.23, in the third zone - 0.17. Find the probability that the shooter hits the target and the probability that the shooter misses the target.

Solution: Find the probability that the shooter will hit the target:

Find the probability that the shooter misses the target:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

Addition of probabilities of mutually joint events

Two random events are said to be joint if the occurrence of one event does not preclude the occurrence of a second event in the same observation. For example, when throwing a dice, the event BUT is considered to be the occurrence of the number 4, and the event AT- dropping an even number. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of the occurrence of one of the mutually joint events.

The theorem of addition of probabilities for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events is as follows:

Because the events BUT and AT compatible, event BUT+ AT occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:

Event BUT occurs if one of two incompatible events occurs: or AB. However, the probability of occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:

Similarly:

Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:

When using formula (8), it should be taken into account that the events BUT and AT can be:

mutually independent;
mutually dependent.

Probability formula for mutually independent events:

Probability formula for mutually dependent events:

If events BUT and AT are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is as follows:

Example 3 In auto racing, when driving in the first car, the probability of winning, when driving in the second car. To find:

the probability that both cars will win;
the probability that at least one car will win;

1) The probability that the first car will win does not depend on the result of the second car, so the events BUT(first car wins) and AT(second car wins) - independent events. Find the probability that both cars win:

2) Find the probability that one of the two cars will win:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

Solve the problem of addition of probabilities yourself, and then look at the solution

Example 4 Two coins are thrown. Event A- loss of coat of arms on the first coin. Event B- loss of coat of arms on the second coin. Find the probability of an event C = A + B .

Probability multiplication

Multiplication of probabilities is used when the probability of a logical product of events is to be calculated.

In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.

Probability multiplication theorem for independent events. The probability of the simultaneous occurrence of two independent events BUT and AT is equal to the product of the probabilities of these events and is calculated by the formula:

Example 5 The coin is tossed three times in a row. Find the probability that the coat of arms will fall out all three times.

Decision. The probability that the coat of arms will fall on the first toss of a coin, the second time, and the third time. Find the probability that the coat of arms will fall out all three times:

Solve problems for multiplying probabilities yourself, and then look at the solution

Example 6 There is a box with nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, they do not distinguish between played and unplayed balls. What is the probability that after three games there will be no unplayed balls in the box?

Example 7 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the letters will form the word "end".

Example 8 From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards are of the same suit.

Example 9 The same problem as in example 8, but each card is returned to the deck after being drawn.

More complex tasks, in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events, on the page "Various tasks for addition and multiplication of probabilities" .

The probability that at least one of the mutually independent events will occur can be calculated by subtracting the product of the probabilities of opposite events from 1, that is, by the formula.

Choosing the right bet depends not only on intuition, sports knowledge, betting odds, but also on the odds of the event. The ability to calculate such an indicator in betting is the key to success in predicting the upcoming event on which the bet is supposed to be made.
In bookmakers, there are three types of odds (for more details, see the article), the variety of which determines how to calculate the probability of an event for a player.

Decimal Odds

The calculation of the probability of an event in this case occurs according to the formula: 1/coefficient of event. = v.i, where the coefficient of sob. is the coefficient of the event, and c.i is the probability of the outcome. For example, we take an event odds of 1.80 at a bet of one dollar, performing a mathematical action according to the formula, the player gets that the probability of the outcome of the event according to the bookmaker is 0.55 percent.

Fractional Odds

When using fractional odds, the probability calculation formula will be different. So with a coefficient of 7/2, where the first digit means the possible amount of net profit, and the second is the size of the required rate, to obtain this profit, the equation will look like this: . Here zn.coef is the denominator of the coefficient, chs.coef is the numerator of the coefficient, s.i is the probability of the outcome. Thus, for a fractional odds of 7/2, the equation looks like 2 / (7+2) = 2 / 9 = 0.22, therefore, 0.22 percent of the probability of the outcome of the event according to the bookmaker.

American odds

American odds are not very popular among bettors and are usually used exclusively in the USA, having a complex and intricate structure. To answer the question: "How to calculate the probability of an event in this way?", you need to know that such coefficients can be negative and positive.

An odd with a “-” sign, such as -150, indicates that a player needs to wager $150 to make a net profit of $100. The probability of an event is calculated based on the formula where you need to divide the negative odds by the sum of the negative odds and 100. This looks like the example of a bet of -150, so (-(-150)) / ((-(-150)) + 100) = 150 / (150 + 100) = 150 / 250 = 0.6, where 0.6 is multiplied by 100 and the outcome of the event is 60 percent. The same formula applies to positive American odds.

What is a probability?

Faced with this term for the first time, I would not understand what it is. So I'll try to explain in an understandable way.

Probability is the chance that the desired event will occur.

For example, you decided to visit a friend, remember the entrance and even the floor on which he lives. But I forgot the number and location of the apartment. And now you are standing on the stairwell, and in front of you are the doors to choose from.

What is the chance (probability) that if you ring the first doorbell, your friend will open it for you? Whole apartment, and a friend lives only behind one of them. With equal chance, we can choose any door.

But what is this chance?

Doors, the right door. Probability of guessing by ringing the first door: . That is, one time out of three you will guess for sure.

We want to know by calling once, how often will we guess the door? Let's look at all the options:

you called to 1st a door
you called to 2nd a door
you called to 3rd a door

And now consider all the options where a friend can be:

a. Behind 1st door
b. Behind 2nd door
in. Behind 3rd door

Let's compare all the options in the form of a table. A tick indicates the options when your choice matches the location of a friend, a cross - when it does not match.

How do you see everything possibly options friend's location and your choice of which door to ring.

BUT favorable outcomes of all . That is, you will guess the times from by ringing the door once, i.e. .

This is the probability - the ratio of a favorable outcome (when your choice coincided with the location of a friend) to the number of possible events.

The definition is the formula. Probability is usually denoted p, so:

It is not very convenient to write such a formula, so let's take for - the number of favorable outcomes, and for - the total number of outcomes.

The probability can be written as a percentage, for this you need to multiply the resulting result by:

Probably, the word “outcomes” caught your eye. Since mathematicians call various actions (for us, such an action is a doorbell) experiments, it is customary to call the result of such experiments an outcome.

Well, the outcomes are favorable and unfavorable.

Let's go back to our example. Let's say we rang at one of the doors, but a stranger opened it for us. We didn't guess. What is the probability that if we ring one of the remaining doors, our friend will open it for us?

If you thought that, then this is a mistake. Let's figure it out.

We have two doors left. So we have possible steps:

1) Call to 1st a door
2) Call 2nd a door

A friend, with all this, is definitely behind one of them (after all, he was not behind the one we called):

a) a friend 1st door
b) a friend for 2nd door

Let's draw the table again:

As you can see, there are all options, of which - favorable. That is, the probability is equal.

Why not?

The situation we have considered is example of dependent events. The first event is the first doorbell, the second event is the second doorbell.

And they are called dependent because they affect the following actions. After all, if a friend opened the door after the first ring, what would be the probability that he was behind one of the other two? Correctly, .

But if there are dependent events, then there must be independent? True, there are.

A textbook example is tossing a coin.

We toss a coin. What is the probability that, for example, heads will come up? That's right - because the options for everything (either heads or tails, we will neglect the probability of a coin to stand on edge), but only suits us.
But the tails fell out. Okay, let's do it again. What is the probability of coming up heads now? Nothing has changed, everything is the same. How many options? Two. How much are we satisfied with? One.

And let tails fall out at least a thousand times in a row. The probability of falling heads at once will be the same. There are always options, but favorable ones.

Distinguishing dependent events from independent events is easy:

If the experiment is carried out once (once a coin is tossed, the doorbell rings once, etc.), then the events are always independent.
If the experiment is carried out several times (a coin is tossed once, the doorbell is rung several times), then the first event is always independent. And then, if the number of favorable or the number of all outcomes changes, then the events are dependent, and if not, they are independent.

Let's practice a little to determine the probability.

Example 1

The coin is tossed twice. What is the probability of getting heads up twice in a row?

Decision:

Consider all possible options:

eagle eagle
tails eagle
tails-eagle
Tails-tails

As you can see, all options. Of these, we are satisfied only. That is the probability:

If the condition asks simply to find the probability, then the answer must be given as a decimal fraction. If it were indicated that the answer must be given as a percentage, then we would multiply by.

Answer:

Example 2

In a box of chocolates, all candies are packed in the same wrapper. However, from sweets - with nuts, cognac, cherries, caramel and nougat.

What is the probability of taking one candy and getting a candy with nuts. Give your answer in percentage.

Decision:

How many possible outcomes are there? .

That is, taking one candy, it will be one of those in the box.

And how many favorable outcomes?

Because the box contains only chocolates with nuts.

Answer:

Example 3

In a box of balls. of which are white and black.

What is the probability of drawing a white ball?
We added more black balls to the box. What is the probability of drawing a white ball now?

Decision:

a) There are only balls in the box. of which are white.

The probability is:

b) Now there are balls in the box. And there are just as many whites left.

Answer:

Full Probability

The probability of all possible events is ().

For example, in a box of red and green balls. What is the probability of drawing a red ball? Green ball? Red or green ball?

Probability of drawing a red ball

Green ball:

Red or green ball:

As you can see, the sum of all possible events is equal to (). Understanding this point will help you solve many problems.

Example 4

There are felt-tip pens in the box: green, red, blue, yellow, black.

What is the probability of drawing NOT a red marker?

Decision:

Let's count the number favorable outcomes.

NOT a red marker, that means green, blue, yellow, or black.

Probability of all events. And the probability of events that we consider unfavorable (when we pull out a red felt-tip pen) is .

Thus, the probability of drawing NOT a red felt-tip pen is -.

Answer:

The probability that an event will not occur is minus the probability that the event will occur.

Rule for multiplying the probabilities of independent events

You already know what independent events are.

And if you need to find the probability that two (or more) independent events will occur in a row?

Let's say we want to know what is the probability that by tossing a coin once, we will see an eagle twice?

We have already considered - .

What if we toss a coin? What is the probability of seeing an eagle twice in a row?

Total possible options:

Eagle-eagle-eagle
Eagle-head-tails
Head-tails-eagle
Head-tails-tails
tails-eagle-eagle
Tails-heads-tails
Tails-tails-heads
Tails-tails-tails

I don't know about you, but I made this list wrong once. Wow! And only option (the first) suits us.

For 5 rolls, you can make a list of possible outcomes yourself. But mathematicians are not as industrious as you.

Therefore, they first noticed, and then proved, that the probability of a certain sequence of independent events decreases each time by the probability of one event.

In other words,

Consider the example of the same, ill-fated, coin.

Probability of coming up heads in a trial? . Now we are tossing a coin.

What is the probability of getting tails in a row?

This rule does not only work if we are asked to find the probability that the same event will occur several times in a row.

If we wanted to find the TAILS-EAGLE-TAILS sequence on consecutive flips, we would do the same.

The probability of getting tails - , heads - .

The probability of getting the sequence TAILS-EAGLE-TAILS-TAILS:

You can check it yourself by making a table.

The rule for adding the probabilities of incompatible events.

So stop! New definition.

Let's figure it out. Let's take our worn out coin and flip it once.
Possible options:

Eagle-eagle-eagle
Eagle-head-tails
Head-tails-eagle
Head-tails-tails
tails-eagle-eagle
Tails-heads-tails
Tails-tails-heads
Tails-tails-tails

So here are incompatible events, this is a certain, given sequence of events. are incompatible events.

If we want to determine what is the probability of two (or more) incompatible events, then we add the probabilities of these events.

You need to understand that the loss of an eagle or tails is two independent events.

If we want to determine what is the probability of a sequence falling out) (or any other), then we use the rule of multiplying probabilities.
What is the probability of getting heads on the first toss and tails on the second and third?

But if we want to know what is the probability of getting one of several sequences, for example, when it comes up heads exactly once, i.e. options and, then we must add the probabilities of these sequences.

Total options suits us.

We can get the same thing by adding up the probabilities of occurrence of each sequence:

Thus, we add probabilities when we want to determine the probability of some, incompatible, sequences of events.

There is a great rule to help you not get confused when to multiply and when to add:

Let's go back to the example where we tossed a coin times, and want to know the probability of seeing heads times.
What is going to happen?

Should drop:
(heads AND tails AND tails) OR (tails AND heads AND tails) OR (tails AND tails AND heads).
And so it turns out:

Let's look at a few examples.

Example 5

There are pencils in the box. red, green, orange and yellow and black. What is the probability of drawing red or green pencils?

Decision:

What is going to happen? We have to pull out (red OR green).

Now it’s clear, we add up the probabilities of these events:

Answer:

Example 6

A die is thrown twice, what is the probability that a total of 8 will come up?

Decision.

How can we get points?

(and) or (and) or (and) or (and) or (and).

The probability of falling out of one (any) face is .

We calculate the probability:

Answer:

Workout.

I think now it has become clear to you when you need to how to count the probabilities, when to add them, and when to multiply them. Is not it? Let's get some exercise.

Tasks:

Let's take a deck of cards in which the cards are spades, hearts, 13 clubs and 13 tambourines. From to Ace of each suit.

What is the probability of drawing clubs in a row (we put the first drawn card back into the deck and shuffle)?
What is the probability of drawing a black card (spades or clubs)?
What is the probability of drawing a picture (jack, queen, king or ace)?
What is the probability of drawing two pictures in a row (we remove the first card drawn from the deck)?
What is the probability, taking two cards, to collect a combination - (Jack, Queen or King) and Ace The sequence in which the cards will be drawn does not matter.

Answers:

In a deck of cards of each value, it means:
The events are dependent, since after the first card drawn, the number of cards in the deck has decreased (as well as the number of "pictures"). Total jacks, queens, kings and aces in the deck initially, which means the probability of drawing the “picture” with the first card:
Since we are removing the first card from the deck, it means that there is already a card left in the deck, of which there are pictures. Probability of drawing a picture with the second card:
Since we are interested in the situation when we get from the deck: “picture” AND “picture”, then we need to multiply the probabilities:
Answer:
After the first card is drawn, the number of cards in the deck will decrease. Thus, we have two options:
1) With the first card we take out Ace, the second - jack, queen or king
2) With the first card we take out a jack, queen or king, the second - an ace. (ace and (jack or queen or king)) or ((jack or queen or king) and ace). Don't forget about reducing the number of cards in the deck!

If you were able to solve all the problems yourself, then you are a great fellow! Now tasks on the theory of probability in the exam you will click like nuts!

PROBABILITY THEORY. MIDDLE LEVEL

Consider an example. Let's say we throw a die. What kind of bone is this, do you know? This is the name of a cube with numbers on the faces. How many faces, so many numbers: from to how many? Before.

So we roll a die and want it to come up with an or. And we fall out.

In probability theory they say what happened favorable event(not to be confused with good).

If it fell out, the event would also be auspicious. In total, only two favorable events can occur.

How many bad ones? Since all possible events, then the unfavorable of them are events (this is if it falls out or).

Definition:

Probability is the ratio of the number of favorable events to the number of all possible events.. That is, the probability shows what proportion of all possible events are favorable.

They denote the probability with a Latin letter (apparently, from the English word probability - probability).

It is customary to measure the probability as a percentage (see topics and). To do this, the probability value must be multiplied by. In the dice example, probability.

And in percentage: .

Examples (decide for yourself):

What is the probability that the toss of a coin will land on heads? And what is the probability of a tails?
What is the probability that an even number will come up when a dice is thrown? And with what - odd?
In a drawer of plain, blue and red pencils. We randomly draw one pencil. What is the probability of pulling out a simple one?

Solutions:

How many options are there? Heads and tails - only two. And how many of them are favorable? Only one is an eagle. So the probability
Same with tails: .
Total options: (how many sides a cube has, so many different options). Favorable ones: (these are all even numbers :).
Probability. With odd, of course, the same thing.
Total: . Favorable: . Probability: .

Full Probability

All pencils in the drawer are green. What is the probability of drawing a red pencil? There are no chances: probability (after all, favorable events -).

Such an event is called impossible.

What is the probability of drawing a green pencil? There are exactly as many favorable events as there are total events (all events are favorable). So the probability is or.

Such an event is called certain.

If there are green and red pencils in the box, what is the probability of drawing a green or a red one? Yet again. Note the following thing: the probability of drawing green is equal, and red is .

In sum, these probabilities are exactly equal. I.e, the sum of the probabilities of all possible events is equal to or.

Example:

In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of not drawing green?

Decision:

Remember that all probabilities add up. And the probability of drawing green is equal. This means that the probability of not drawing green is equal.

Remember this trick: The probability that an event will not occur is minus the probability that the event will occur.

Independent events and the multiplication rule

You flip a coin twice and you want it to come up heads both times. What is the probability of this?

Let's go through all the possible options and determine how many there are:

Eagle-Eagle, Tails-Eagle, Eagle-Tails, Tails-Tails. What else?

The whole variant. Of these, only one suits us: Eagle-Eagle. So, the probability is equal.

Good. Now let's flip a coin. Count yourself. Happened? (answer).

You may have noticed that with the addition of each next throw, the probability decreases by a factor. The general rule is called multiplication rule:

The probabilities of independent events change.

What are independent events? Everything is logical: these are those that do not depend on each other. For example, when we toss a coin several times, each time a new toss is made, the result of which does not depend on all previous tosses. With the same success, we can throw two different coins at the same time.

More examples:

A die is thrown twice. What is the probability that it will come up both times?
A coin is tossed times. What is the probability of getting heads first and then tails twice?
The player rolls two dice. What is the probability that the sum of the numbers on them will be equal?

Answers:

The events are independent, which means that the multiplication rule works: .
The probability of an eagle is equal. Tails probability too. We multiply:
12 can only be obtained if two -ki fall out: .

Incompatible events and the addition rule

Incompatible events are events that complement each other to full probability. As the name implies, they cannot happen at the same time. For example, if we toss a coin, either heads or tails can fall out.

Example.

In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of drawing green or red?

Decision .

The probability of drawing a green pencil is equal. Red - .

Auspicious events of all: green + red. So the probability of drawing green or red is equal.

The same probability can be represented in the following form: .

This is the addition rule: the probabilities of incompatible events add up.

Mixed tasks

Example.

The coin is tossed twice. What is the probability that the result of the rolls will be different?

Decision .

This means that if heads come up first, tails should be second, and vice versa. It turns out that there are two pairs of independent events here, and these pairs are incompatible with each other. How not to get confused about where to multiply and where to add.

There is a simple rule for such situations. Try to describe what should happen by connecting the events with the unions "AND" or "OR". For example, in this case:

Must roll (heads and tails) or (tails and heads).

Where there is a union "and", there will be multiplication, and where "or" is addition:

Try it yourself:

What is the probability that two coin tosses come up with the same side both times?
A die is thrown twice. What is the probability that the sum will drop points?

Solutions:

(Heads up and heads up) or (tails up and tails up): .
What are the options? and. Then:
Rolled (and) or (and) or (and): .

Another example:

We toss a coin once. What is the probability that heads will come up at least once?

Decision:

Oh, how I don’t want to sort through the options ... Head-tails-tails, Eagle-heads-tails, ... But you don’t have to! Let's talk about full probability. Remembered? What is the probability that the eagle will never drop? It's simple: tails fly all the time, that means.

PROBABILITY THEORY. BRIEFLY ABOUT THE MAIN

Probability is the ratio of the number of favorable events to the number of all possible events.

Independent events

Two events are independent if the occurrence of one does not change the probability of the other occurring.

Full Probability

The probability of all possible events is ().

The probability that an event will not occur is minus the probability that the event will occur.

Rule for multiplying the probabilities of independent events

The probability of a certain sequence of independent events is equal to the product of the probabilities of each of the events

Incompatible events

Incompatible events are those events that cannot possibly occur simultaneously as a result of an experiment. A number of incompatible events form a complete group of events.

The probabilities of incompatible events add up.

Having described what should happen, using the unions "AND" or "OR", instead of "AND" we put the sign of multiplication, and instead of "OR" - addition.