Probability theory on the exam in mathematics. Probability of an event. Classic definition
When a coin is tossed, you can say that it will land heads up, or probability this is 1/2. Of course, this does not mean that if a coin is tossed 10 times, it will necessarily land on heads 5 times. If the coin is "fair" and if it is tossed many times, then heads will land very close half the time. Thus, there are two types of probabilities: experimental And theoretical .
Experimental and theoretical probability
If we flip a coin a large number of times - say 1000 - and count how many times it lands on heads, we can determine the probability that it lands on heads. If heads are thrown 503 times, we can calculate the probability of it landing:
503/1000, or 0.503.
This experimental determination of probability. This definition of probability comes from observation and study of data and is quite common and very useful. Here, for example, are some probabilities that were determined experimentally:
1. The probability that a woman will develop breast cancer is 1/11.
2. If you kiss someone who has a cold, then the probability that you will also get a cold is 0.07.
3. A person who has just been released from prison has an 80% chance of returning to prison.
If we consider tossing a coin and taking into account that it is just as likely that it will come up heads or tails, we can calculate the probability of getting heads: 1/2. This is a theoretical definition of probability. Here are some other probabilities that have been determined theoretically using mathematics:
1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding year) is 0.706.
2. During a trip, you meet someone, and during the conversation you discover that you have a mutual friend. Typical reaction: “This can’t be!” In fact, this phrase is not suitable, because the probability of such an event is quite high - just over 22%.
Thus, experimental probabilities are determined through observation and data collection. Theoretical probabilities are determined through mathematical reasoning. Examples of experimental and theoretical probabilities, such as those discussed above, and especially those that we do not expect, lead us to the importance of studying probability. You may ask, "What is true probability?" In fact, there is no such thing. Probabilities within certain limits can be determined experimentally. They may or may not coincide with the probabilities that we obtain theoretically. There are situations in which it is much easier to determine one type of probability than another. For example, it would be sufficient to find the probability of catching a cold using theoretical probability.
Calculation of experimental probabilities
Let us first consider the experimental definition of probability. The basic principle we use to calculate such probabilities is as follows.
Principle P (experimental)
If in an experiment in which n observations are made, a situation or event E occurs m times in n observations, then the experimental probability of the event is said to be P (E) = m/n.
Example 1 Sociological survey. An experimental study was conducted to determine the number of left-handed people, right-handed people and people whose both hands are equally developed. The results are shown in the graph.
a) Determine the probability that the person is right-handed.
b) Determine the probability that the person is left-handed.
c) Determine the probability that a person is equally fluent in both hands.
d) Most Professional Bowling Association tournaments are limited to 120 players. Based on the data from this experiment, how many players could be left-handed?
Solution
a)The number of people who are right-handed is 82, the number of left-handers is 17, and the number of those who are equally fluent in both hands is 1. The total number of observations is 100. Thus, the probability that a person is right-handed is P
P = 82/100, or 0.82, or 82%.
b) The probability that a person is left-handed is P, where
P = 17/100, or 0.17, or 17%.
c) The probability that a person is equally fluent in both hands is P, where
P = 1/100, or 0.01, or 1%.
d) 120 bowlers, and from (b) we can expect that 17% are left-handed. From here
17% of 120 = 0.17.120 = 20.4,
that is, we can expect about 20 players to be left-handed.
Example 2 Quality control
. It is very important for a manufacturer to keep the quality of its products at a high level. In fact, companies hire quality control inspectors to ensure this process. The goal is to produce the minimum possible number of defective products. But since the company produces thousands of products every day, it cannot afford to test every product to determine whether it is defective or not. To find out what percentage of products are defective, the company tests far fewer products.
The USDA requires that 80% of the seeds sold by growers must germinate. To determine the quality of the seeds that an agricultural company produces, 500 seeds from those that were produced are planted. After this, it was calculated that 417 seeds sprouted.
a) What is the probability that the seed will germinate?
b) Do the seeds meet government standards?
Solution a) We know that out of 500 seeds that were planted, 417 sprouted. Probability of seed germination P, and
P = 417/500 = 0.834, or 83.4%.
b) Since the percentage of seeds germinated has exceeded 80% as required, the seeds meet government standards.
Example 3 Television ratings. According to statistics, there are 105,500,000 households with televisions in the United States. Every week, information about viewing programs is collected and processed. In one week, 7,815,000 households tuned in to the hit comedy series "Everybody Loves Raymond" on CBS and 8,302,000 households tuned in to the hit series "Law & Order" on NBC (Source: Nielsen Media Research). What is the probability that one household's TV is tuned to "Everybody Loves Raymond" during a given week? to "Law & Order"?
Solution The probability that the TV in one household is tuned to "Everybody Loves Raymond" is P, and
P = 7,815,000/105,500,000 ≈ 0.074 ≈ 7.4%.
The chance that a household's TV was tuned to Law & Order is P, and
P = 8,302,000/105,500,000 ≈ 0.079 ≈ 7.9%.
These percentages are called ratings.
Theoretical probability
Suppose we are conducting an experiment, such as throwing a coin or darts, drawing a card from a deck, or testing products for quality on an assembly line. Each possible result of such an experiment is called Exodus . The set of all possible outcomes is called outcome space . Event it is a set of outcomes, that is, a subset of the space of outcomes.
Example 4 Throwing darts. Suppose that in a dart throwing experiment, a dart hits a target. Find each of the following:
b) Outcome space
Solution
a) The outcomes are: hitting black (B), hitting red (R) and hitting white (B).
b) The space of outcomes is (hitting black, hitting red, hitting white), which can be written simply as (H, K, B).
Example 5 Throwing dice.
A die is a cube with six sides, each with one to six dots on it. 
Suppose we are throwing a die. Find
a) Outcomes
b) Outcome space
Solution
a) Outcomes: 1, 2, 3, 4, 5, 6.
b) Outcome space (1, 2, 3, 4, 5, 6).
We denote the probability that an event E occurs as P(E). For example, “the coin will land on heads” can be denoted by H. Then P(H) represents the probability that the coin will land on heads. When all outcomes of an experiment have the same probability of occurrence, they are said to be equally likely. To see the differences between events that are equally likely and events that are not, consider the target shown below. 
For target A, the events of hitting black, red and white are equally probable, since the black, red and white sectors are the same. However, for target B, the zones with these colors are not the same, that is, hitting them is not equally likely.
Principle P (Theoretical)
If an event E can happen in m ways out of n possible equally probable outcomes from the outcome space S, then theoretical probability
events, P(E) is
P(E) = m/n.
Example 6 What is the probability of rolling a die to get a 3?
Solution There are 6 equally probable outcomes on a dice and there is only one possibility of rolling the number 3. Then the probability P will be P(3) = 1/6.
Example 7 What is the probability of rolling an even number on a die?
Solution The event is the throwing of an even number. This can happen in 3 ways (if you roll a 2, 4 or 6). The number of equally probable outcomes is 6. Then the probability P(even) = 3/6, or 1/2.
We will use a number of examples involving a standard 52 card deck. This deck consists of the cards shown in the figure below. 
Example 8 What is the probability of drawing an Ace from a well-shuffled deck of cards?
Solution There are 52 outcomes (the number of cards in the deck), they are equally likely (if the deck is well shuffled), and there are 4 ways to draw an Ace, so according to the P principle, the probability
P(draw an ace) = 4/52, or 1/13.
Example 9 Suppose we choose, without looking, one ball from a bag with 3 red balls and 4 green balls. What is the probability of choosing a red ball?
Solution There are 7 equally probable outcomes of drawing any ball, and since the number of ways to draw a red ball is 3, we get
P(red ball selection) = 3/7.
The following statements are results from Principle P.
Properties of Probability
a) If event E cannot happen, then P(E) = 0.
b) If event E is certain to happen then P(E) = 1.
c) The probability that event E will occur is a number from 0 to 1: 0 ≤ P(E) ≤ 1.
For example, in a coin toss, the event that the coin lands on its edge has zero probability. The probability that a coin is either heads or tails has a probability of 1.
Example 10 Let's assume that 2 cards are drawn from a 52-card deck. What is the probability that both of them are peaks?
Solution The number n of ways to draw 2 cards from a well-shuffled deck of 52 cards is 52 C 2 . Since 13 of the 52 cards are spades, the number of ways m to draw 2 spades is 13 C 2 . Then,
P(pulling 2 peaks) = m/n = 13 C 2 / 52 C 2 = 78/1326 = 1/17.
Example 11 Suppose 3 people are randomly selected from a group of 6 men and 4 women. What is the probability that 1 man and 2 women will be selected?
Solution The number of ways to select three people from a group of 10 people is 10 C 3. One man can be chosen in 6 C 1 ways, and 2 women can be chosen in 4 C 2 ways. According to the fundamental principle of counting, the number of ways to choose 1 man and 2 women is 6 C 1. 4 C 2 . Then, the probability that 1 man and 2 women will be selected is
P = 6 C 1 . 4 C 2 / 10 C 3 = 3/10.
Example 12 Throwing dice. What is the probability of rolling a total of 8 on two dice?
Solution Each dice has 6 possible outcomes. The outcomes are doubled, meaning there are 6.6 or 36 possible ways in which the numbers on the two dice can appear. (It’s better if the cubes are different, say one is red and the other is blue - this will help visualize the result.) 
The pairs of numbers that add up to 8 are shown in the figure below. There are 5 possible ways to obtain a sum equal to 8, hence the probability is 5/36.
I understand that everyone wants to know in advance how the sporting event will end, who will win and who will lose. With this information, you can bet on sporting events without fear. But is it even possible, and if so, how to calculate the probability of an event?
Probability is a relative quantity, therefore it cannot speak with certainty about any event. This value allows you to analyze and evaluate the need to place a bet on a particular competition. Determining probabilities is a whole science that requires careful study and understanding.
Probability coefficient in probability theory
In sports betting, there are several options for the outcome of the competition:
- first team victory;
- victory of the second team;
- draw;
- total
Each outcome of the competition has its own probability and frequency with which this event will occur, provided that the initial characteristics are maintained. As we said earlier, it is impossible to accurately calculate the probability of any event - it may or may not coincide. Thus, your bet can either win or lose.
There cannot be a 100% accurate prediction of the results of the competition, since many factors influence the outcome of the match. Naturally, bookmakers do not know the outcome of the match in advance and only assume the result, making decisions using their analysis system and offering certain odds for betting.
How to calculate the probability of an event?
Let’s assume that the bookmaker’s odds are 2.1/2 – we get 50%. It turns out that coefficient 2 is equal to the probability of 50%. Using the same principle, you can get a break-even probability coefficient - 1/probability.
Many players think that after several repeated defeats, a win will definitely happen - this is a mistaken opinion. The probability of winning a bet does not depend on the number of losses. Even if you flip several heads in a row in a coin game, the probability of flipping tails remains the same - 50%.
Our answer
Choosing the right bet depends not only on intuition, sports knowledge, bookmaker odds, but also on the probability coefficient of the event. The ability to calculate such an indicator in betting is the key to success in predicting the upcoming event on which a bet is supposed to be placed.
In bookmakers there are three types of odds (more details in the article), the type of which determines how to calculate the probability of an event for a player.
Decimal odds
The probability of an event in this case is calculated using the formula: 1/coefficient. = v.i, where coefficient. is the event coefficient, and v.i is the probability of the outcome. For example, we take an event odd of 1.80 with a bet of one dollar, performing a mathematical operation according to the formula, the player receives that the probability of the outcome of the event according to the bookmaker is 0.55 percent.
Fractional odds
When using fractional odds, the formula for calculating the probability will be different. So, with a coefficient of 7/2, where the first figure means the possible amount of net profit, and the second the size of the required bet to obtain this profit, the equation will look like this: zn.od/ for the sum of zn.od and chs.od = v.i . Here zn.coef is the denominator of the coefficient, chs.coef is the numerator of the coefficient, v.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 probability of the outcome of the event according to the bookmaker.
American odds
American odds are not very popular among players and, as a rule, are used exclusively in the USA, having a complex and confusing 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.
A coefficient with a “-” sign, for example -150, shows that the player needs to place a bet of $150 to receive a net profit of $100. The probability of an event is calculated based on the formula where you need to divide the negative coefficient by the sum of the negative coefficient and 100. This looks like using 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 probability of the event is 60 percent. The same formula is also suitable for positive American odds.
The need to act on probabilities occurs when the probabilities of some events are known, and it is necessary to calculate the probabilities of other events that are associated with these events.
Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.
Sum of events A And B denote 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. This means that A + B– an event that occurs if and only if the event occurred during observation A or event B, or simultaneously A And B.
If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.
Probability addition theorem. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:
For example, while hunting, two shots are fired. Event A– hitting a duck with the first shot, event IN– hit from the second shot, event ( A+ IN) – a hit from the first or second shot or from two shots. So, if two events A And IN– incompatible events, then A+ IN– the occurrence of at least one of these events or two events.
Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.
Solution. Let us assume that the event A- “the red ball is taken”, and the event IN- “The blue ball was taken.” Then the event is “a colored (not white) ball is taken.” Let's find the probability of the event A:
and events IN:
Events A And IN– mutually incompatible, since if one ball is taken, then it is impossible to take balls of different colors. Therefore, we use the addition of probabilities:
The theorem for adding probabilities for several incompatible events. If events constitute a 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.
Probabilities of opposite events are usually indicated 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 shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the 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 will hit the target and the probability that the shooter will miss the target.
Solution: Find the probability that the shooter will hit the target:
Let's find the probability that the shooter will miss the target:
For more complex problems, in which you need to use both addition and multiplication of probabilities, see the page "Various problems involving addition and multiplication of probabilities".
Addition of probabilities of mutually simultaneous events
Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a die the event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since 4 is an even number, the two events are compatible. In practice, there are problems of calculating the probabilities of the occurrence of one of the mutually simultaneous events.
Probability addition theorem 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 has the following form:
Since events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:
Event A will occur if one of two incompatible events occurs: or AB. However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:
Likewise:
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 events A And IN may be:
- mutually independent;
- mutually dependent.
Probability formula for mutually independent events:
Probability formula for mutually dependent events:
If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is:
Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive the second car. Find:
- the probability that both cars will win;
- the probability that at least one car will win;
1) The probability that the first car wins does not depend on the result of the second car, so the events A(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:
2) Find the probability that one of the two cars will win:
For more complex problems, in which you need to use both addition and multiplication of probabilities, see the page "Various problems involving addition and multiplication of probabilities".
Solve the addition of probabilities problem yourself, and then look at the solution
Example 4. Two coins are tossed. Event A- loss of the coat of arms on the first coin. Event B- loss of the coat of arms on the second coin. Find the probability of an event C = A + B .
Multiplying Probabilities
Probability multiplication is used when the probability of a logical product of events must 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. Probability of simultaneous occurrence of two independent events A And IN 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 appear all three times.
Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:
Solve probability multiplication problems on your own and then look at the solution
Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games there will be no unplayed balls left in the box?
Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. 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 will be of different suits.
Example 9. The same task as in example 8, but each card after being removed is returned to the deck.
More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".
The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula:
Example 10. Cargo is delivered by three modes of transport: river, rail and road transport. The probability that the cargo will be delivered by river transport is 0.82, by railway transport is 0.87, and by road transport is 0.90. Find the probability that the cargo will be delivered by at least one of the three modes of transport.
Want to know the mathematical odds of your bet being successful? Then there is two good news for you. First: to calculate cross-country ability, you don’t 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 can easily calculate the probability of any of your transactions passing.
To correctly determine cross-country ability, you need to take three steps:
- Calculate the percentage of probability of the outcome of an event according to the bookmaker's office;
- Calculate the probability using statistical data yourself;
- Find out the value of the bet, taking into account both probabilities.
Let's look at each of the steps in detail, using not only formulas, but also examples.
Quick Jump
Calculating the probability included in bookmaker odds
The first step is to find out with what probability the bookmaker himself estimates the chances of a particular outcome. It’s clear that bookmakers don’t set odds just like that. To do 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 that the odds for London Arsenal’s victory in the match against Bayern Munich are 4. This means that the probability of their victory is assessed by the bookmaker as (1/4)*100%=25%. Or Djokovic plays against Youzhny. The multiplier for Novak's victory is 1.2, his chances are (1/1.2)*100%=83%.
This is how the bookmaker itself evaluates the chances of success of 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 and game tone, we will use a simplified model and use only statistics from previous meetings. To calculate the statistical probability of an outcome, we use the formula:
PAND=(UM/M)*100%,
WherePAND– probability of an event according to the player;
UM – the number of successful matches in which such an event occurred;
M – total number of matches.
To make it clearer, let's give examples. Andy Murray and Rafael Nadal played 14 matches between themselves. In 6 of them the total was less than 21 in games, in 8 the total was more. You need to find out the probability that the next match will be played with a higher total: (8/14)*100=57%. Valencia played 74 matches against Atlético at Mestalla, in which they won 29 victories. Probability of Valencia winning: (29/74)*100%=39%.
And we learn all this only thanks to the statistics of previous games! Naturally, it will not be possible to calculate such a probability for any new team or player, so this betting strategy is only suitable for matches in which the opponents meet more than once. Now we know how to determine the bookmaker's and our own probabilities of outcomes, and we have all the knowledge to move on to the last step.
Determining the value of a bet
The value (value) of a bet and the passability have a direct connection: the higher the value, the higher the chance of passing. The value is calculated as follows:
V=PAND*K-100%,
where V is value;
P I – probability of outcome according to the bettor;
K – bookmaker odds for the outcome.
Let’s say we want to bet on Milan’s victory in the match against Roma and we calculate that the probability of the “red-blacks” winning is 45%. The bookmaker offers us odds of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V=45%*2.5-100%=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, the probability of which, according to our calculations, is 60%. Bookmakers offer a 1.5 multiplier for this outcome. We determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be avoided.
