Events a and b are called independent if. Dependent and independent events. Conditional probability. Dependent and independent random events. Basic formulas for adding and multiplying probabilities

There are dependent and independent events. Two events are called independent if the occurrence of one of them does not change the probability of the occurrence of the other. For example, if there are two automatic lines operating in a workshop, which are not interconnected due to production conditions, then stops of these lines are independent events.

Several events are called collectively independent, if any of them does not depend on any other event and on any combination of the others.

The events are called dependent, if one of them affects the probability of the other. For example, two production plants are connected by a single technological cycle. Then the probability of failure of one of them depends on the state of the other. The probability of one event B, calculated assuming the occurrence of another event A, is called conditional probability events B and are denoted by P(A|B).

The condition for the independence of event B from event A is written as P(B|A)=P(B), and the condition for its dependence as P(B|A)≠P(B).

Probability of an event in Bernoulli tests. Poisson's formula.

Repeated independent tests, Bernoulli tests or Bernoulli circuit such tests are called if for each test there are only two outcomes - the occurrence of event A or and the probability of these events remains unchanged for all tests. This simple random trial design has great importance in probability theory.

The most famous example of Bernoulli tests is the experiment with sequential throwing of a fair (symmetrical and uniform) coin, where event A is the loss of, for example, a “coat of arms” (“tails”).

Let in some experiment the probability of event A be equal to P(A)=p, then , where p+q=1. Let us perform the experiment n times, assuming that the individual trials are independent, which means that the outcome of any of them is not related to the outcomes of previous (or subsequent) trials. Let's find the probability of occurrence of events A exactly k times, say only in the first k trials. Let be the event that in n trials event A will appear exactly k times in the first trials. The event can be represented as

Since we assumed the experiments to be independent, then

41)[page2] If we ask the question about the occurrence of event A k times in n trials in random order, then the event can be represented in the form

The number of different terms on the right side of this equality is equal to the number of trials from n to k, therefore the probability of events, which we will denote, is equal to

The sequence of events forms a complete group of independent events . Indeed, from the independence of events we obtain

If, when an event occurs, the probability of the event does not change, then events And are called independent.

Theorem:Probability of co-occurrence of two independent events And (works And ) is equal to the product of the probabilities of these events.

Indeed, since events And are independent, then
. In this case, the formula for the probability of events occurring is And takes on the form.

Events
are called pairwise independent, if any two of them are independent.

Events
are called jointly independent (or simply independent), if every two of them are independent and every event and all possible products of the others are independent.

Theorem:Probability of the product of a finite number of independently independent events
is equal to the product of the probabilities of these events.

Let us illustrate the difference in the application of formulas for the probability of a product of events for dependent and independent events using examples

Example 1. The probability of the first shooter hitting the target is 0.85, the second 0.8. The guns fired one shot each. What is the probability that at least one shell hit the target?

Solution: P(A+B) =P(A) +P(B) –P(AB) Since the shots are independent, then

P(A+B) = P(A) +P(B) –P(A)*P(B) = 0.97

Example 2. The urn contains 2 red and 4 black balls. 2 balls are taken out of it in a row. What is the probability that both balls are red?

Solution: 1 case. Event A is the appearance of a red ball on the first draw, event B on the second. Event C – the appearance of two red balls.

P(C) =P(A)*P(B/A) = (2/6)*(1/5) = 1/15

Case 2. The first ball drawn is returned to the basket

P(C) =P(A)*P(B) = (2/6)*(2/6) = 1/9

Total probability formula.

Let the event can only happen with one of the incompatible events
, forming a complete group. For example, a store receives the same products from three enterprises and in different quantities. The likelihood of producing low-quality products at these enterprises varies. One of the products is randomly selected. It is required to determine the probability that this product is of poor quality (event ). Events here
– this is the selection of a product from the products of the corresponding enterprise.

In this case, the probability of the event can be considered as the sum of the products of events
.

Using the theorem for adding the probabilities of incompatible events, we obtain
. Using the probability multiplication theorem, we find

.

The resulting formula is called total probability formula.

Bayes formula

Let the event occurs simultaneously with one of incompatible events
, the probabilities of which
(
) are known before experiment ( a priori probabilities). An experiment is carried out, as a result of which the occurrence of an event is registered , and it is known that this event had certain conditional probabilities
(
). We need to find the probabilities of events
if it is known that the event happened ( a posteriori probabilities).

The problem is that, having new information (event A occurred), we need to reestimate the probabilities of events
.

Based on the theorem on the probability of the product of two events

.

The resulting formula is called Bayes formulas.

Basic concepts of combinatorics.

When solving a number of theoretical and practical problems, it is necessary to create various combinations from a finite set of elements according to given rules and count the number of all possible such combinations. Such tasks are usually called combinatorial.

When solving problems, combinatorists use the rules of sum and product.

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

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

Problems of a different type. Several events are given, for example, a coin is tossed three times. You need to find the probability that either the coat of arms will appear all three times, or that the coat of arms will appear at least once. This is a probability multiplication problem.

Addition of probabilities of incompatible 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. It 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:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on 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 dice 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 can 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 will win 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:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on 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.

Wherein 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 Are there any 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 full deck 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.

Dependent and independent random events.
Basic formulas for adding and multiplying probabilities

The concepts of dependence and independence of random events. Conditional probability. Formulas for adding and multiplying probabilities for dependent and independent random events. Total probability formula and Bayes formula.

Probability addition theorems

Let us find the probability of the sum of events and (assuming their compatibility or incompatibility).

Theorem 2.1. The probability of the sum of a finite number of incompatible events is equal to the sum of their probabilities:

Example 1. The probability that a store will sell a pair of size 44 men's shoes is 0.12; 45th - 0.04; 46th and above - 0.01. Find the probability that a pair of men's shoes of at least size 44 will be sold.

Solution. The desired event will occur if a pair of shoes of size 44 (event) or 45 (event) or at least size 46 (event) is sold, i.e. the event is the sum of events. Events are incompatible. Therefore, according to the sum of probabilities theorem, we obtain

Example 2. Under the conditions of Example 1, find the probability that the next pair of shoes smaller than size 44 will be sold.

Solution. The events “the next pair of shoes smaller than size 44 will be sold” and “a pair of shoes no smaller than size 44 will be sold” are opposite. Therefore, according to formula (1.2), the probability of the occurrence of the desired event

because as was found in example 1.

Theorem 2.1 of addition of probabilities is valid only for incompatible events. Using it to find the probability of joint events can lead to incorrect and sometimes absurd conclusions, as is clearly seen in the following example. Let the execution of an order on time by Electra Ltd be estimated with a probability of 0.7. What is the probability that out of three orders the company will complete at least one on time? We denote the events that the company will complete the first, second, and third orders on time, respectively. If we apply Theorem 2.1 of addition of probabilities to find the desired probability, we obtain . The probability of the event turned out to be greater than one, which is impossible. This is because the events are joint. Indeed, fulfilling the first order on time does not exclude the fulfillment of the other two on time.

Let us formulate a theorem for adding probabilities in the case of two joint events (the probability of their joint occurrence will be taken into account).

Theorem 2.2. The probability of the sum of two joint events is equal to the sum of the probabilities of these two events without the probability of their joint occurrence:

Dependent and independent events. Conditional probability

There are dependent and independent events. Two events are called independent if the occurrence of one of them does not change the probability of the occurrence of the other. For example, if there are two automatic lines operating in a workshop, which are not interconnected due to production conditions, then stops of these lines are independent events.

Example 3. The coin is tossed twice. The probability of the “coat of arms” appearing in the first trial (event ) does not depend on the appearance or non-appearance of the “coat of arms” in the second trial (event ). In turn, the probability of the “coat of arms” appearing in the second test does not depend on the result of the first test. Thus, events are both independent.

Several events are called collectively independent, if any of them does not depend on any other event and on any combination of the others.

The events are called dependent, if one of them affects the probability of the other. For example, two production plants are connected by a single technological cycle. Then the probability of failure of one of them depends on the state of the other. The probability of one event calculated assuming the occurrence of another event is called conditional probability events and is denoted by .

The condition of independence of an event from an event is written in the form , and the condition of its dependence - in the form . Let's consider an example of calculating the conditional probability of an event.

Example 4. The box contains 5 cutters: two worn and three new. Two sequential extractions of the incisors are performed. Determine the conditional probability of a worn cutter appearing during the second extraction, provided that the cutter removed the first time is not returned to the box.

Solution. Let's denote the extraction of a worn-out cutter in the first case, and - the extraction of a new one. Then . Since the removed cutter is not returned to the box, the ratio between the quantities of worn and new cutters changes. Consequently, the probability of removing a worn cutter in the second case depends on what event took place before.

Let us denote the event that means the removal of a worn cutter in the second case. The probabilities of this event could be:

Therefore, the probability of an event depends on whether the event occurred or not.

Probability multiplication formulas

Let the events be independent, and the probabilities of these events are known. Let's find the probability of combining the events and .

Theorem 2.3. The probability of the joint occurrence of two independent events is equal to the product of the probabilities of these events:

Corollary 2.1. The probability of the joint occurrence of several events that are independent in the aggregate is equal to the product of the probabilities of these events:

Example 5. Three boxes contain 10 parts. The first box contains 8 standard parts, the second – 7, and the third – 9. One part is taken out at random from each box. Find the probability that all three parts taken out will be standard.

Solution. The probability that a standard part is taken from the first box (event), . The probability that a standard part is taken from the second box (event), . The probability that a standard part is taken from the third box (event), . Since the events , and are independent in the aggregate, the desired probability (by the multiplication theorem)

Let the events be dependent, and the probabilities are known. Let us find the probability of the occurrence of these events, i.e. the probability that both the event and the event will appear.

Theorem 2.4. The probability of the joint occurrence of two dependent events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the assumption that the first event has already occurred:

Corollary 2.2. The probability of the joint occurrence of several dependent events is equal to the product of the probability of one of them and the conditional probabilities of all the others, and the probability of each subsequent event is calculated under the assumption that all previous events have already occurred.

Example 6. The urn contains 5 white balls, 4 black and 3 blue. Each test consists of drawing one ball at random without returning it to the urn. Find the probability that a white ball will appear on the first trial (event), a black ball on the second (event) and a blue ball on the third (event).

Solution. Probability of occurrence white ball at the first test. The probability of a black ball appearing on the second trial, calculated under the assumption that a white ball appeared on the first trial, i.e. conditional probability. Probability of occurrence blue ball on the third trial, calculated under the assumption that a white ball appeared on the first trial, and a black one on the second, . Required probability

Total Probability Formula

Theorem 2.5. If an event occurs only under the condition of the occurrence of one of the events forming a complete group of incompatible events, then the probability of the event is equal to the sum of the products of the probabilities of each of the events by the corresponding conditional probability of the event:

In this case, events are called hypotheses, and probabilities are called a priori. This formula is called the total probability formula.

Example 7. The assembly line receives parts from three machines. The productivity of the machines is not the same. The first machine produces 50% of all parts, the second - 30%, and the third - 20%. The probability of a high-quality assembly when using a part manufactured on the first, second and third machine is 0.98, 0.95 and 0.8, respectively. Determine the probability that the assembly coming off the assembly line is of high quality.

Solution. Let us denote the event that signifies the validity of the assembled unit; , and are events meaning that the parts are made on the first, second and third machine, respectively. Then

Required probability

Bayes formula

This formula is used when solving practical problems when an event appearing together with any of the events forming a complete group of events has occurred and it is necessary to carry out a quantitative re-estimation of the probabilities of hypotheses. A priori (before experiment) probabilities are known. It is required to calculate posterior (after experience) probabilities, i.e., essentially, you need to find conditional probabilities. For a hypothesis, Bayes' formula looks like this.

Definition 1. Event A is said to be dependent on event B if the probability of occurrence of event A depends on whether event B occurred or did not occur. The probability that event A occurred given that event B occurred will be denoted and called the conditional probability of event A subject to B.

Example 1. There are 3 white balls and 2 black balls in the urn. One ball is drawn from the urn (first draw) and then a second ball (second draw). Event B is the appearance of a white ball during the first draw. Event A is the appearance of a white ball during the second draw.

Obviously, the probability of event A, if event B occurs, will be

The probability of event A, provided that event B did not occur (a black ball appeared during the first draw), will be

We see that

Theorem 1. The probability of combining two events is equal to the product of the probability of one of them by the conditional probability of the second, calculated under the condition that the first event occurred, i.e.

Proof. We present the proof for events that reduce to the urn pattern (i.e., in the case when the classical definition of probability is applicable).

Let there be balls in the urn, white and black. Suppose that among the white balls there are balls marked “asterisk”, the rest are pure white (Fig. 408).

One ball is drawn from the urn. What is the probability of the event of taking out a white ball marked “star”?

Let B be an event consisting in the appearance of a white ball, A be an event consisting in the appearance of a ball marked with an asterisk. Obviously,

The probability of a white ball with an asterisk appearing, given that a white ball appears, will be

The probability of a white ball with a star appearing is P (A and B). Obviously,

Substituting the left-hand sides of expressions (2), (3) and (4) into (5), we obtain

Equality (1) has been proven.

If the events under consideration do not fit into the classical scheme, then formula (1) serves to determine the conditional probability. Namely, the conditional probability of event A given the occurrence of event B is determined using

Remark 1. Apply the last formula to the expression:

In equalities (1) and (6), the left sides are equal, since this is the same probability; therefore, the right sides are also equal. Therefore we can write the equality

Example 2. For the case of example 1 given at the beginning of this section, we have According to formula (1) we obtain Probability P(A and B) is easily calculated and directly.

Example 3. The probability of producing a suitable product using this machine is 0.9. The probability of a 1st grade product appearing among suitable products is 0.8. Determine the probability of producing a 1st grade product using this machine.

Solution. Event B is the production of a suitable product using this machine, event A is the appearance of a 1st grade product. Here, substituting into formula (1), we obtain the desired probability

Theorem 2. If event A can only occur if one of the events that form a complete group of incompatible events occurs, then the probability of event A is calculated by the formula

Formula (8) is called the total probability formula. Proof. Event A can occur when any of the combined events occur

Therefore, by the theorem on the addition of probabilities we obtain

Replacing the terms on the right side according to formula (1), we obtain equality (8).

Example 4. Three consecutive shots are fired at the target. Probability of hitting with the first shot with the second with the third With one hit, the probability of hitting the target with two hits, with three hits Determine the probability of hitting the target with three shots (event A).

Solution. Let's consider the complete group of incompatible events:

There was one hit;