How is probability calculated by the formula? Probability theory: formulas and examples of problem solving

To calculate the probability P A of event A, it is necessary to construct a mathematical model of the object under study, which contains event A. The basis of the model is the probability space (,?,P), where is the space of elementary events, ? - a class of events with composition operations introduced on them,

The probability of any event A that makes sense in and is included in the class of events? 25. If, for example,

then from axiom 3, probabilities, it follows that

Thus, calculating the probability of event A is reduced to calculating the probabilities of the elementary events that make it up, and since they are “basic”, the methods for calculating them do not have to depend on the axiomatics of probability theory.

Three approaches to calculating the probabilities of elementary events are considered here:

classical;

geometric;

statistical or frequency.

Classic method of calculating probabilities

From the axiomatic definition of probability it follows that probability exists for any event A, but nothing is said about how to calculate it, although it is known that for each elementary event i there is a probability pi such that the sum of the probabilities of all elementary events in space is equal to one, that is

The classical method of calculating the probabilities of random events is based on the use of this fact, which, due to its specificity, provides a way to find the probabilities of these events directly from the axioms.

Let a fixed probability space (,?,P) be given, in which:

a) consists of a finite number n of elementary events,
b) each elementary event i is associated with a probability

Consider event A, which consists of m elementary events:

then from axiom 3 of probabilities, due to the incompatibility of elementary events, it follows that

Thus we have the formula

which can be interpreted as follows: the probability of event A to occur is equal to the ratio of the number of elementary events favorable to the occurrence of event A to the number of all elementary events from.

That's the point classical method calculating event probabilities.

Comment. Having assigned the same probability to each of the elementary events of space, we, on the one hand, having a probabilistic space and relying on the axioms of probability theory, have received a rule for calculating the probabilities of any random events from space according to formula (2), on the other hand, this gives us reason to consider all elementary events are equally possible and the calculation of the probabilities of any random events from the reduction to the “urn” scheme, regardless of the axioms.

From formula (2) it follows that the probability of event A depends only on the number of elementary events of which it consists and does not depend on their specific content. Thus, to use formula (2), it is necessary to find the number of points in space and the number of points that make up the event A, but then this is already a task of combinatorial analysis.

Let's look at a few examples.

Example 8. An urn of n balls contains k red and (n - k) black balls. We draw r balls at random without returning r balls. What is the probability that in a sample of r balls, s balls are red?

Solution. Let event (A) (in a sample of r balls s be red). The required probability is found according to the classical scheme, formula (2):

where is the number of possible samples of volume r that differ in at least one ball number, and m is the number of samples of volume r in which s balls are red. For, obviously, the number of possible sampling options is equal, and m, as follows from Example 7, is equal to

Thus, the required probability is equal to

Let the set be given in pairs not joint events As,

forming a complete group, then

In this case we say that we have a probability distribution of events As.

Probability distributions are one of the fundamental concepts of modern probability theory and form the basis of Kolmagorov's axioms.

Definition. Probability distribution

the hypergeometric distribution is determined.

Borovkov A.A. in his book, using formula (3) as an example, he explains the nature of problems in probability theory and mathematical statistics as follows: knowing the composition of the general population, we can use the hypergeometric distribution to find out what the composition of the sample may be - this is a typical problem in probability theory (direct problem). In the natural sciences, they solve the inverse problem: using the composition of samples, they determine the nature of general populations - this is the inverse problem, and, figuratively speaking, it constitutes the content of mathematical statistics.

A generalization of binomial coefficients (combinations) are polynomial coefficients, which owe their name to the expansion of a polynomial of the form

by powers of terms.

Polynomial coefficients (4) are often used when solving combinatorial problems.

Theorem. Let there be k different boxes into which numbered balls are placed. Then the number of balls to be placed in boxes such that the box number r contains ri balls,

is determined by polynomial coefficients (4).

Proof. Since the order of the boxes is important, but the balls in the boxes are not important, combinations can be used to count the placement of balls in any box.

In the first box r1 balls from n can be selected in ways, in the second box r2 balls, from the remaining (n - r1) can be chosen in ways and so on, in the (k - 1) box rk-1 balls we select

ways; in box k - the remaining ones

The balls fall automatically, in one way.

Thus, the total placements will be

Example. n balls are randomly distributed into n boxes. Assuming that the boxes and balls are distinguishable, find the probabilities of the following events:

a) all boxes are not empty = A0;
b) one box is empty = A1;
c) two empty boxes = A2;
d) three empty boxes = A3;
e) (n-1) - the box is empty = A4.

Solve the problem for the case n = 5.

Solution. It follows from the condition that the distribution of balls among the boxes is simple random selection, therefore, all options are nn.

This sequence means that the first, second and third boxes have three balls each, the fourth and fifth boxes have two balls each, and the remaining (n - 5) boxes have one ball each. The total number of such placements of balls into boxes will be

Since the balls are actually distinguishable, then for each such combination we will have

ball placements. Thus, there will be a total of options

Let's move on to the solution point by point in the example:

a) since each box contains one ball, we have the sequence 111...11, for which the number of placements is n!/ n! = 1. If the balls are distinguishable, then we have n!/ 1! placements, therefore, the total number of options is m = 1n!= n!, hence

b) if one box is empty, then some box contains two balls, then we have the sequence 211...10, for which the number of placements is n! (n-2)!. Since the balls are distinguishable, for each such combination we have n!/ 2! placements. Total options

c) if two boxes are empty, then we have two sequences: 311...100 and 221...100. For the first, the number of placements is equal to

n!/ (2! (n - 3)!).

For each such combination we have n!/ 3! ball placements. So, for the first sequence, the number of options is

For the second sequence, the total options will be

Finally we have

d) for three empty boxes there will be three sequences: 411...1000, or 3211...1000, or 22211...1000.

For the first sequence we have

For the second sequence

For the third sequence we get

Total options

m = k1 + k2 + k3,

The required probability is equal to

e) if (n -1) the box is empty, then all the balls must be in one of the boxes. Obviously, the number of combinations is equal to

The probability corresponding to this event is equal to

For n = 5, we have

Note that for n = 5 events Аi should form a complete group, which is true. Indeed

Initially, being just a collection of information and empirical observations about the game of dice, the theory of probability became a thorough science. The first to give it a mathematical framework were Fermat and Pascal.

From thinking about the eternal to the theory of probability

The two individuals to whom probability theory owes many of its fundamental formulas, Blaise Pascal and Thomas Bayes, are known as deeply religious people, the latter being a Presbyterian minister. Apparently, the desire of these two scientists to prove the fallacy of the opinion about a certain Fortune giving good luck to her favorites gave impetus to research in this area. After all, in fact, any gambling with its wins and losses, it is just a symphony of mathematical principles.

Thanks to the passion of the Chevalier de Mere, who was equally a gambler and a man not indifferent to science, Pascal was forced to find a way to calculate probability. De Mere was interested in the following question: “How many times do you need to throw two dice in pairs so that the probability of getting 12 points exceeds 50%?” The second question, which was of great interest to the gentleman: “How to divide the bet between the participants in the unfinished game?” Of course, Pascal successfully answered both questions of de Mere, who became the unwitting initiator of the development of probability theory. It is interesting that the person of de Mere remained known in this area, and not in literature.

Previously, no mathematician had ever attempted to calculate the probabilities of events, since it was believed that this was only a guessing solution. Blaise Pascal gave the first definition of the probability of an event and showed that it is a specific figure that can be justified mathematically. Probability theory has become the basis for statistics and is widely used in modern science.

What is randomness

If we consider a test that can be repeated an infinite number of times, then we can define a random event. This is one of the likely outcomes of the experiment.

Experience is the implementation of specific actions under constant conditions.

To be able to work with the results of the experiment, events are usually designated by the letters A, B, C, D, E...

Probability of a random event

In order to begin the mathematical part of probability, it is necessary to define all its components.

The probability of an event is a numerical measure of the possibility of some event (A or B) occurring as a result of an experience. The probability is denoted as P(A) or P(B).

In probability theory they distinguish:

reliable the event is guaranteed to occur as a result of the experience P(Ω) = 1;
impossible the event can never happen P(Ø) = 0;
random an event lies between reliable and impossible, that is, the probability of its occurrence is possible, but not guaranteed (the probability of a random event is always within the range 0≤Р(А)≤ 1).

Relationships between events

Both one and the sum of events A+B are considered, when the event is counted when at least one of the components, A or B, or both, A and B, is fulfilled.

In relation to each other, events can be:

Equally possible.
Compatible.
Incompatible.
Opposite (mutually exclusive).
Dependent.

If two events can happen with equal probability, then they equally possible.

If the occurrence of event A does not reduce to zero the probability of the occurrence of event B, then they compatible.

If events A and B never occur simultaneously in the same experience, then they are called incompatible. Coin toss - good example: the appearance of heads is automatically the non-appearance of heads.

The probability for the sum of such incompatible events consists of the sum of the probabilities of each of the events:

P(A+B)=P(A)+P(B)

If the occurrence of one event makes the occurrence of another impossible, then they are called opposite. Then one of them is designated as A, and the other - Ā (read as “not A”). The occurrence of event A means that Ā did not happen. These two events form a complete group with a sum of probabilities equal to 1.

Dependent events have mutual influence, decreasing or increasing the probability of each other.

Relationships between events. Examples

Using examples it is much easier to understand the principles of probability theory and combinations of events.

The experiment that will be carried out consists of taking balls out of a box, and the result of each experiment is an elementary outcome.

An event is one of the possible outcomes of an experiment - a red ball, a blue ball, a ball with number six, etc.

Test No. 1. There are 6 balls involved, three of which are blue with odd numbers on them, and the other three are red with even numbers.

Test No. 2. 6 balls involved of blue color with numbers from one to six.

Based on this example, we can name combinations:

Reliable event. In Spanish No. 2 the event “get the blue ball” is reliable, since the probability of its occurrence is equal to 1, since all the balls are blue and there can be no miss. Whereas the event “get the ball with the number 1” is random.
Impossible event. In Spanish No. 1 with blue and red balls, the event “getting the purple ball” is impossible, since the probability of its occurrence is 0.
Equally possible events. In Spanish No. 1, the events “get the ball with the number 2” and “get the ball with the number 3” are equally possible, and the events “get the ball with an even number” and “get the ball with the number 2” have different probabilities.
Compatible Events. Getting a six twice in a row while throwing a die is a compatible event.
Incompatible events. In the same Spanish No. 1, the events “get a red ball” and “get a ball with an odd number” cannot be combined in the same experience.
Opposite events. Most shining example This is coin tossing, where drawing heads is equivalent to not drawing tails, and the sum of their probabilities is always 1 (full group).
Dependent Events. So, in Spanish No. 1, you can set the goal of drawing the red ball twice in a row. Whether or not it is retrieved the first time affects the likelihood of being retrieved the second time.

It can be seen that the first event significantly affects the probability of the second (40% and 60%).

Event probability formula

The transition from fortune-telling to precise data occurs through the translation of the topic into a mathematical plane. That is, judgments about a random event such as “high probability” or “minimal probability” can be translated into specific numerical data. It is already permissible to evaluate, compare and enter such material into more complex calculations.

From a calculation point of view, determining the probability of an event is the ratio of the number of elementary positive outcomes to the number of all possible outcomes of experience regarding a certain event. Probability is denoted by P(A), where P stands for the word “probabilite”, which is translated from French as “probability”.

So, the formula for the probability of an event is:

Where m is the number of favorable outcomes for event A, n is the sum of all outcomes possible for this experience. In this case, the probability of an event always lies between 0 and 1:

0 ≤ P(A)≤ 1.

Calculation of the probability of an event. Example

Let's take Spanish. No. 1 with balls, which was described earlier: 3 blue balls with the numbers 1/3/5 and 3 red balls with the numbers 2/4/6.

Based on this test, several different problems can be considered:

A - red ball falling out. There are 3 red balls, and there are 6 options in total. This is simplest example, in which the probability of the event is equal to P(A)=3/6=0.5.
B - rolling an even number. There are 3 even numbers (2,4,6), and the total number of possible numerical options is 6. The probability of this event is P(B)=3/6=0.5.
C - the occurrence of a number greater than 2. There are 4 such options (3,4,5,6) out of a total number of possible outcomes of 6. The probability of event C is equal to P(C)=4/6=0.67.

As can be seen from the calculations, event C has a higher probability, since the number of probable positive outcomes is higher than in A and B.

Incompatible events

Such events cannot appear simultaneously in the same experience. As in Spanish No. 1 it is impossible to get a blue and a red ball at the same time. That is, you can get either a blue or a red ball. In the same way, an even and an odd number cannot appear in a dice at the same time.

The probability of two events is considered as the probability of their sum or product. The sum of such events A+B is considered to be an event that consists of the occurrence of event A or B, and the product of them AB is the occurrence of both. For example, the appearance of two sixes at once on the faces of two dice in one throw.

The sum of several events is an event that presupposes the occurrence of at least one of them. The production of several events is the joint occurrence of them all.

In probability theory, as a rule, the use of the conjunction “and” denotes a sum, and the conjunction “or” - multiplication. Formulas with examples will help you understand the logic of addition and multiplication in probability theory.

Probability of the sum of incompatible events

If probability is considered incompatible events, then the probability of the sum of events is equal to the addition of their probabilities:

P(A+B)=P(A)+P(B)

For example: let's calculate the probability that in Spanish. No. 1 with blue and red balls, a number between 1 and 4 will appear. We will calculate not in one action, but by the sum of the probabilities of the elementary components. So, in such an experiment there are only 6 balls or 6 of all possible outcomes. The numbers that satisfy the condition are 2 and 3. The probability of getting the number 2 is 1/6, the probability of getting the number 3 is also 1/6. The probability of getting a number between 1 and 4 is:

The probability of the sum of incompatible events of a complete group is 1.

So, if in an experiment with a cube we add up the probabilities of all numbers appearing, the result will be one.

This is also true for opposite events, for example in the experiment with a coin, where one side is the event A, and the other is the opposite event Ā, as is known,

P(A) + P(Ā) = 1

Probability of incompatible events occurring

Probability multiplication is used when considering the occurrence of two or more incompatible events in one observation. The probability that events A and B will appear in it simultaneously is equal to the product of their probabilities, or:

P(A*B)=P(A)*P(B)

For example, the probability that in Spanish No. 1, as a result of two attempts, a blue ball will appear twice, equal to

That is, the probability of an event occurring when, as a result of two attempts to extract balls, only blue balls are extracted is 25%. It is very easy to do practical experiments on this problem and see if this is actually the case.

Joint events

Events are considered joint when the occurrence of one of them can coincide with the occurrence of another. Despite the fact that they are joint, the probability of independent events is considered. For example, throwing two dice can give a result when the number 6 appears on both of them. Although the events coincided and appeared at the same time, they are independent of each other - only one six could fall out, the second die has no influence on it.

The probability of joint events is considered as the probability of their sum.

Probability of the sum of joint events. Example

The probability of the sum of events A and B, which are joint in relation to each other, is equal to the sum of the probabilities of the event minus the probability of their occurrence (that is, their joint occurrence):

R joint (A+B)=P(A)+P(B)- P(AB)

Let's assume that the probability of hitting the target with one shot is 0.4. Then event A is hitting the target in the first attempt, B - in the second. These events are joint, since it is possible that you can hit the target with both the first and second shots. But events are not dependent. What is the probability of the event of hitting the target with two shots (at least with one)? According to the formula:

0,4+0,4-0,4*0,4=0,64

The answer to the question is: “The probability of hitting the target with two shots is 64%.”

This formula for the probability of an event can also be applied to incompatible events, where the probability of the joint occurrence of an event P(AB) = 0. This means that the probability of the sum of incompatible events can be considered a special case of the proposed formula.

Geometry of probability for clarity

Interestingly, the probability of the sum of joint events can be represented as two areas A and B, which intersect with each other. As can be seen from the picture, the area of their union is equal to the total area minus the area of their intersection. This geometric explanation makes the seemingly illogical formula more understandable. Note that geometric solutions are not uncommon in probability theory.

Determining the probability of the sum of many (more than two) joint events is quite cumbersome. To calculate it, you need to use the formulas that are provided for these cases.

Dependent Events

Events are called dependent if the occurrence of one (A) of them affects the probability of the occurrence of another (B). Moreover, the influence of both the occurrence of event A and its non-occurrence is taken into account. Although events are called dependent by definition, only one of them is dependent (B). Ordinary probability was denoted as P(B) or the probability of independent events. In the case of dependent events, a new concept is introduced - conditional probability P A (B), which is the probability of a dependent event B, subject to the occurrence of event A (hypothesis), on which it depends.

But event A is also random, so it also has a probability that needs and can be taken into account in the calculations performed. The following example will show how to work with dependent events and a hypothesis.

An example of calculating the probability of dependent events

A good example for calculating dependent events would be a standard deck of cards.

Using a deck of 36 cards as an example, let’s look at dependent events. We need to determine the probability that the second card drawn from the deck will be of diamonds if the first card drawn is:

Bubnovaya.
A different color.

Obviously, the probability of the second event B depends on the first A. So, if the first option is true, that there is 1 card (35) and 1 diamond (8) less in the deck, the probability of event B:

R A (B) =8/35=0.23

If the second option is true, then the deck has 35 cards, and the full number of diamonds (9) is still retained, then the probability of the following event B:

R A (B) =9/35=0.26.

It can be seen that if event A is conditioned on the fact that the first card is a diamond, then the probability of event B decreases, and vice versa.

Multiplying dependent events

Guided by the previous chapter, we accept the first event (A) as a fact, but in essence, it is of a random nature. The probability of this event, namely drawing a diamond from a deck of cards, is equal to:

P(A) = 9/36=1/4

Since the theory does not exist on its own, but is intended to serve for practical purposes, it is fair to note that what is most often needed is the probability of producing dependent events.

According to the theorem on the product of probabilities of dependent events, the probability of occurrence of jointly dependent events A and B is equal to the probability of one event A, multiplied by the conditional probability of event B (dependent on A):

P(AB) = P(A) *P A(B)

Then, in the deck example, the probability of drawing two cards with the suit of diamonds is:

9/36*8/35=0.0571, or 5.7%

And the probability of extracting not diamonds first, and then diamonds, is equal to:

27/36*9/35=0.19, or 19%

It can be seen that the probability of event B occurring is greater provided that the first card drawn is of a suit other than diamonds. This result is quite logical and understandable.

Total probability of an event

When a problem with conditional probabilities becomes multifaceted, it cannot be calculated using conventional methods. When there are more than two hypotheses, namely A1, A2,…, A n, ..forms a complete group of events provided:

P(A i)>0, i=1,2,…
A i ∩ A j =Ø,i≠j.
Σ k A k =Ω.

So, the formula for the total probability for event B at full group random events A1,A2,…,And n is equal to:

A look into the future

The probability of a random event is extremely necessary in many areas of science: econometrics, statistics, physics, etc. Since some processes cannot be described deterministically, since they themselves are probabilistic in nature, it is necessary special methods work. The theory of event probability can be used in any technological field as a way to determine the possibility of an error or malfunction.

We can say that by recognizing probability, we in some way take a theoretical step into the future, looking at it through the prism of formulas.

So, let's talk about a topic that interests a lot of people. In this article I will answer the question of how to calculate the probability of an event. I will give formulas for such a calculation and several examples to make it clearer how this is done.

What is probability

Let's start with the fact that the probability that this or that event will occur is a certain amount of confidence in the eventual occurrence of some result. For this calculation, a total probability formula has been developed that allows you to determine whether the event you are interested in will occur or not, through the so-called conditional probabilities. This formula looks like this: P = n/m, the letters can change, but this does not affect the essence itself.

Examples of probability

Using a simple example, let's analyze this formula and apply it. Let's say you have a certain event (P), let it be a throw of a dice, that is, an equilateral die. And we need to calculate what is the probability of getting 2 points on it. To do this, you need the number of positive events (n), in our case - the loss of 2 points, for the total number of events (m). A roll of 2 points can only happen in one case, if there are 2 points on the dice, since otherwise the sum will be greater, it follows that n = 1. Next, we count the number of rolls of any other numbers on the dice, per 1 dice - these are 1, 2, 3, 4, 5 and 6, therefore, there are 6 favorable cases, that is, m = 6. Now, using the formula, we make a simple calculation P = 1/6 and we find that the roll of 2 points on the dice is 1/6, that is, the probability of the event is very low.

Let's also look at an example using colored balls that are in a box: 50 white, 40 black and 30 green. You need to determine what is the probability of drawing a green ball. And so, since there are 30 balls of this color, that is, there can only be 30 positive events (n = 30), the number of all events is 120, m = 120 (based on the total number of all balls), using the formula we calculate that the probability of drawing a green ball is will be equal to P = 30/120 = 0.25, that is, 25% of 100. In the same way, you can calculate the probability of drawing a ball of a different color (black it will be 33%, white 42%).

When a coin is tossed, we 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 you toss a coin a large number of times - say 1000 - and count the number of times heads are thrown, we can determine the probability of heads being thrown. 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 theoretical definition probabilities. 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. Total observations - 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 maintain the quality of its products at 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.
Ministry Agriculture The US 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. Within one week, 7,815,000 households were tuned in to the popular comedy series"Everybody Loves Raymond" on CBS and 8,302,000 households tuned in popular 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 the 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 occurring, 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 probable.

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 On dice There are 6 equally probable outcomes and there is only one possibility of throwing out 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.

With a variety of rules, victory conditions, prizes, however, there are general principles calculating the probability of winning, which can be adapted to the conditions of a particular lottery. But first, it is advisable to define the terminology.

So, probability is a calculated estimate of the likelihood that a certain event will occur, which is most often expressed in the form of the ratio of the number of desired events to total number outcomes. For example, the probability of getting heads when tossing a coin is one in two.

Based on this, it is obvious that the probability of winning is the ratio of the number winning combinations to the number of all possible ones. However, we must not forget that the criteria and definitions of the concept of “winning” can also be different. For example, most lotteries use the definition of “winning”. The requirements for winning the third class are lower than for winning the first, so the probability of winning the first class is the lowest. As a rule, this winning is a jackpot.

Another significant moment in the calculations is that the probability of two related events is calculated by multiplying the probabilities of each of them. Simply put, if you flip a coin twice, the chance of getting heads each time is one in two, but the chance of getting heads both times is only one in four. In the case of three tosses, the chance will generally drop to one in eight.

Calculation of odds

Thus, to calculate the chance of winning a jackpot in an abstract lottery, where you need to correctly guess several dropped values from a certain number of balls (for example, 6 out of 36), you need to calculate the probability of each of the six balls falling out and multiply them together. Please note that as the number of balls remaining in the drum decreases, the probability of getting the desired ball changes. If for the first ball the probability that the right one will come out is 6 in 36, that is, 1 in 6, then for the second the chance is 5 in 35 and so on. In this example, the probability that the ticket will be a winner is 6x5x4x3x2x1 to 36x35x34x33x32x31, that is, 720 to 1402410240, which is equal to 1 to 1947792.

Despite these scary numbers, people win regularly all over the world. Don't forget that even if you don't take Grand Prize, there are also second and third classes, the likelihood of which is much higher. Moreover, it is obvious that the best strategy is the purchase of several tickets of the same circulation, since each additional ticket increases your chances multiple times. For example, if you buy not one ticket, but two, then the probability of winning will be twice as high: two out of 1.95 million, that is, approximately 1 in 950 thousand.