Probability theory. Problem Solving (2020). Rule for adding probabilities of incompatible events

We already know that probability is a numerical measure of the possibility of a random event occurring, i.e. an event that may or may not occur if a certain set of conditions are met. When a set of conditions changes, the probability of a random event may change. As an additional condition, we can consider the occurrence of another event. So, if to the set of conditions under which a random event occurs A, add one more, consisting in the occurrence of a random event IN, then the probability of the event occurring A will be called conditional.

Conditional probability of event A- the probability of occurrence of event A, provided that event B occurs. Conditional probability is denoted by (A).

Example 16. There are 7 white and 5 black balls in the box, differing only in color. The experiment consists of randomly taking out one ball and, without putting it back, take out another ball. What is the probability that the second ball drawn is black if the first ball drawn is white?

Solution.

Before us are two random events: event A– the first ball drawn turned out to be white, IN– the second ball drawn is black. A and B are incompatible events, let’s use the classical definition of probability. The number of elementary outcomes when drawing the first ball is 12, and the number of favorable outcomes of getting the white ball is 7. Therefore, the probability P(A) = 7/12.

If the first ball turns out to be white, then the conditional probability of the event IN- the appearance of the second black ball (provided that the first ball was white) is equal to (IN)= 5/11, since before the second ball is taken out there are 11 balls left, of which 5 are black.

Note that the probability of a black ball appearing on the second draw would not depend on the color of the first ball taken out if, after removing the first ball, we put it back in the box.

Consider two random events A and B. Let the probabilities P(A) and (B) be known. Let us determine the probability of occurrence of both event A and event B, i.e. products of these events.

Probability multiplication theorem. The probability of two events occurring is equal to the product of the probability of one of them and the conditional probability of the other, calculated under the condition that the first event occurred:

P(A×B) = P(A)×(B) .

Since for calculating the probability of a product it does not matter which of the considered events A And IN which was the first and which was the second, we can write:

P(A×B) = P(A) × (B) = P(B) × (A).

The theorem can be extended to a product of n events:

P(A 1 A 2 . A p) = P(A x) P(A 2 /A 1) .. P(A p /A 1 A 2 ... A p-1).

Example 17. For the conditions of the previous example, calculate the probability of drawing two balls: a) the white ball first, and the black ball second; b) two black balls.

Solution.

a) From the previous example, we know the probabilities of getting the white ball out of the box first and the black ball second, provided that the white ball was pulled out first. To calculate the probability of both events occurring together, we use the probability multiplication theorem: P(A×B) = P(A) × (B)= .

b) Similarly, we calculate the probability of drawing two black balls. Probability of getting the black ball first . The probability of drawing a black ball a second time, provided that we do not put the first black ball taken out back into the box (there are 4 black balls left, and there are 11 balls in total). The resulting probability can be calculated using the formula P(A×B)= P(A) × (B) 0,152.

The probability multiplication theorem has a simpler form if events A and B are independent.

Event B is said to be independent of event A if the probability of event B does not change whether event A occurs or not. If event B is independent of event A, then its conditional probability (B) is equal to the usual probability P(B):

It turns out that if the event IN will be independent of the event A, then the event A will be independent of IN, i.e. (A)= P(A).

Let's prove it. Let's substitute the equality from the definition of independence of an event IN from the event A to the probability multiplication theorem: P(A×B) = P(A)× (B)= P(A)× (B). But in other way P(A×B)= P(B) × (A). Means P(A) × (B)= P(B) × (A) And (A)= P(A).

Thus, the property of independence (or dependence) of events is always mutual and the following definition can be given: two events are called independent, if the appearance of one of them does not change the probability of the appearance of the other.

It should be noted that the independence of events is based on the independence physical nature their origin. This means that the sets of random factors leading to one or another outcome of testing one and another random event are different. So, for example, hitting a target with one shooter does not affect in any way (unless, of course, you come up with any exotic reasons) on the probability of hitting the target with a second shooter. In practice, independent events occur very often, since the causal relationship of phenomena in many cases is absent or insignificant.

Probability multiplication theorem for independent events. The probability of the product of two independent events is equal to the product of the probability of these events: P(A×B) = P(A) × P(B).

The following corollary follows from the probability multiplication theorem for independent events.

If events A and B are incompatible and P(A)¹0, P(B)¹0, then they are dependent.

Let's prove this by contradiction. Let us assume that incompatible events A And IN independent. Then P(A×B) = P(A) ×P(B). And since P(A)¹0, P(B)¹0, i.e. events A And IN are not impossible, then P(A×B)¹0. But, on the other hand, the event Až IN is impossible as a product of incompatible events (this was discussed above). Means P(A×B)=0. got a contradiction. Thus, our initial assumption is incorrect. Events A And IN– dependent.

Example 18. Let us now return to the unsolved problem of two shooters shooting at the same target. Let us recall that if the probability of hitting the target by the first shooter is 0.8, and the second is 0.7, it is necessary to find the probability of hitting the target.

Events A And IN– hitting the target by the first and second shooter, respectively, are joint, therefore, to find the probability of the sum of events A + IN– hitting the target with at least one shooter – you must use the formula: P(A+B)=P(A)+P(B)–P(Až IN). Events A And IN independent, therefore P(A×B) = P(A) × P(B).

So, P(A+B) = P(A) + P(B) - P(A) × P(B).

P(A+B)= 0.8 + 0.7 – 0.8×0.7 = 0.94.

Example 19.

Two independent shots are fired at the same target. The probability of a hit on the first shot is 0.6, and on the second - 0.8. Find the probability of hitting the target with two shots.

1) Let us denote a hit on the first shot as an event
A 1, with the second - as the event A 2.

Hitting the target requires at least one hit: either only with the first shot, or only with the second, or with both the first and second. Therefore, the problem requires determining the probability of the sum of two joint events A 1 and A 2:

P(A 1 + A 2) = P(A 1) + P(A 2) - P(A 1 A 2).

2) Since the events are independent, then P(A 1 A 2) = P(A 1) P(A 2).

3) We get: P(A 1 + A 2) = 0.6 + 0.8 - 0.6 0.8 = 0.92.
If the events are incompatible, then P(A B) = 0 and P(A + B) = = P(A) + P(B).

Example 20.

The urn contains 2 white, 3 red and 5 blue balls of the same size. What is the probability that a ball drawn at random from an urn will be colored (not white)?

1) Let event A be the removal of a red ball from the urn,
event B - drawing the blue ball. Then event (A + B)
there is the extraction of a colored ball from an urn.

2) P(A) = 3/10, P(B) = 5/10.

3) Events A and B are incompatible, since only
one ball. Then: P(A + B) = P(A) + P(B) = 0.3 + 0.5 = 0.8.

Example 21.

The urn contains 7 white and 3 black balls. What is the probability of: 1) drawing a white ball from the urn (event A); 2) removing a white ball from the urn after removing one ball from it, which is white (event B); 3) removing a white ball from the urn after removing one ball from it, which is black (event C)?

1) P(A) = = 0.7 (see classical probability).

2)P B (A) = = 0,(6).

3) R S (A) = | = 0,(7).

Example 22.

The mechanism is assembled from three identical parts and is considered inoperative if all three parts fail. There are 15 parts left in the assembly shop, of which 5 are non-standard (defective). What is the probability that a mechanism assembled from the remaining parts taken at random will be inoperative?

1) Let us denote the desired event by A, the choice of the first non-standard part by A 1, the second by A 2, the third by A 3

2) Event A will occur if both event A 1, event A 2, and event A 3 occur, i.e.

A = A 1 A 2 A 3,

since the logical “and” corresponds to a product (see section “Propositional Algebra. Logical Operations”).

3) Events A 1, A 2, A 3 are dependent, therefore P(A 1 A 2 A 3) =
= P(A 1) P(A 2 /A 1) P(A 3 /A 1 A 2).

4)P(A 1) = ,P(A 2 /A 1) = ,P(A 3 /A 1 A 2)= . Then

P(A 1 A 2 A 3) = 0.022.

For independent events: P(A B) = P(A) P(B).

Based on the above, the criterion for the independence of two events A and B:

P(A) = P B (A) = P (A), P (B) = P A (B) = P (B).

Example 23.

The probability of hitting the target by the first shooter (event A) is 0.9, and the probability of hitting the target by the second shooter (event B) is 0.8. What is the probability that the target will be hit by at least one shooter?

1) Let C be the event of interest to us; the opposite event is that both shooters miss.

3) Since when shooting one shooter does not interfere with the other, the events are independent.

We have: P() = P() P() = =(1 - 0.9) (1 - 0.8) =

0,1 0,2 = 0,02.

4) P(C) = 1 -P() = 1 -0.02 = 0.98.

Total Probability Formula

Let event A can occur as a result of the manifestation of one and only one event H i (i = 1,2,... n) from some complete group of incompatible events H 1, H 2,... H n. Events in this group are usually called hypotheses.

Total probability formula. The probability of event A is equal to the sum of paired products of the probabilities of all hypotheses forming a complete group by the corresponding conditional probabilities of a given event A:

P(A) = , where = 1.

Example 24.

There are 3 identical urns. The first urn contains 2 white and 1 black ball, the second urn contains 3 white and 1 black ball, and the third urn contains 2 white and 2 black balls. 1 ball is selected from a randomly selected urn. What is the probability that he will be white?

All urns are considered the same, therefore, the probability of choosing the i-th urn is

Р(H i) = 1/3, where i = 1, 2, 3.

2) Probability of drawing a white ball from the first urn: (A) = .

Probability of drawing a white ball from the second urn: (A) = .

Probability of drawing a white ball from the third urn: (A) = .

3) The required probability:

P(A) = =0.63(8)

Example 25.

The store receives products for sale from three factories, the relative shares of which are: I - 50%, II - 30%, III - 20%. For factory products, defects are respectively: I - 2%, P - 2%, III - 5%. What is the probability that a product of this product, accidentally purchased in a store, turns out to be of good quality (event A)?

1) The following three hypotheses are possible here: H 1, H 2, H 3 -
the purchased item was produced respectively at factories I, II, III; the system of these hypotheses is complete.

Probabilities: P(H 1) = 0.5; P(H 2) = 0.3; P(H 3) = 0.2.

2) The corresponding conditional probabilities of event A are: (A) = 1-0.02 = 0.98; (A) = 1-0.03 = 0.97; (A) = = 1-0.05 = 0.95.

3) According to the total probability formula, we have: P(A) = 0.5 0.98 + + 0.3 0.97 + 0.2 0.95 = 0.971.

Posterior probability formula (Bayes formula)

Let's consider the situation.

Available full group inconsistent hypotheses H 1, H 2, ... H n, the probabilities of which (i = 1, 2, ... n) are known before experiment (a priori probabilities). An experiment (test) is carried out, as a result of which the occurrence of event A is recorded, and it is known that our hypotheses assigned certain probabilities to this event (i = 1, 2, ...n). What will be the probabilities of these hypotheses after the experiment (a posteriori probabilities)?

The answer to a similar question is given by the posterior probability formula (Bayes formula):

, where i=1,2, ...p.

Example 26.

The probability of hitting an aircraft with a single shot for the 1st missile system (event A) is 0.2, and for the 2nd (event B) - 0.1. Each of the complexes fires one shot, and one hit is recorded on the plane (event C). What is the probability that the successful shot belongs to the first missile system?

Solution.

1) Before the experiment, four hypotheses are possible:

H 1 = A B - the plane is hit by the 1st complex and the plane is hit by the 2nd complex (the product corresponds to the logical “and”),

H 2 = A B - the plane is hit by the 1st complex and the plane is not hit by the 2nd complex,

H 3 = A B - the plane is not hit by the 1st complex and the plane is hit by the 2nd complex,

H 4 = A B - the plane is not hit by the 1st complex and the plane is not hit by the 2nd complex.

These hypotheses form a complete group of events.

2) Corresponding probabilities (with independent action complexes):

P(H 1) = 0.2 0.1 = 0.02;

P(H2) = 0.2 (1-0.1) = 0.18;

P(H 3) = (1-0.2) 0.1 = 0.08;

P(H 4) = (1-0.2) (1-0.1) = 0.72.

3) Since the hypotheses form a complete group of events, the equality = 1 must be satisfied.

We check: P(H 1) + P(H 2) + P(H 3) + P(H 4) = 0.02 + 0.18 + + 0.08 + 0.72 = 1, thus the group in question hypothesis is correct.

4) Conditional probabilities for the observed event C under these hypotheses will be: (C) = 0, since according to the conditions of the problem one hit was registered, and hypothesis H 1 assumes two hits:

(C) = 1; (C) = 1.

(C) = 0, since according to the conditions of the problem one hit was registered, and hypothesis H 4 assumes the absence of hits. Therefore, hypotheses H 1 and H 4 are eliminated.

5) We calculate the probabilities of hypotheses H 2 and H 3 using the Bayes formula:

0,7, 0,3.

Thus, with a probability of approximately 70% (0.7) it can be stated that the successful shot belongs to the first missile system.

5.4. Random variables. Discrete distribution law random variable

Quite often in practice, such tests are considered, as a result of which a certain number is randomly obtained. For example, when you throw a dice, you get a number of points from 1 to 6; when you take 6 cards from a deck, you get from 0 to 4 aces. Over a certain period of time (say, a day or a month), a certain number of crimes are registered in the city, a certain number of road accidents occur. A gun is fired. The projectile's flight range also takes on a random value.

In all of the tests listed above, we are faced with so-called random variables.

A numerical quantity that takes on one value or another as a result of a random test is called random variable.

The concept of a random variable plays a very important role in probability theory. If the “classical” theory of probability studied mainly random events, then modern theory Probabilities primarily deals with random variables.

In what follows, we will denote random variables by capital Latin letters X, Y, Z, etc., and their possible values ​​by the corresponding lowercase letters x, y, z. For example, if a random variable has three possible values, then we will denote them as follows: , , .

So, examples of random variables could be:

1) the number of points rolled on the top face of the die:

2) the number of aces, when taking 6 cards from the deck;

3) the number of registered crimes per day or month;

4) the number of hits on the target with four shots from a pistol;

5) the distance that a projectile will travel when fired from a gun;

6) the height of a random person.

You can notice that in the first example the random variable can take one of six possible values: 1, 2, 3, 4, 5 and 6. In the second and fourth examples, the number of possible values ​​of the random variable is five: 0, 1, 2, 3, 4 In the third example, the value of the random variable can be any (theoretically) natural number or 0. In the fifth and sixth examples, the random variable can take any real value from a certain interval ( A, b).

If a random variable can take a finite or countable set of values, then it is called discrete(discretely distributed).

Continuous A random variable is a random variable that can take all values ​​from a certain finite or infinite interval.

To specify a random variable, it is not enough to list its possible values. For example, in the second and third examples, the random variables could take the same values: 0, 1, 2, 3 and 4. However, the probabilities with which these random variables take their values ​​will be completely different. Therefore, to specify a discrete random variable, in addition to the list of all possible values, you also need to indicate their probabilities.

The correspondence between possible values ​​of a random variable and their probabilities is called law of distribution discrete random variable. , …, X=

The distribution polygon, as well as the distribution series, completely characterizes the random variable. It is one of the forms of the law of distribution.

Example 27. A coin is tossed randomly. Construct a row and a polygon for the distribution of the number of dropped coats of arms.

A random variable equal to the number of dropped coats of arms can take two values: 0 and 1. Value 1 corresponds to the event - the loss of a coat of arms, value 0 - the loss of heads. The probabilities of getting a coat of arms and getting a tail are the same and equal. Those. the probabilities with which a random variable takes the values ​​0 and 1 are equal. The distribution series looks like:

X
p

Individual tasks in mathematics

Problem 1

There are 6 white balls and 11 black balls in the urn. Two balls are drawn at random at the same time. Find the probability that both balls will be:

Solution

1) The probability that one of the drawn balls will be white is equal to the number of chances of drawing a white ball from the total sum of balls in the urn. These chances are exactly the same as the number of white balls in the urn, and the sum of all chances is equal to the sum of the white and black balls.

The probability that the second ball drawn will also be white is equal to

Since one of the white balls has already been drawn.

Thus, the probability that both balls drawn from the urn will be white is equal to the product of these probabilities, since these possibilities are independent:

.

or two black balls:

.

3) The probability that both drawn balls will be different colors is the probability that the first ball will be white and the second black or that the first ball will be black and the second ball will be white. It is equal to the sum of the corresponding probabilities.

.

Answer: 1)

2) 3) .

Problem 2

The first urn has 6 white balls, 11 black, the second urn has 5 white and 2 black. A ball is drawn at random from each urn. Find the probability that both balls will be:

1) white, 2) the same color, 3) different colors.

Solution

1) The probability that both balls will be white is equal to the product of the probability that the ball drawn from the first urn will be white by the probability that the ball drawn from the second urn will also be white:

2) The probability that both balls drawn will be the same color is the probability that both balls will be either white or black. It is equal to the sum of the probabilities - drawing two white balls or two black balls:

.

3) The probability that a ball drawn from the first urn will be white, and a ball drawn from the second urn will be black, or on the contrary, the first ball will be black and the second will be white, equal to the sum of the corresponding probabilities:

Answer: 1)

2) 3) .

Problem 3

Among the 24 lottery tickets, there are 11 winning ones. Find the probability that at least one of the 2 tickets purchased will be a winner.

Solution

The probability that at least one of the 24 tickets purchased will be a winner is equal to the difference between one and the probability that none of the tickets purchased will be a winner. And the probability that none of the purchased tickets will be winning is equal to the product of the probability that the first ticket will not be winning by the probability that the second ticket will not be winning:

Hence, the probability that at least one of the 24 tickets purchased will be a winner:

Answer:

Problem 4

The box contains 6 parts of the first grade, 5 of the second and 2 of the third. Two details are taken at random. What is the probability that they will both be of the same variety?

Solution

The required probability is the probability that both parts will be either 1st, 2nd or 3rd grade and is equal to the sum of the corresponding probabilities:

The probability that both parts taken will be first grade:

The probability that both parts taken will be second grade:

The probability that both parts taken will be third grade:

Hence the probability of pulling out 2 parts of the same type is equal to:

Answer:

Problem 5

During the hour 0 ≤ t ≤ 1 (t is time in hours), one and only one bus arrives at the stop.

Solution

The bus can arrive at any moment t, where 0 ≤ t ≤ 1 (where t is time in hours) or, which is the same, 0 ≤ t ≤ 60 (where t is time in minutes).

The passenger arrives at t = 0 and waits no more than 28 minutes.

The chances of a bus arriving at the station during this time or during the remaining 32 minutes are equally probable, so the probability that a passenger arriving at this stop at time t = 0 will have to wait for a bus no more than 28 minutes is

.

Answer:

Problem 8

The probability of the first shooter hitting the target is 0.2, the second – 0.2 and the third – 0.2. All three shooters fired simultaneously. Find the probability that:

1) only one shooter will hit the target;

2) two shooters will hit the target;

3) at least one will hit the target.

Solution

1) The probability that only one shooter will hit the target is equal to the probability of the first shooter hitting the target and the second and third missing or hitting the target by the second shooter and missing by the first and third or hitting the target by the third shooter and missing by the first and second, and therefore equal to the sum of the corresponding probabilities.

The probability that the first shooter will hit the target, and the second and third will miss, is equal to the product of these probabilities:

.

Similar probabilities of the second shooter hitting the target and the first and third missing, as well as the third hitting the target and the first and second missing:

, .

Hence, the desired probability:

.

2) The probability that two shooters will hit the target is equal to the probability of the first and second shooter hitting the target and the third missing or hitting the target by the first and third shooter and missing by the second or hitting the target by the second and third shooter and missing by the first, which means it is equal to the sum of the corresponding probabilities.

The probability that the first and second shooters will hit the target, and the third will miss, is equal to the product of these probabilities:

.

Similar probabilities of the first and third shooter hitting the target and the second missing, as well as the second and third hitting the target and missing the first.

The probability that the required part is not in any box is equal to:

The required probability is equal to

Total probability formula.

Suppose that some event A can occur together with one of the incompatible events that make up the complete group of events. Let the probabilities of these events and the conditional probabilities of the occurrence of event A be known when the event occurs H i .

Theorem. The probability of event A, which can occur together with one of the events, is equal to the sum of the paired products of the probabilities of each of these events by the corresponding conditional probabilities of the occurrence of event A.

In fact this formula full probability has already been used in solving the examples given above, for example, in the problem with a revolver.

Proof.

Because events form a complete group of events, then event A can be represented as the following sum:

Because events are incompatible, then events AH i are also incompatible. Then we can apply the theorem on adding the probabilities of incompatible events:

Wherein

Finally we get:

The theorem has been proven.

Example. One of the three shooters fires two shots. The probability of hitting the target with one shot for the first shooter is 0.4, for the second – 0.6, for the third – 0.8. Find the probability that the target will be hit twice.

The probability that the shots are fired by the first, second or third shooter is equal to .

The probability that one of the shooters firing hits the target twice is equal to:

For the first shooter:

For the second shooter:

For the third shooter:

The required probability is:

LECTURE 2.

Bayes formula. (hypothesis formula)

Let there be a complete group of inconsistent hypotheses with known probabilities of their occurrence. Let event A occur as a result of the experiment, the conditional probabilities of which for each of the hypotheses are known, i.e. the probabilities are known.

It is required to determine what probabilities the hypotheses have regarding event A, i.e. conditional probabilities.

Theorem. The probability of a hypothesis after the test is equal to the product of the probability of the hypothesis before the test and the corresponding conditional probability of the event that occurred during the test, divided by the total probability of this event.

This formula is called Bayes formula.

Proof.

Using the Probability Multiplication Theorem we get:

Then if .

To find the probability P(A) we use the total probability formula.

If before the test all hypotheses are equally probable with probability , then Bayes’ formula takes the form:

Repetition of tests.

Bernoulli's formula.

If a certain number of tests are performed, as a result of which event A may or may not occur, and the probability of the occurrence of this event in each of the tests does not depend on the results of other tests, then such tests are called independent with respect to event A.

Let us assume that event A occurs in each trial with probability P(A)=p. Let's determine the probability R t, p that as a result P test event A occurred exactly T once.

This probability can, in principle, be calculated using the theorems of addition and multiplication of probabilities, as was done in the examples discussed above. However, with a sufficiently large number of tests, this leads to very large calculations. Thus, there is a need to develop a general approach to solving the problem. This approach is implemented in Bernoulli's formula. (Jacob Bernoulli (1654 – 1705) – Swiss mathematician)

Let as a result P independent tests conducted under identical conditions, event A occurs with probability P(A) = p, and the opposite event with probability .

Let's denote A i– occurrence of event A in trial number i. Because the experimental conditions are the same, then these probabilities are equal.

If as a result P experiments, event A occurs exactly T once, then the rest p-t since this event does not occur. Event A may occur T once every P tests in various combinations, the number of which is equal to the number of combinations of P elements by T. This number of combinations is found by the formula:

The probability of each combination is equal to the product of the probabilities:

Applying the theorem for adding the probabilities of incompatible events, we obtain Bernoulli formula:

Bernoulli's formula is important because it is valid for any number of independent tests, i.e. the very case in which the laws of probability theory are most clearly manifested.

Example. 5 shots are fired at the target. The hit probability for each shot is 0.4. Find the probability that the target was hit at least three times.

The probability of at least three hits is the sum of the probability of five hits, four hits and three hits.

Because shots are independent, then we can apply Bernoulli’s formula for the probability that in T trials event in probability R comes exactly P once.

In case of five hits out of five possible:

Four hits out of five shots:

Three hits out of five:

Finally, we get the probability of at least three hits out of five shots:

Random variables.

Above, we considered random events that are a qualitative characteristic of a random result of an experiment. To obtain a quantitative characteristic, the concept of a random variable is introduced.

Definition. Random variable is a quantity that, as a result of experiment, can take on one or another value, and which one is known in advance.

Random variables can be divided into two categories.

Definition. Discrete random variable is a quantity that, as a result of experience, can take on certain values ​​with a certain probability, forming a countable set (a set whose elements can be numbered).

This set can be either finite or infinite.

For example, the number of shots before the first hit on the target is a discrete random variable, because this quantity can take on an infinite, albeit countable, number of values.

Definition. Continuous random variable is a quantity that can take any value from some finite or infinite interval.

Obviously, the number of possible values ​​of a continuous random variable is infinite.

To specify a random variable, it is not enough to simply indicate its value; you must also indicate the probability of this value.

Distribution law of a discrete random variable.

Definition. The relationship between possible values ​​of a random variable and their probabilities is called discrete distribution law random variable.

The distribution law can be specified analytically, in the form of a table or graphically.

The table of correspondence between the values ​​of a random variable and their probabilities is called near distribution.

The graphical representation of this table is called distribution polygon. In this case, the sum of all ordinates of the distribution polygon represents the probability of all possible values ​​of the random variable, and, therefore, is equal to one.

Example. 5 shots are fired at the target. The hit probability for each shot is 0.4. Find the probability of the number of hits and construct a distribution polygon.

The probabilities of five hits out of five possible, four out of five and three out of five were found above using the Bernoulli formula and are equal, respectively:

Similarly we find:

Let us graphically represent the dependence of the number of hits on their probabilities.

When constructing a distribution polygon, one must remember that the connection of the resulting points is conditional. In the intervals between the values ​​of a random variable, probability does not take on any significance. The dots are connected for clarity only.

Example. The probability of at least one hit by a shooter with three shots is 0.875. Find the probability of hitting the target with one shot.

If we designate R– the probability of a shooter hitting the target with one shot, then the probability of missing with one shot is obviously equal to (1 – R).

The probability of three misses out of three shots is (1 – R) 3 . This probability is equal to 1 – 0.875 = 0.125, i.e. They don't hit the target even once.

We get:

Example. The first box contains 10 balls, 8 of which are white; The second box contains 20 balls, 4 of which are white. One ball is drawn at random from each box, and then one ball is drawn at random from these two balls. Find the probability that this ball is white.

The probability that the ball taken from the first box is white is that it is not white.

The probability that the ball taken from the second box is white - that it is not white -

The probability that a ball drawn from the first box is reselected and the probability that a ball drawn from the second box is reselected are 0.5.

The probability that a ball drawn from the first box is re-selected and it is white is

The probability that the ball drawn from the second box is re-selected and it is white is

The probability that a white ball will be selected again is

Example. There are five rifles, three of which are equipped with a telescopic sight. The probability that a shooter will hit the target when firing from a rifle with an optical sight is 0.95; for a rifle without an optical sight, this probability is 0.7. Find the probability that the target will be hit if the shooter fires one shot from a randomly selected rifle.

We denote the probability that a rifle with an optical sight is selected, and the probability that a rifle without an optical sight is selected is denoted by .

The probability that a rifle with an optical sight was chosen, and the target was hit, where R(PC/O) – the probability of hitting a target with a rifle with an optical sight.

Similarly, the probability that a rifle without an optical sight was selected and the target was hit is , where R(PC/BO) – the probability of hitting a target with a rifle without an optical sight.

The final probability of hitting the target is equal to the sum of the probabilities P 1 And R 2, because To hit the target, it is enough for one of these incompatible events to occur.

Example. Three hunters simultaneously shot at the bear, which was killed by one bullet. Determine the probability that the bear was killed by the first shooter if the hit probabilities for these shooters are 0.3, 0.4, 0.5, respectively.

In this problem, you need to determine the probability of a hypothesis after the event has already occurred. To determine the desired probability, you need to use Bayes' formula. In our case it looks like:

In this formula N 1, N 2, N 3– hypotheses that the bear will be killed by the first, second and third shooter, respectively. Before the shots are fired, these hypotheses are equally probable and their probability is equal.

P(H 1 /A)– the probability that the bear was killed by the first shooter, provided that the shots have already been fired (event A).

The probabilities that the bear will be killed by the first, second or third shooter, calculated before the shots are fired, are equal respectively:

Here q 1= 0,7; q 2 = 0,6; q 3= 0.5 – miss probabilities for each shooter, calculated as q = 1 – p, Where R– hit probability for each shooter.

Let's substitute these values ​​into Bayes' formula:

Example. Four radio signals are sent sequentially. The probabilities of receiving each of them do not depend on whether the other signals are received or not. The probabilities of receiving signals are 0.2, 0.3, 0.4, 0.5, respectively. Determine the probability of receiving three radio signals.

The event of receiving three signals out of four is possible in four cases:

To receive three signals, one of the events A, B, C or D must occur. Thus, we find the required probability:

Example. Twenty exam papers contain two questions that are not repeated. The examinee only knows the answers to 35 questions. Determine the probability that the exam will be passed if it is enough to answer two questions on one ticket or one question on one ticket and the specified additional question on another ticket.

There are a total of 40 questions (2 in each of the 20 tickets). The probability that a question comes up for which the answer is known is obviously equal to .

In order to pass the exam, one of three events must occur:

1) Event A - answered the first question (probability) and answered the second question (probability). Because After successfully answering the first question, there are still 39 questions left, 34 of which the answers are known.

2) Event B - the first question was answered (probability), the second - no (probability), the third - answered (probability).

3) Event C - the first question was not answered (probability), the second was answered (probability), the third was answered (probability).

The probability that, under given conditions, the exam will be passed is:

Example. There are two batches of similar parts. The first batch consists of 12 parts, 3 of which are defective. The second batch consists of 15 parts, 4 of which are defective. Two parts are removed from the first and second batches. What is the probability that there are no defective parts among them?

The probability of being not defective for the first part removed from the first batch is equal to , for the second part removed from the first batch, provided that the first part was not defective - .

The probability of being not defective for the first part removed from the second batch is equal to , for the second part removed from the second batch, provided that the first part was not defective - .

The probability that there are no defective parts among the four extracted parts is equal to:

Let's look at the same example, but with a slightly different condition.

Example. There are two batches of similar parts. The first batch consists of 12 parts, 3 of which are defective. The second batch consists of 15 parts, 4 of which are defective. 5 parts are selected at random from the first batch, and 7 parts from the second batch. These details form new batch. What is the probability of getting a defective part from them?

In order for a part selected at random to be defective, one of two incompatible conditions must be met:

1) The selected part was from the first batch (probability - ) and at the same time it was defective (probability - ). Finally:

2) The selected part was from the second batch (probability - ) and at the same time it was defective (probability - ). Finally:

Finally, we get: .

Example. There are 3 white and 5 black balls in the urn. Two balls are drawn at random from the urn. Find the probability that these balls are not the same color.

The event that the selected balls of different colors will occur in one of two cases:

1) The first ball is white (probability - ), and the second is black (probability - ).

2) The first ball is black (probability - ), and the second is white (probability - ).

Finally we get:

Binomial distribution.

If produced P independent trials, in each of which event A can occur with the same probability R in each of the trials, then the probability that the event will not appear is equal to q = 1 – p.

Let us take the number of occurrences of an event in each test as a certain random variable X.

To find the distribution law of this random variable, it is necessary to determine the values ​​of this variable and their probabilities.

The values ​​are quite easy to find. Obviously, as a result P tests, the event may not appear at all, appear once, twice, three times, etc. before P once.

The probability of each value of this random variable can be found using Bernoulli's formula.

This formula analytically expresses the desired distribution law. This distribution law is called binomial.

Example. The batch contains 10% non-standard parts. 4 parts were selected at random. Write a binomial distribution law for a discrete random variable X - the number of non-standard parts among the four selected ones and construct a polygon of the resulting distribution.

The probability of a non-standard part appearing in each case is 0.1.

Let us find the probabilities that among the selected parts:

1) There are no non-standard ones at all.

2) One is non-standard.

3) Two non-standard parts.

4) Three non-standard parts.

5) Four non-standard parts.

Let's construct a distribution polygon.

Example. Two dice are thrown 2 times at the same time. Write a binomial law for the distribution of a discrete random variable X - the number of occurrences of an even number of points on two dice.

Each die has three options for even points - 2, 4 and 6 out of six possible, so the probability of getting an even number of points on one die is 0.5.

The probability of getting even points on two dice at the same time is 0.25.

The probability of getting even points on both dice in two trials is equal.

Example 1. In the first urn: three red, one white balls. In the second urn: one red, three white balls. A coin is tossed at random: if it is a coat of arms, it is chosen from the first urn, otherwise, from the second.
Solution:
a) the probability that a red ball was drawn
A – got a red ball
P 1 – the coat of arms fell, P 2 - otherwise

b) The red ball is selected. Find the probability that it is taken from the first urn from the second urn.
B 1 – from the first urn, B 2 – from the second urn
,

Example 2. There are 4 balls in a box. Can be: only white, only black or white and black. (Composition unknown).
Solution:
A – probability of a white ball appearing
a) All white:
(the probability that you got one of the three options where there are white ones)
(probability of a white ball appearing where everyone is white)

b) Pulled out where everyone is black



c) pulled out the option where everyone is white and/or black

- at least one of them is white

P a +P b +P c =

Example 3. There are 5 white and 4 black balls in an urn. 2 balls are taken out of it in a row. Find the probability that both balls are white.
Solution:
5 white, 4 black balls
P(A 1) – the white ball was taken out

P(A 2) – probability that the second ball is also white

P(A) – white balls chosen in a row

Example 3a. The pack contains 2 fake and 8 real banknotes. 2 bills were pulled out of the pack in a row. Find the probability that both of them are fake.
Solution:
P(2) = 2/10*1/9 = 1/45 = 0.022

Example 4. There are 10 bins. There are 9 urns with 2 black and 2 white balls. There are 5 whites and 1 black in 1 urn. A ball was drawn from an urn taken at random.
Solution:
P(A) - ? a white ball is taken from an urn containing 5 white
B – probability of being drawn from an urn containing 5 whites
, - taken out from others
C 1 – probability of a white ball appearing at level 9.

C 2 – probability of a white ball appearing, where there are 5 of them

P(A 0)= P(B 1) P(C 1)+P(B 2) P(C 2)

Example 5. 20 cylindrical rollers and 15 cone-shaped ones. The picker takes 1 roller, and then another.
Solution:
a) both rollers are cylindrical
P(C 1)=; P(Ts 2)=
C 1 – first cylinder, C 2 – second cylinder
P(A)=P(Ts 1)P(Ts 2) =
b) At least one cylinder
K 1 – first cone-shaped.
K 2 - second cone-shaped.
P(B)=P(Ts 1)P(K 2)+P(Ts 2)P(K 1)+P(Ts 1)P(Ts 2)
;

c) the first cylinder, but not the second
P(C)=P(C 1)P(K 2)

e) Not a single cylinder.
P(D)=P(K 1)P(K 2)

e) Exactly 1 cylinder
P(E)=P(C 1)P(K 2)+P(K 1)P(K 2)

Example 6. There are 10 standard parts and 5 defective parts in a box.
Three parts are drawn at random
a) One of them is defective
P n (K)=C n k ·p k ·q n-k ,
P – probability of defective products

q – probability of standard parts

n=3, three parts


b) two out of three parts are defective P(2)
c) at least one standard
P(0) - no defective

P=P(0)+ P(1)+ P(2) - probability that at least one part will be standard

Example 7. The 1st urn contains 3 white and black balls, and the 2nd urn contains 3 white and 4 black balls. 2 balls are transferred from the 1st urn to the 2nd without looking, and then 2 balls are drawn from the 2nd. What is the probability that they are different colors?
Solution:
When moving balls from the first urn, the following options are possible:
a) took out 2 white balls in a row
P BB 1 =
In the second step there will always be one less ball, since in the first step one ball was already taken out.
b) took out one white and one black ball
The situation when the white ball is drawn first, and then the black one
P warhead =
The situation when the black ball was drawn first, and then the white one
P BW =
Total: P warhead 1 =
c) took out 2 black balls in a row
P HH 1 =
Since 2 balls were transferred from the first urn to the second urn, the total number of balls in the second urn will be 9 (7 + 2). Accordingly, we will look for all possible options:
a) first a white and then a black ball was taken from the second urn

P BB 2 P BB 1 - means the probability that first a white ball was drawn, then a black ball, provided that 2 white balls were drawn from the first urn in a row. That is why the number of white balls in this case is 5 (3+2).
P BC 2 P BC 1 - means the probability that first a white ball was drawn, then a black ball, provided that white and black balls were drawn from the first urn. That is why the number of white balls in this case is 4 (3+1), and the number of black balls is five (4+1).
P BC 2 P BC 1 - means the probability that first a white ball was drawn, then a black ball, provided that both black balls were drawn from the first urn in a row. That is why the number of black balls in this case is 6 (4+2).

The probability that 2 balls drawn will be of different colors is equal to:

Answer: P = 0.54

Example 7a. From the 1st urn containing 5 white and 3 black balls, 2 balls were randomly transferred to the 2nd urn containing 2 white and 6 black balls. Then 1 ball was drawn at random from the 2nd urn.
1) What is the probability that the ball drawn from the 2nd urn turns out to be white?
2) The ball taken from the 2nd urn turned out to be white. Calculate the probability that balls of different colors were moved from the 1st urn to the 2nd.
Solution.
1) Event A - the ball drawn from the 2nd urn turns out to be white. Let's consider the following options for the occurrence of this event.
a) Two white balls were placed from the first urn into the second: P1(bb) = 5/8*4/7 = 20/56.
There are a total of 4 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(4) = 20/56*(2+2)/(6+2) = 80/448
b) White and black balls were placed from the first urn into the second: P1(bch) = 5/8*3/7+3/8*5/7 = 30/56.
There are a total of 3 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(3) = 30/56*(2+1)/(6+2) = 90/448
c) Two black balls were placed from the first urn into the second: P1(hh) = 3/8*2/7 = 6/56.
There are a total of 2 white balls in the second urn. Then the probability of drawing a white ball from the second urn is P2(2) = 6/56*2/(6+2) = 12/448
Then the probability that the ball drawn from the 2nd urn turns out to be white is:
P(A) = 80/448 + 90/448 + 12/448 = 13/32

2) The ball taken from the 2nd urn turned out to be white, i.e. the total probability is P(A)=13/32.
Probability that balls of different colors (black and white) were placed in the second urn and white was chosen: P2(3) = 30/56*(2+1)/(6+2) = 90/448
P = P2(3)/ P(A) = 90/448 / 13/32 = 45/91

Example 7b. The first urn contains 8 white and 3 black balls, the second urn contains 5 white and 3 black balls. One ball is chosen at random from the first, and two balls from the second. After this, one ball is taken at random from the selected three balls. This last ball turned out to be black. Find the probability that a white ball is drawn from the first urn.
Solution.
Let's consider all variants of event A - out of three balls, the drawn ball turns out to be black. How could it happen that among the three balls there was a black one?
a) A black ball was taken from the first urn, and two white balls were taken from the second urn.
P1 = (3/11)(5/8*4/7) = 15/154
b) A black ball was taken from the first urn, two black balls were taken from the second urn.
P2 = (3/11)(3/8*2/7) = 9/308
c) A black ball was taken from the first urn, one white and one black ball were taken from the second urn.
P3 = (3/11)(3/8*5/7+5/8*3/7) = 45/308
d) A white ball was taken from the first urn, and two black balls were taken from the second urn.
P4 = (8/11)(3/8*2/7) = 6/77
e) A white ball was taken from the first urn, one white and one black ball were taken from the second urn.
P5 = (8/11)(3/8*5/7+5/8*3/7) = 30/77
The total probability is: P = P1+P2+ P3+P4+P5 = 15/154+9/308+45/308+6/77+30/77 = 57/77
The probability that a white ball is drawn from a white urn is:
Pb(1) = P4 + P5 = 6/77+30/77 = 36/77
Then the probability that a white ball was chosen from the first urn, given that a black ball was chosen from three balls, is equal to:
Pch = Pb(1)/P = 36/77 / 57/77 = 36/57

Example 7c. The first urn contains 12 white and 16 black balls, the second urn contains 8 white and 10 black balls. At the same time, a ball is drawn from the 1st and 2nd urns, mixed and returned one to each urn. Then a ball is drawn from each urn. They turned out to be the same color. Determine the probability that there are as many white balls left in the 1st urn as there were at the beginning.

Solution.
Event A - a ball is drawn simultaneously from the 1st and 2nd urns.
Probability of drawing a white ball from the first urn: P1(B) = 12/(12+16) = 12/28 = 3/7
Probability of drawing a black ball from the first urn: P1(H) = 16/(12+16) = 16/28 = 4/7
Probability of drawing a white ball from the second urn: P2(B) = 8/18 = 4/9
Probability of drawing a black ball from the second urn: P2(H) = 10/18 = 5/9

Event A happened. Event B - a ball is drawn from each urn. After shuffling, the probability of a white or black ball returning to the urn is ½.
Let's consider the options for event B - they turned out to be the same color.

For the first urn
1) a white ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BB/A=B) = ½ * 12/28 * 3/7 = 9/98
2) a white ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BB/A=H) = ½ * 13/28 * 4/7 = 13/98
3) a white ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(BC/A=B) = ½ * 16/28 * 3/7 = 6/49
4) a white ball was placed in the first urn and a black one was pulled out, provided that a black ball was pulled out earlier, P1(BC/A=H) = ½ * 15/28 * 4/7 = 15/98
5) a black ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BW/A=B) = ½ * 11/28 * 3/7 = 33/392
6) a black ball was placed in the first urn and a white ball was drawn, provided that a black ball was drawn earlier, P1(B/A=H) = ½ * 12/28 * 4/7 = 6/49
7) a black ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(HH/A=B) = ½ * 17/28 * 3/7 = 51/392
8) a black ball was placed in the first urn and a black one was pulled out, provided that a black ball was drawn earlier, P1(HH/A=H) = ½ * 16/28 * 4/7 = 8/49

For the second urn
1) a white ball was placed in the first urn and a white ball was drawn, provided that a white ball was previously drawn, P1(BB/A=B) = ½ * 8/18 * 3/7 = 2/21
2) a white ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BB/A=H) = ½ * 9/18 * 4/7 = 1/7
3) a white ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(BC/A=B) = ½ * 10/18 * 3/7 = 5/42
4) a white ball was placed in the first urn and a black one was pulled out, provided that a black ball was pulled out earlier, P1(BC/A=H) = ½ * 9/18 * 4/7 = 1/7
5) a black ball was put into the first urn and a white ball was pulled out, provided that a white ball was pulled out earlier, P1(BW/A=B) = ½ * 7/18 * 3/7 = 1/12
6) a black ball was placed in the first urn and a white ball was pulled out, provided that a black ball was pulled out earlier, P1(BW/A=H) = ½ * 8/18 * 4/7 = 8/63
7) a black ball was placed in the first urn and a black one was pulled out, provided that a white ball was pulled out earlier, P1(HH/A=B) = ½ * 11/18 * 3/7 = 11/84
8) a black ball was placed in the first urn and a black one was drawn, provided that a black ball had been drawn earlier, P1(HH/A=H) = ½ * 10/18 * 4/7 = 10/63

The balls turned out to be the same color:
a) white
P1(B) = P1(BB/A=B) + P1(BB/A=B) + P1(BW/A=B) + P1(BW/A=B) = 9/98 + 13/98 + 33 /392 + 6/49 = 169/392
P2(B) = P1(BB/A=B) + P1(BB/A=B) + P1(BW/A=B) + P1(BW/A=B) = 2/21+1/7+1 /12+8/63 = 113/252
b) black
P1(H) = P1(HH/A=B) + P1(HH/A=H) + P1(HH/A=B) + P1(HH/A=H) = 6/49 + 15/98 + 51 /392 + 8/49 = 223/392
P2(H) = P1(HH/A=B) + P1(HH/A=H) + P1(HH/A=B) + P1(HH/A=H) =5/42+1/7+11 /84+10/63 = 139/252

P = P1(B)* P2(B) + P1(H)* P2(H) = 169/392*113/252 + 223/392*139/252 = 5/42

Example 7d. The first box contains 5 white and 4 blue balls, the second contains 3 and 1, and the third contains 4 and 5, respectively. A box was chosen at random and a ball pulled out of it turned out to be blue. What is the probability that this ball is from the second box?

Solution.
A - event of drawing a blue ball. Let's consider all the possible outcomes of such an event.
H1 - the ball drawn from the first box,
H2 - the ball pulled out from the second box,
H3 - a ball drawn from the third box.
P(H1) = P(H2) = P(H3) = 1/3
According to the conditions of the problem, the conditional probabilities of event A are equal to:
P(A|H1) = 4/(5+4) = 4/9
P(A|H2) = 1/(3+1) = 1/4
P(A|H3) = 5/(4+5) = 5/9
P(A) = P(H1)*P(A|H1) + P(H2)*P(A|H2) + P(H3)*P(A|H3) = 1/3*4/9 + 1 /3*1/4 + 1/3*5/9 = 5/12
The probability that this ball is from the second box is:
P2 = P(H2)*P(A|H2) / P(A) = 1/3*1/4 / 5/12 = 1/5 = 0.2

Example 8. Five boxes with 30 balls each contain 5 red balls (this is a box of composition H1), six other boxes with 20 balls each contain 4 red balls (this is a box of composition H2). Find the probability that a red ball taken at random is contained in one of the first five boxes.
Solution: The problem is to apply the total probability formula.

The probability that any the taken ball is contained in one of the first five boxes:
P(H 1) = 5/11
The probability that any the taken ball is contained in one of six boxes:
P(H2) = 6/11
The event happened - the red ball was pulled out. Therefore, this could happen in two cases:
a) pulled out from the first five boxes.
P 5 = 5 red balls * 5 boxes / (30 balls * 5 boxes) = 1/6
P(P 5 /H 1) = 1/6 * 5/11 = 5/66
b) pulled out from six other boxes.
P 6 = 4 red balls * 6 boxes / (20 balls * 6 boxes) = 1/5
P(P 6 /H 2) = 1/5 * 6/11 = 6/55
Total: P(P 5 /H 1) + P(P 6 /H 2) = 5/66 + 6/55 = 61/330
Therefore, the probability that a red ball drawn at random is contained in one of the first five boxes is:
P k.sh. (H1) = P(P 5 /H 1) / (P(P 5 /H 1) + P(P 6 /H 2)) = 5/66 / 61/330 = 25/61

Example 9. The urn contains 2 white, 3 black and 4 red balls. Three balls are drawn at random. What is the probability that at least two balls will be the same color?
Solution. There are three possible outcomes:
a) among the three drawn balls there were at least two white ones.
P b (2) = P 2b
The total number of possible elementary outcomes for these tests is equal to the number of ways in which 3 balls can be extracted from 9:

Let's find the probability that among the 3 selected balls, 2 are white.

Number of options to choose from 2 white balls:

Number of options to choose from 7 other balls third ball:

b) among the three drawn balls there were at least two black ones (i.e. either 2 black or 3 black).
Let's find the probability that among the selected 3 balls, 2 are black.

Number of options to choose from 3 black balls:

Number of options to choose from 6 other balls of one ball:


P 2h = 0.214
Let's find the probability that all the selected balls are black.

P h (2) = 0.214+0.0119 = 0.2259

c) among the three drawn balls there were at least two red ones (i.e., either 2 red or 3 red).
Let's find the probability that among the 3 selected balls, 2 are red.

Number of options to choose from 4 black balls:

Number of options to choose from: 5 white balls, remaining 1 white:


Let's find the probability that all the selected balls are red.

P to (2) = 0.357 + 0.0476 = 0.4046
Then the probability that at least two balls will be the same color is equal to: P = P b (2) + P h (2) + P k (2) = 0.0833 + 0.2259 + 0.4046 = 0.7138

Example 10. The first urn contains 10 balls, 7 of them white; The second urn contains 20 balls, 5 of which are white. One ball is drawn at random from each urn, and then one ball is drawn at random from these two balls. Find the probability that the white ball is drawn.
Solution. The probability that a white ball is drawn from the first urn is P(b)1 = 7/10. Accordingly, the probability of drawing a black ball is P(h)1 = 3/10.
The probability that a white ball is drawn from the second urn is P(b)2 = 5/20 = 1/4. Accordingly, the probability of drawing a black ball is P(h)2 = 15/20 = 3/4.
Event A - a white ball is taken from two balls
Let's consider the possible outcome of event A.

1. A white ball was drawn from the first urn, and a white ball was drawn from the second urn. Then a white ball was drawn from these two balls. P1 = 7/10*1/4 = 7/40
2. A white ball was drawn from the first urn and a black ball was drawn from the second urn. Then a white ball was drawn from these two balls. P2 = 7/10*3/4 = 21/40
3. A black ball was drawn from the first urn, and a white ball was drawn from the second urn. Then a white ball was drawn from these two balls. P3 = 3/10*1/4 = 3/40

Thus, the probability can be found as the sum of the above probabilities.
P = P1 + P2 + P3 = 7/40 + 21/40 + 3/40 = 31/40

Example 11. There are n tennis balls in the box. Of these, m were played. For the first game, two balls were taken at random and put back after the game. For the second game we also took two balls at random. What is the probability that the second game will be played with new balls?
Solution. Consider event A - the game was played for the second time with new balls. Let's see what events can lead to this.
Let us denote by g = n-m the number of new balls before being pulled out.
a) for the first game two new balls were pulled out.
P1 = g/n*(g-1)/(n-1) = g(g-1)/(n(n-1))
b) for the first game, they pulled out one new ball and one already played one.
P2 = g/n*m/(n-1) + m/n*g/(n-1) = 2mg/(n(n-1))
c) for the first game, two played balls were pulled out.
P3 = m/n*(m-1)/(n-1) = m(m-1)/(n(n-1))

Let's look at the events of the second game.
a) Two new balls were drawn, under condition P1: since new balls had already been drawn for the first game, then for the second game their number decreased by 2, g-2.
P(A/P1) = (g-2)/n*(g-2-1)/(n-1)*P1 = (g-2)/n*(g-2-1)/(n- 1)*g(g-1)/(n(n-1))
b) Two new balls were drawn, under condition P2: since one new ball had already been drawn for the first game, then for the second game their number decreased by 1, g-1.
P(A/P2) =(g-1)/n*(g-2)/(n-1)*P2 = (g-1)/n*(g-2)/(n-1)*2mg /(n(n-1))
c) Two new balls were drawn, under condition P3: since previously no new balls were used for the first game, their number did not change for the second game g.
P(A/P3) = g/n*(g-1)/(n-1)*P3 = g/n*(g-1)/(n-1)*m(m-1)/(n (n-1))

Total probability P(A) = P(A/P1) + P(A/P2) + P(A/P3) = (g-2)/n*(g-2-1)/(n-1)* g(g-1)/(n(n-1)) + (g-1)/n*(g-2)/(n-1)*2mg/(n(n-1)) + g/n *(g-1)/(n-1)*m(m-1)/(n(n-1)) = (n-2)(n-3)(n-m-1)(n-m)/(( n-1)^2*n^2)
Answer: P(A)=(n-2)(n-3)(n-m-1)(n-m)/((n-1)^2*n^2)

Example 12. The first, second and third boxes contain 2 white and 3 black balls, the fourth and fifth boxes contain 1 white and 1 black ball. A box is randomly selected and a ball is drawn from it. What is the conditional probability that the fourth or fifth box is chosen if the ball drawn is white?
Solution.
The probability of choosing each box is P(H) = 1/5.
Let us consider the conditional probabilities of event A - drawing the white ball.
P(A|H=1) = 2/5
P(A|H=2) = 2/5
P(A|H=3) = 2/5
P(A|H=4) = ½
P(A|H=5) = ½
Total probability of drawing a white ball:
P(A) = 2/5*1/5 + 2/5*1/5 +2/5*1/5 +1/2*1/5 +1/2*1/5 = 0.44
Conditional probability that the fourth box is selected
P(H=4|A) = 1/2*1/5 / 0.44 = 0.2273
Conditional probability that the fifth box is selected
P(H=5|A) = 1/2*1/5 / 0.44 = 0.2273
In total, the conditional probability that the fourth or fifth box is selected is
P(H=4, H=5|A) = 0.2273 + 0.2273 = 0.4546

Example 13. There were 7 white and 4 red balls in the urn. Then another ball of white or red or black color was put into the urn and after mixing one ball was taken out. It turned out to be red. What is the probability that a) a red ball was placed? b) black ball?
Solution.
a) red ball
Event A - the red ball is drawn. Event H - the red ball is placed. Probability that a red ball was placed in the urn P(H=K) = 1 / 3
Then P(A|H=K)= 1 / 3 * 5 / 12 = 5 / 36 = 0.139
b) black ball
Event A - the red ball is drawn. Event H - a black ball is placed.
Probability that a black ball was placed in the urn P(H=H) = 1/3
Then P(A|H=H)= 1 / 3 * 4 / 12 = 1 / 9 = 0.111

Example 14. There are two urns with balls. One has 10 red and 5 blue balls, the second has 5 red and 7 blue balls. What is the probability that a red ball will be drawn at random from the first urn and a blue ball from the second?
Solution. Let the event A1 be a red ball drawn from the first urn; A2 - a blue ball is drawn from the second urn:
,
Events A1 and A2 are independent. The probability of the joint occurrence of events A1 and A2 is equal to

Example 15. There is a deck of cards (36 pieces). Two cards are drawn at random in a row. What is the probability that both cards drawn will be red?
Solution. Let event A 1 be the first red card drawn. Event A 2 - the second red card drawn. B - both cards taken out are red. Since both event A 1 and event A 2 must occur, then B = A 1 · A 2 . Events A 1 and A 2 are dependent, therefore, P(B) :
,
From here

Example 16. Two urns contain balls that differ only in color, and in the first urn there are 5 white balls, 11 black and 8 red balls, and in the second there are 10, 8, 6 balls, respectively. One ball is drawn at random from both urns. What is the probability that both balls are the same color?
Solution. Let index 1 mean white, index 2 mean black; 3 - red color. Let the event A i be that a ball of the i-th color is drawn from the first urn; event B j - a ball of color j is drawn from the second urn; event A - both balls are the same color.
A = A 1 · B 1 + A 2 · B 2 + A 3 · B 3. Events A i and B j are independent, and A i · B i and A j · B j are incompatible for i ≠ j. Hence,
P(A)=P(A 1) P(B 1)+P(A 2) P(B 2)+P(A 3) P(B 3) =

Example 17. From an urn with 3 white and 2 black balls, balls are drawn one at a time until black appears. Find the probability that 3 balls will be drawn from the urn? 5 balls?
Solution.
1) the probability that 3 balls will be drawn from the urn (i.e. the third ball will be black, and the first two will be white).
P=3/5*2/4*2/3=1/5
2) the probability that 5 balls will be drawn from the urn
This situation is not possible, because only 3 white balls.
P=0

Probability shows the possibility of a particular event given a certain number of repetitions. It is the number of possible outcomes with one or more outcomes divided by the total number of possible events. The probability of multiple events is calculated by dividing the problem into individual probabilities and then multiplying these probabilities.

Steps

Probability of a single random event

1. Select an event with mutually exclusive results. Probability can only be calculated if the event in question either occurs or does not occur. It is impossible to simultaneously obtain an event and its opposite result. Examples of such events are rolling a 5 on a dice or winning a certain horse at a race. Five will either come up or it won't; a certain horse will either come first or not.

   • For example, it is impossible to calculate the probability of such an event: with one throw of the die, 5 and 6 will appear at the same time.

2. Identify all possible events and outcomes that could occur. Suppose you need to determine the probability that when throwing a game die with 6 numbers you will get a three. "Rolling a three" is an event, and since we know that any of the 6 numbers can be rolled, the number of possible outcomes is six. Thus, we know that in in this case there are 6 possible outcomes and one event whose probability we want to determine. Below are two more examples.

   • Example 1. In this case, the event is “choosing a day that falls on the weekend,” and the number of possible outcomes is equal to the number of days of the week, that is, seven.
   • Example 2. The event is “draw a red ball”, and the number of possible outcomes is equal to the total number of balls, that is, twenty.

3. Divide the number of events by the number of possible outcomes. This way you will determine the probability of a single event. If we consider the case of rolling a die as a 3, the number of events is 1 (the 3 is on only one side of the die) and the total number of outcomes is 6. The result is a ratio of 1/6, 0.166, or 16.6%. The probability of an event for the two examples above is found as follows:

   • Example 1. What is the probability that you randomly select a day that falls on a weekend? The number of events is 2, since there are two days off in one week, and the total number of outcomes is 7. Thus, the probability is 2/7. The result obtained can also be written as 0.285 or 28.5%.
   • Example 2. The box contains 4 blue, 5 red and 11 white balls. If you take a random ball out of a box, what is the probability that it will be red? The number of events is 5, since there are 5 red balls in the box, and the total number of outcomes is 20. We find the probability: 5/20 = 1/4. The result obtained can also be written as 0.25 or 25%.

4. Add up the probabilities of all possible events and see if the sum totals 1. The total probability of all possible events must be 1, or 100%. If you don't get 100%, you most likely made a mistake and missed one or more possible events. Check your calculations and make sure you have considered all possible outcomes.

   • For example, the probability of getting a 3 when rolling a dice is 1/6. In this case, the probability of any other number falling out of the remaining five is also equal to 1/6. As a result, we get 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6, that is, 100%.
   • If, for example, you forget about the number 4 on the die, adding up the probabilities will give you only 5/6, or 83%, which is not equal to one and indicates an error.

5. Express the probability of an impossible outcome as 0. This means that the given event cannot happen and its probability is 0. This way you can account for impossible events.

   • For example, if you were to calculate the probability that Easter will fall on a Monday in 2020, you would get 0 because Easter is always celebrated on Sunday.

   Probability of several random events

   1. When considering independent events, calculate each probability separately. Once you determine what the probabilities of events are, they can be calculated separately. Suppose we want to know the probability of rolling a die twice in a row and getting a 5. We know that the probability of getting one 5 is 1/6, and the probability of getting a second 5 is also 1/6. The first outcome is not related to the second.

      • Several rolls of fives are called independent events, since what happens the first time does not affect the second event.

   2. Consider the influence of previous outcomes when calculating the probability for dependent events. If the first event affects the probability of the second outcome, we talk about calculating the probability dependent events. For example, if you select two cards from a 52-card deck, after drawing the first card, the composition of the deck changes, which affects the selection of the second card. To calculate the probability of the second of two dependent events, you need to subtract 1 from the number of possible outcomes when calculating the probability of the second event.

      • Example 1. Consider the following event: Two cards are drawn randomly from the deck, one after another. What is the probability that both cards will be of clubs? The probability that the first card will be a club suit is 13/52, or 1/4, since there are 13 cards of the same suit in the deck.
        • After this, the probability that the second card will be a club suit is 12/51, since one club card is no longer there. This is because the first event influences the second. If you draw the Three of Clubs and don't put it back, there will be one less card in the deck (51 instead of 52).
      • Example 2. There are 4 blue, 5 red and 11 white balls in the box. If three balls are drawn at random, what is the probability that the first is red, the second is blue, and the third is white?
        • The probability that the first ball will be red is 5/20, or 1/4. The probability that the second ball will be blue is 4/19, since there is one less ball left in the box, but still 4 blue ball. Finally, the probability that the third ball will be white is 11/18 since we have already drawn two balls.

   3. Multiply the probabilities of each individual event. Regardless of whether you are dealing with independent or dependent events, or the number of outcomes (there could be 2, 3, or even 10), you can calculate the overall probability by multiplying the probabilities of all the events in question by each other. As a result, you will get the probability of several events, the following one after another. For example, the task is Find the probability that when rolling a die twice in a row you will get a 5. These are two independent events, the probability of each of which is 1/6. Thus, the probability of both events is 1/6 x 1/6 = 1/36, that is, 0.027, or 2.7%.

      • Example 1. Two cards are drawn from the deck at random, one after another. What is the probability that both cards will be of clubs? The probability of the first event is 13/52. The probability of the second event is 12/51. We find the total probability: 13/52 x 12/51 = 12/204 = 1/17, that is, 0.058, or 5.8%.
      • Example 2. The box contains 4 blue, 5 red and 11 white balls. If three balls are drawn at random from a box one after the other, what is the probability that the first is red, the second is blue, and the third is white? The probability of the first event is 5/20. The probability of the second event is 4/19. The probability of the third event is 11/18. So the total probability is 5/20 x 4/19 x 11/18 = 44/1368 = 0.032, or 3.2%.