Probability of an event. Determining the probability of an event. Independence of events. Probability multiplication theorem How to find the probability of independent events

P(A)= 1 - 0,3 = 0,7.

3. Theorem for adding the probabilities of opposite events

Opposite name two incompatible events that form a complete group. If one of two opposing events is indicated by A, something else is usually denoted . Opposite event consists in the non-occurrence of an event A.

Theorem. The sum of the probabilities of opposite events is equal to one:

P(A)+P()= 1.

Example 4. The box contains 11 parts, of which 8 are standard. Find the probability that among 3 randomly extracted parts there is at least one defective one.

Solution. The problem can be solved in two ways.

1 way. The events “among the extracted parts there is at least one defective one” and “there is not a single defective part among the extracted parts” are opposite. Let us denote the first event by A, and the second through :

P(A) =1 - P( ) .

We'll find R(). The total number of ways in which 3 parts can be extracted from 11 parts is equal to the number of combinations
. The number of standard parts is 8 ; from this number of parts it is possible
ways to extract 3 standard parts. Therefore, the probability that among the extracted 3 parts there is not a single non-standard part is equal to:

According to the theorem of addition of probabilities of opposite events, the desired probability is equal to: P(A)=1 - P()=

Method 2. Event A- “among the extracted parts there is at least one defective one” - can be realized as the appearance of:

or events IN- “1 defective and 2 non-defective parts were removed”,

or events WITH- “2 defective and 1 non-defective parts were removed”,

or events D - “3 defective parts were removed.”

Then A= B+ C+ D. Since events B, C And D inconsistent, then we can apply the theorem for adding the probabilities of incompatible events:

4. Theorem for multiplying probabilities of independent events

The product of two eventsA AndIN call the event C=AB, consisting in the joint appearance (combination) of these events.

The product of several events call an event consisting of the joint occurrence of all these events. For example, event ABC consists of combining events A, B And WITH.

Two events are called independent, if the probability of one of them does not depend on the appearance or non-appearance of the other.

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

P(AB)=P(A) P(B).

Consequence. 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 :

P(A 1 A 2 ... A n ) = P(A 1 ) P(A 2 )...P(A n ).

Example 5. Find the probability of the coat of arms appearing together when two coins are tossed.

Solution. Let's denote the events: A - the appearance of the coat of arms on the first coin, IN - the appearance of the coat of arms on the second coin, WITH- appearance of the coat of arms on two coins C=AB.

Probability of the appearance of the coat of arms of the first coin :

P(A) =.

Probability of the appearance of the coat of arms of the second coin :

P(B) =.

Since events A And IN independent, then the required probability according to the multiplication theorem is equal to:

P(C)=P(AB) = P(A) P(B) = =.

Example 6. There are 3 boxes containing 10 parts each. The first box contains 8, the second 7 and the third 9 standard parts. 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 removed from the first box (event A):

P(A) =

The probability that a standard part is removed from the second box (event IN):

The probability that a standard part is removed from the third box (event WITH):

P(C)=

Since events A, B And WITH independent in the aggregate, then the desired probability (by the multiplication theorem) is equal to:

P(ABC)=P(A) P(B) P(C)= 0,8 0,70,9 = 0,504.

Example 7. Probabilities of occurrence of each of two independent events A 1 And A 2 respectively equal R 1 And R 2. Find the probability of occurrence of only one of these events.

Solution. Let us introduce event designations:

IN 1 – only the event appeared A 1 ; IN 2 – only the event appeared A 2 .

Event occurrence IN 1 is equivalent to the occurrence of an event A 1 2 (the first event appeared and the second did not appear), i.e. IN 1 = A 1 2 .

Event occurrence IN 2 is equivalent to the occurrence of an event 1 A 2 (the first event did not appear and the second one appeared), i.e. IN 1 = 1 A 2 .

Thus, to find the probability of occurrence of only one of the events A 1 or A 2 , it is enough to find the probability of the occurrence of one, no matter which of the events IN 1 And IN 2 . Events IN 1 And IN 2 are inconsistent, therefore the theorem of addition of incompatible events applies:

P(B 1 +B 2 ) = P(B 1 ) + P(B 2 ) .

Theorem

The probability of two events occurring is equal to the product of the probabilities of one of them and the conditional probability of the other, calculated under the condition that the first took place.

$P(A B)=P(A) \cdot P(B | A)$

Event $A$ is called event independent$B$ if the probability of event $A$ does not depend on whether event $B$ occurred or not. Event $A$ is called event dependent$B$ if the probability of event $A$ changes depending on whether event $B$ occurs or not.

The probability of event $A$, calculated given that another event $B$ took place, is called conditional probability of an event$A$ and is denoted by $P(A | B)$ .

The condition for the independence of event $A$ from event $B$ can be written as:

$$P(A | B)=P(A)$$

and the dependence condition is in the form:

$$P(A | B) \neq P(A)$$

Corollary 1. If event $A$ does not depend on event $B$, then event $B$ does not depend on event $A$.

Corollary 2. The probability of the product of two independent events is equal to the product of the probabilities of these events:

$$P(A B)=P(A) \cdot P(B)$$

The probability multiplication theorem can be generalized to the case of an arbitrary number of events. In general terms, it is formulated as follows.

The probability of several events occurring is equal to the product of the probabilities of these events, and the probability of each subsequent event in order is calculated provided that all previous ones took place:

$$P\left(A_(1) A_(2) \ldots A_(n)\right)=P\left(A_(1)\right) \cdot P\left(A_(2) | A_(1) \right) \cdot P\left(A_(3) | A_(1) A_(2)\right) \cdots \cdots P\left(A_(n) | A_(1) A_(2) \ldots A_( n-1)\right)$$

In the case of independent events, the theorem simplifies and takes the form:

$$P\left(A_(1) A_(2) \ldots A_(n)\right)=P\left(A_(1)\right) \cdot P\left(A_(2)\right) \cdot P\left(A_(3)\right) \cdot \ldots \cdot P\left(A_(n)\right)$$

that is, the probability of producing independent events is equal to the product of the probabilities of these events:

$$P\left(\prod_(i=1)^(n) A_(i)\right)=\prod_(i=1)^(n) P\left(A_(i)\right)$$

Examples of problem solving

Example

Exercise. There are 2 white and 3 black balls in the urn. Two balls are taken out of the urn in a row and are not returned.

Solution. Find the probability that both balls are white.

Let event $A$ be the appearance of two white balls. This event is the product of two events:

$$A=A_(1) A_(2)$$

where event $A_1$ is the appearance of a white ball during the first removal, $A_2$ is the appearance of a white ball during the second removal. Then, by the probability multiplication theorem

$$P(A)=P\left(A_(1) A_(2)\right)=P\left(A_(1)\right) \cdot P\left(A_(2) | A_(1)\ right)=\frac(2)(5) \cdot \frac(1)(4)=\frac(1)(10)=0.1$$ $0,1$

Example

Exercise. Answer.

Solution. There are 2 white and 3 black balls in the urn. Two balls are drawn in a row from the urn. After the first draw, the ball is returned to the urn and the balls in the urn are mixed. Find the probability that both balls are white.

In this case, the events $A_1$ and $A_2$ are independent, and then the required probability

$$P(A)=P\left(A_(1) A_(2)\right)=P\left(A_(1)\right) \cdot P\left(A_(2)\right)=\frac (2)(5) \cdot \frac(2)(5)=\frac(4)(25)=0.16$$

Classic definition of probability.

An event that can be represented as a collection (sum) of several elementary events is called composite.

An event that cannot be broken down into simpler ones is called elementary.

An event is called impossible if it never occurs under the conditions of a given experiment (test).

Certain and impossible events are not random.

Joint events– several events are called joint if, as a result of the experiment, the occurrence of one of them does not exclude the occurrence of others.

Incompatible events– several events are called incompatible in a given experiment if the occurrence of one of them excludes the occurrence of others. The two events are called opposite, if one of them occurs if and only if the other does not occur.

The probability of event A is P(A) – is called the number ratio m elementary events (outcomes) favorable to the occurrence of the event A, to the number n of all elementary events under the conditions of a given probabilistic experiment.

The following properties of probability follow from the definition:

1. The probability of a random event is a positive number between 0 and 1:

2. The probability of a certain event is 1: (3)

3. If an event is impossible, then its probability is equal to

4. If events are incompatible, then

5. If events A and B are joint, then the probability of their sum is equal to the sum of the probabilities of these events without the probability of their joint occurrence:

P(A+B) = P(A) + P(B) - P(AB)(6)

6. If and are opposite events, then (7)

7. Sum of event probabilities A 1, A 2, …, A n, forming a complete group, is equal to 1:

P(A 1) + P(A 2) + …+ P(A n) = 1.(8)

In economic studies, the values and in the formula may be interpreted differently. At statistical definition The probability of an event is the number of observations of the experimental results in which the event occurred exactly once. In this case the relation is called relative frequency (frequency) of an event

Events A, B are called independent, if the probability of each of them does not depend on whether another event occurred or not. The probabilities of independent events are called unconditional.

Events A, B are called dependent, if the probability of each of them depends on whether another event occurred or not. The probability of event B, calculated under the assumption that another event A has already occurred, is called conditional probability.

If two events A and B are independent, then the equalities are true:

P(B) = P(B/A), P(A) = P(A/B) or P(B/A) – P(B) = 0(9)

The probability of the product of two dependent events A, B is equal to the product of the probability of one of them by the conditional probability of the other:

P(AB) = P(B) ∙ P(A/B) or P(AB) = P(A) ∙ P(B/A) (10)

Probability of event B given the occurrence of event A:

Probability of the product of two independent events A, B is equal to the product of their probabilities:

P(AB) = P(A) ∙ P(B)(12)

If several events are pairwise independent, then it does not follow that they are independent in the aggregate.

Events A 1, A 2, …, A n (n>2) are called independent in the aggregate if the probability of each of them does not depend on whether any of the other events occurred or not.

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:

P(A 1 ∙A 2 ∙A 3 ∙…∙A n) = P(A 1)∙P(A 2)∙P(A 3)∙…∙P(A n). (13)

If two events A and B are independent, then the equalities are true:

P(B) = P(B/A), P(A) = P(A/B) or P(B/A) – P(B) = 0(9)

The probability of the product of two dependent events A, B is equal to the product of the probability of one of them by the conditional probability of the other:

P(AB) = P(B) ∙ P(A/B) or P(AB) = P(A) ∙ P(B/A) (10)

Probability of event B given the occurrence of event A:

Probability of the product of two independent events A, B is equal to the product of their probabilities:

P(AB) = P(A) ∙ P(B)(12)

If several events are pairwise independent, then it does not follow that they are independent in the aggregate.

Events A 1, A 2, …, A n (n>2) are called independent in the aggregate if the probability of each of them does not depend on whether any of the other events occurred or not.

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:

P(A 1 ∙A 2 ∙A 3 ∙…∙A n) = P(A 1)∙P(A 2)∙P(A 3)∙…∙P(A n). (13)

End of work -

This topic belongs to the section:

Lecture notes: basic concepts of probability theory and statistics used in econometrics

Kazan State.. Financial and Economic Institute.. Department of Statistics and Econometrics..

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All topics in this section:

Discrete random variable
The most complete, exhaustive description of a discrete variable is its distribution law. The distribution law of a random variable is any relation established

Continuous random variable
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The general population is the set of all possible values or realizations of the studied SV X under a given real set of conditions.

Sampling
Calculation of sample characteristics

For any SV X, in addition to determining its distribution function, it is desirable to indicate numerical characteristics, the most important of which are: - mathematical expectation;
The normal distribution (Gaussian distribution) is an extreme case of almost all real probability distributions. Therefore, it is used in a very large number of real applications of the theory

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Let SV U ~ N (0,1), SV V is a quantity independent of U, distributed according to the χ2 law with n degrees of freedom. Then the value

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Let V and W be independent SVs distributed according to the χ2 law with degrees of freedom v1 = m and v2 = n, respectively. Then the value

Point estimates and their properties
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Confidence interval for the variance of normal SV
Let X ~ N (m, σ2) and and are unknown. Let for evaluation

Verification criteria. Critical region
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Events A, B, C... are called dependent from each other if the probability of the occurrence of at least one of them changes depending on the occurrence or non-occurrence of other events. The events are called independent, if the probabilities of the appearance of each of them do not depend on the appearance or non-appearance of the others.

Conditional probability(PA (B) - conditional probability of event B relative to A) is the probability of event B, calculated under the assumption that event A has already occurred. example of conditional probability The conditional probability of event B, provided that event A has already occurred, by definition, is equal to PA (B) = P (AB) / P (A) (P (A)>0).

Multiplying the probabilities of dependent events: the probability of the joint occurrence of two 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:
P (AB) = P (A) PA (B)

Example. The collector has 3 conical and 7 elliptical rollers. The picker took one roller, and then the second. Find the probability that the first of the taken rollers is conical, and the second is elliptical.

Solution: The probability that the first roller turns out to be conical (event A), P (A) = 3 / 10. The probability that the second roller turns out to be elliptical (event B), calculated under the assumption that the first roller is conical, i.e. conditional probability RA (B) = 7 / 9.
According to the multiplication formula, the desired probability is P (AB) = P (A) PA (B) = (3 / 10) * (7 / 9) = 7 / 30. Note that, keeping the notation, we can easily find: P (B) = 7 / 10, РB (A) = 3/9, Р (В) РB (А) = 7 / 30

Condition for the independence of events. Multiplying the probabilities of independent events. Examples.

Event B does not depend on event A if

P(B/A) = P(B) i.e. The probability of event B does not depend on whether event A occurs.

In this case, event A does not depend on event B, that is, the property of independence of events is mutual.

The probability of the product of two independent events is equal to the product of their probabilities:

P(AB) = P(A)P(B) .

Example 1: A device operating for time t consists of three nodes, each of which, independently of the others, can fail (fail) during time t. Failure of at least one node leads to failure of the device as a whole. During time t, the reliability (probability of failure-free operation) of the first node is p 1 = 0.8; second p 2 = 0.9 third p 3 = 0.7. Find the reliability of the device as a whole.

Solution. Denoting:

A – trouble-free operation of devices,

A 1 - trouble-free operation of the first node,

A 2 - trouble-free operation of the second node,

A 3 - trouble-free operation of the third node,

whence, by the multiplication theorem for independent events

P(A) = P(A 1)P(A 2)P(A 3) = p 1 p 2 p 3 = 0.504

Example 2. Find the probability of a number appearing together when two coins are tossed in one.

Solution. Probability of the appearance of the digit of the first coin (event A) P(A) = 1/2; the probability of the appearance of the digit of the second coin (event B) is P(B) = 1/2.

Events A and B are independent, so we will find the required probability

according to the formula:

P(AB) = P(A)P(B) = 1/2 *1/2 = 1/4

Compatibility and incompatibility of events. Adding the probabilities of two joint events. Examples.

The two events are called joint, if the appearance of one of them does not affect or exclude the appearance of the other. Joint events can occur simultaneously, such as, for example, the appearance of a number on one dice or

does not in any way affect the appearance of numbers on another die. Events are incompatible, if in one phenomenon or during one test they cannot be realized simultaneously and the appearance of one of them excludes the appearance of the other (hitting the target and missing are incompatible).

The probability of the occurrence of at least one of two joint events A or B is equal to the sum of the probabilities of these events without the probability of their joint occurrence:

P(A+B) = P(A)+P(B)-P(AB).

Example. The probability of hitting the target for the first athlete is 0.85, and for the second - 0.8. Athletes independently of each other

fired one shot each. Find the probability that at least one athlete will hit the target?

Solution. Let us introduce the following notations: events A - “hit by the first athlete”, B - “hit by the second athlete”, C - “hit by at least one of the athletes”. Obviously, A + B = C, and events A and B are simultaneous. In accordance with the formula we get:

P(C) = P(A) + P(B) - P(AB)

P(C) = P(A)+ P(B)-P(A)P(B),

since A and B are independent events. Substituting these values P(A) = 0.85, P(B) = 0.8 into the formula for P(C), we find the desired probability

P(C) = (0.85 + 0.8) - 0.85·0.8 = 0.97