Theoretical material on the modules "Probability Theory and Mathematical Statistics".

The classical definition of probability reduces the concept of probability to the concept of equiprobability (equal possibility) of events, which is considered the main one and is not subject to a formal definition. This definition is applicable in cases where it is possible to single out a complete group of incompatible and equally probable events - elementary outcomes. For example, consider an urn with balls.

Let an urn contain 7 identical, carefully mixed balls, 2 of them red, 1 blue and 4 white. The test will consist in the fact that one ball is taken at random from the urn. Each event that can occur in the ongoing trial is an elementary outcome. AT this example seven elementary outcomes, which we will denote E 1 , E 2 ,..., E 7. outcomes E 1 , E 2 - the appearance of a red ball, E 3 - the appearance of a blue ball, E 4 , E 5 , E 6 , E 7 - appearance white ball. In our example, the events E 1 , E 2 ,... E 7 - pairwise incompatible. In addition, they are also equally likely in this test. Let the event BUT is that a ball taken at random from the urn turned out to be colored (red or blue).

Those elementary outcomes in which the event of interest to us BUT comes, is called favorable outcomes event BUT. In our example, outcomes favoring the event BUT, are outcomes E 1 , E 2 and E 3 . Reasonable as a measure of the possibility of an event occurring BUT, that is, the probabilities R(BUT), accept a number equal to the ratio of outcomes that favor the occurrence of the event BUT, to all possible outcomes. In our example

R The above example led us to the definition of probability, which is commonly called classic .

Probability of an event BUT called the ratio of the number m favorable outcomes for this event to total number n all elementary outcomes:

R(BUT) = . (1.4.4)

The classical definition of probability serves as a good mathematical model those random experiments, the number of outcomes of which is finite, and the outcomes themselves are equally probable.

EXAMPLE 2. rushes dice. Find the probability of getting no more than four points.

Decision. Total number of elementary outcomes n= 6 (can roll 1, 2, 3, 4, 5, 6). Among these outcomes favor the event BUT(no more than four points will fall) only four outcomes m= 4. Therefore, the desired probability

EXAMPLE 3. What is the probability of guessing 4 numbers by filling in the sports lotto card "6" out of "49"?

Decision. The total number of elementary outcomes of experience is equal to the number of ways in which 6 numbers out of 49 can be crossed out, that is n = C. Let's find a number outcomes favorable to the event of interest to us
BUT= (4 numbers guessed), 4 numbers out of 6 winners can be crossed out C ways, while the remaining two numbers must not be winning. You can cross out 2 wrong numbers out of 43 non-winning numbers C ways. Therefore, the number of favorable outcomes m = C× C. Taking into account that all outcomes of the experiment are incompatible and equally possible, we find the desired probability using the classical probability formula:

P(A) =

EXAMPLE 4. taken at random telephone number consists of 5 digits. How great is the probability that in it: 1) all numbers are different; 2) are all the numbers odd?

Decision. 1. Since each of the five places in a five-digit number can contain any of the numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, then all different five-digit numbers will be 10 5 (00000 - 1 -th, 00001 - 2nd, 00002 -3rd, ..., 99998 - 99999th, and finally 99999 - 100000th). Numbers in which all numbers are different are placements of 10 elements of 5.

Formula for number placements from n elements by k:

K! == n (n - 1) ... (n - k + 1).

Therefore, the number of favorable cases m= = 10× 9× 8× 7× 6 and the desired probability

P(A) = = 0,3024.

2. From 5 odd numbers (1, 3, 5, 7, 9) you can form 5 5 different five-digit numbers. 5 5 is the number of favorable outcomes m . Since all equally possible cases n= 10 5 , then the desired probability

P(A) ====0.03125.

EXAMPLE 5. A full deck of cards (52 sheets) is divided at random into two equal packs of 26 sheets. Find the probabilities of the following events:

BUT- in each of the packs there will be two aces;

AT- in one of the packs there will be no aces, and in the other - all four;

With- in one of the packs there will be one ace, and in the other - three.

Decision. The total number of possible elementary outcomes of the test is equal to the number of ways in which 26 cards out of 52 can be drawn, that is, the number of combinations from 52 to 26, n= . Number favorable event BUT cases
m= (according to the basic rule of combinatorics), where the first factor shows that two aces out of four can be taken in ways, the second factor shows that the remaining 24 cards are taken from 48 cards that do not contain aces in ways. The desired probability is equal to the ratio of the number of outcomes that favor the event BUT, to the total number of all outcomes:

Event AT can be realized in two equally possible ways: either in the first pack there will be all four aces, and in the second - none, or vice versa:

Similarly:

notice, that classic definition probability was introduced for the case when the space elementary events of course, and all outcomes and trials are equally possible and incompatible.

Task 174tv

a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.

Task 176tv

An urn contains 6 black and 5 white balls. 5 balls are drawn randomly. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.

Task 178tv

An urn contains 4 black and 5 white balls. 4 balls are drawn randomly. Find the probability that among them there is:
a) 2 white balls;
b) less than 2 white balls;
c) at least one white ball.

Task 180tv

An urn contains 6 black and 7 white balls. 4 balls are drawn randomly. Find the probability that among them there is:
a) 4 white balls;
b) less than 4 white balls;
c) at least one white ball.

Task 184tv

An urn contains 8 black and 6 white balls. 4 balls are drawn randomly. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.

Task 186tv

An urn contains 4 black and 6 white balls. 4 balls are drawn randomly. Find the probability that among them there is:
a) 3 white balls;
b) less than 3 white balls;
c) at least one white ball.

Task 188tv

An urn contains 5 black and 6 white balls. 5 balls are drawn randomly. Find the probability that among them there is:
a) 4 white balls;
b) less than 4 white balls;
c) at least one white ball.

Example 1. In the first urn: three red, one white balls. In the second urn: one red, three white balls. A coin is thrown at random: if the coat of arms is chosen from the first urn, otherwise, from the second.
Decision:
a) the probability of drawing a red ball
A - got a red ball
P 1 - coat of arms fell out, P 2 - otherwise

b) A red ball is chosen. 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).
Decision:
A is the probability of a white ball appearing
a) All whites:
(probability that one of the three options where there is white is caught)
(probability of a white ball appearing where all are white)

b) Pulled out where everyone is black

c) pulled out a variant where all are white or/and black

- at least one of them is white

P a + P b + P c =

Example 3 . An urn contains 5 white and 4 black balls. 2 balls are taken out of it in a row. Find the probability that both balls are white.
Decision:
5 white, 4 black balls
P(A 1) - drawn a white ball

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

P(A) – White balls selected in a row

Example 3a. There are 2 counterfeit and 8 real banknotes in a pack. 2 banknotes were pulled out of the pack in a row. Find the probability that both are false.
Decision:
P(2) = 2/10*1/9 = 1/45 = 0.022

Example 4. There are 10 urns. 9 urns contain 2 black and 2 white balls. There are 5 whites and 1 black in 1 urn. A ball is drawn from an urn taken at random.
Decision:
P(A)-? a white ball is taken from an urn containing 5 white
B - the probability of being taken out of the urn, where 5 are white
, - taken out from others
C 1 - the probability of the appearance of a white ball in lvl 9.

C 2 - the 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 conical ones. The picker takes 1 roller and then another.
Decision:
a) both rollers are cylindrical
P(C 1)=; P(C 2)=
C 1 - the first cylinder, C 2 - the second cylinder
P(A)=P(C 1)P(C 2) =
b) At least one cylinder
K 1 - the first cone.
K 2 - the second cone.
P(B)=P(C 1)P(K 2)+P(C 2)P(K 1)+P(C 1)P(C 2)
;

c) the first cylinder, and the second is not
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 pieces are drawn at random.
a) One of them is defective
P n (K)=C n k p k q n-k ,
P is the probability of defective products

q is the probability of standard parts

n=3, three parts

b) two of the 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 3 black balls, and the 2nd urn contains 3 white and 4 black. 2 balls are transferred from the 1st urn to the 2nd urn without looking, and then 2 balls are drawn from the 2nd urn. What is the probability that they different colors?
Decision:
When transferring balls from the first urn, the following options are possible:
a) 2 white balls are drawn in a row
P WB 1 =
There will always be one less ball in the second step, since one ball has already been taken out in the first step.
b) one white and one black ball is drawn
The situation when the white ball was drawn first, and then the black one
P BC =
The situation when the black ball was drawn first, and then the white one
P BW =
Total: P CU 1 =
c) 2 black balls are drawn in a row
P HH 1 =
Since 2 balls were transferred from the first urn to the second urn, then total quantity 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 is drawn from the second urn

P BC 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 a white ball was drawn first, 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 taken out of the first urn in a row. That is why the number of black balls in this case is 6 (4+2).

The probability that the drawn 2 balls 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 are randomly transferred to the 2nd urn, containing 2 white and 6 black balls. Then 1 ball is drawn at random from the 2nd urn.
1) What is the probability that the ball drawn from urn 2 is white?
2) The ball drawn from the 2nd urn turned out to be white. Calculate the probability that balls were transferred from urn 1 to urn 2. different color.
Decision.
1) Event A - the ball drawn from the 2nd urn turned out to be white. Consider the following options for the occurrence of this event.
a) Two white balls are placed from the first urn into the second one: P1(bb) = 5/8*4/7 = 20/56.
There are 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 are placed from the first urn into the second one: P1(bc) = 5/8*3/7+3/8*5/7 = 30/56.
There are 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 are placed from the first urn into the second one: P1(hh) = 3/8*2/7 = 6/56.
There are 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 turned out to be white is equal to:
P(A) = 80/448 + 90/448 + 12/448 = 13/32

2) The ball drawn from the 2nd urn turned out to be white, i.e. full probability equals P(A)=13/32.
The probability that balls of different colors (black and white) were transferred to 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. One ball is chosen at random from the first, and two balls from the second. After that, one ball is taken at random from the chosen three balls. This last ball turned out to be black. Find the probability that a white ball was chosen from the first urn.
Decision.
Let's consider all variants of the event A - out of three balls, the drawn ball turned out to be black. How could it happen that among the three balls was black?
a) A black ball is drawn from the first urn, and two white balls are drawn from the second urn.
P1 = (3/11)(5/8*4/7) = 15/154
b) A black ball is drawn from the first urn, and two black balls are drawn from the second urn.
P2 = (3/11)(3/8*2/7) = 9/308
c) A black ball is drawn from the first urn, and one white and one black ball are drawn from the second urn.
P3 = (3/11)(3/8*5/7+5/8*3/7) = 45/308
d) A white ball is drawn from the first urn, and two black balls are taken from the second urn.
P4 = (8/11)(3/8*2/7) = 6/77
e) A white ball was taken out of the first urn, and one white and one black ball were taken out of 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 was chosen 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, provided that a black one was chosen from three balls, is equal to:
Pch \u003d Pb (1) / P \u003d 36/77 / 57/77 \u003d 36/57

Example 7c. The first urn contains 12 white and 16 black balls, the second urn contains 8 white and 10 black. At the same time, a ball is drawn from the 1st and 2nd urn, mixed and returned one at a time 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.

Decision.
Event A - at the same time, a ball is drawn 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 returning the ball to the urn of a white or black ball is ½.
Consider the variants of event B - they turned out to be of the same color.

For the first urn
1) a white ball was placed in the first urn, and a white one 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 drawn, provided that a black ball was drawn earlier, P1(BB/A=W) = ½ * 13/28 * 4/7 = 13/98
3) a white ball was placed in the first urn and a black one was drawn, provided that a white ball was previously drawn, 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 drawn, provided that a black ball was drawn earlier, P1(BC/A=Ch) = 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 drawn earlier, P1(BW/A=B) = ½ * 11/28 * 3/7 = 33/392
6) a black ball was placed in the first urn and a white one was drawn, provided that a black ball was previously drawn, P1(BW/A=W) = ½ * 12/28 * 4/7 = 6/49
7) a black ball was placed in the first urn, and a black one was drawn, provided that a white ball was previously drawn, 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 drawn, 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 one 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 drawn, provided that a black ball was drawn earlier, P1(BB/A=W) = ½ * 9/18 * 4/7 = 1/7
3) a white ball was placed in the first urn and a black one was drawn, provided that a white ball was drawn 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 drawn, provided that a black ball was drawn earlier, P1(BC/A=Ch) = ½ * 9/18 * 4/7 = 1/7
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 * 7/18 * 3/7 = 1/12
6) a black ball was placed in the first urn and a white one was drawn, provided that a black ball was previously drawn, P1(BW/A=W) = ½ * 8/18 * 4/7 = 8/63
7) a black ball was placed in the first urn, and a black one was drawn, provided that a white ball was previously drawn, 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 was 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(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) = 6/49 + 15/98 + 51 /392 + 8/49 = 223/392
P2(H) = P1(WB/A=B) + P1(BH/A=B) + P1(BH/A=B) + P1(BH/A=B) =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 7g. The first box contains 5 white and 4 blue balls, the second 3 and 1, and the third 4 and 5, respectively. A box is chosen at random and a ball drawn from it turns out to be blue. What is the probability that this ball is from the second box?

Decision.
A - blue balloon extraction event. Consider all options for the outcome of such an event.
H1 - drawn ball from the first box,
H2 - drawn ball from the second box,
H3 - the drawn ball from the third box.
P(H1) = P(H2) = P(H3) = 1/3
According to the condition of the problem conditional probabilities events A are:
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 the H1 composition box), six other boxes with 20 balls each contain 4 red balls (this is the H2 composition box). Find the probability that a randomly drawn red ball is contained in one of the first five boxes.
Solution: The task of applying the total probability formula.

The probability that any the ball taken 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(H 2) = 6/11
The event happened - a red ball was drawn. Therefore, this could happen in two cases:
a) pulled out of the first five boxes.
P 5 = 5 red balls * 5 boxes / (30 balls * 5 boxes) = 1/6
P(P 5 / H 1) \u003d 1/6 * 5/11 \u003d 5/66
b) pulled out of six other boxes.
P 6 = 4 red balls * 6 boxes / (20 balls * 6 boxes) = 1/5
P (P 6 / H 2) \u003d 1/5 * 6/11 \u003d 6/55
Total: P(P 5 /H 1) + P(P 6 /H 2) = 5/66 + 6/55 = 61/330
Therefore, the probability that a randomly drawn red ball 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 . An 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 are of the same color?
Decision. There are three possible outcomes of events:
a) among the three balls drawn, at least two are white.
P b (2) = P 2b
The total number of possible elementary outcomes for these trials is equal to the number of ways in which 3 balls can be drawn out of 9:

Find the probability that 2 of the 3 balls 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 balls drawn, at least two are black (i.e. either 2 black or 3 black).
Find the probability that 2 of the 3 balls 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
Find the probability that all the chosen balls are black.

P h (2) = 0.214+0.0119 = 0.2259

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

Find the probability that all the chosen balls are red.

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

Example 10 . The first urn contains 10 balls, of which 7 are 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 a white ball is drawn.
Decision. The probability that a white ball was 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 was 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
Consider the outcome of event A.

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

So 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 a box. Of them played m . For the first game, they took two balls at random and put them back after the game. For the second game, they also took two balls at random. What is the probability that the second game will be played with new balls?
Decision. Consider event A - the game was played for the second time with new balls. Let's see what events can lead to this.
Denote by g = n-m, the number of new balls before pulling out.
a) Two new balls are drawn for the first game.
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.
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))

Consider the events of the second game.
a) Two new balls were drawn, provided P1: since new balls were already 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, subject to P2: since one new ball was already 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) They pulled out two new balls, provided P3: since 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 each, the fourth and fifth boxes each 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 selected if the drawn ball is white?
Decision.
The probability of choosing each box is P(H) = 1/5.
Consider the conditional probabilities of the event A - drawing a 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
So, the conditional probability that the fourth or fifth box is chosen is
P(H=4, H=5|A) = 0.2273 + 0.2273 = 0.4546

Example 13 . An urn contains 7 white and 4 red balls. Then another ball of white or red or black color was placed in the urn and after mixing one ball was taken out. He turned out to be red. What is the probability that a) a red ball was placed? b) black ball?
Decision.
a) red ball
Event A - a red ball is drawn. Event H - put a red ball. 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 - a red ball is drawn. Event H - put a black ball.
The probability that a black ball was placed in the urn is 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 other 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 one from the second?
Decision. Let the event A1 - a red ball is drawn from the first urn; A2 - a blue ball is drawn from the second urn:
,
Events A1 and A2 are independent. The probability of 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. What is the probability that both cards drawn are red?
Decision. Let the event A 1 be the first drawn card of the red suit. Event A 2 - the second drawn card of the red suit. B - both drawn cards of red suit. Since both the event A 1 and the event A 2 must occur, then B = A 1 · A 2 . Events A 1 and A 2 are dependent, hence 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, and in the second 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?
Decision. Let index 1 mean White color, index 2 - black color; 3 - red color. Let the event A i - a ball of the i-th color is drawn from the first urn; event B j - a ball of j -th color was taken from the second urn; event A - both balls are of the same color.
A \u003d A 1 B 1 + A 2 B 2 + A 3 B 3. The events A i and B j are independent, while 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 are drawn one at a time until black appears. What is the probability that 3 balls will be drawn from the urn? 5 balls?
Decision.
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
such a situation is not possible, because only 3 white balls.
P=0

Job #1

random events

6 option.

Task 1.1. Throw three coins. Find the probability that "coat of arms" appears on only two coins.

Investigated event A - only two coins out of three will have a coat of arms. A coin has two sides, which means that there will be 8 events when throwing three coins. In three cases, only two coins will have a coat of arms. The probability of event A is calculated using the formula:

P(A) = m/n = 3/8.

Answer: probability 3/8.

Task 1.2. The word EVENT is made up of cards, each of which has one letter written on it. Then the cards are mixed and taken out without returning one at a time. Find the probability that the letters are taken out in the order of the given word.

The test consists in taking out cards with letters in random order without returning. The elementary event is the received sequence of letters. Event A is to receive the right word EVENT . Elementary events are permutations of 7 letters, which means that according to the formula we have n= 7!

The letters in the word EVENT do not repeat, so permutations are not possible in which the word does not change. Their number is 1.

Thus,

P(A) = 1/7! = 1/5040.

Answer: P(A) = 1/5040.

Task 1.3. As in the previous problem, find the corresponding probability of the case when the given word is the word ANTONOV ILYA.

This problem is solved similarly to the previous one.

n=11!; M = 2!*2! = 4.

P(A) = 4/11 = 4/39916800 = 1/9979200

Answer: P(A)=1/9979200.

Task 1.4. An urn contains 8 black and 6 white balls. 5 balls are drawn randomly. Find the probability that among them there is:

a) 3 white balls;

b) less than 3 white balls;

c) at least one white ball.

8 hours The test will be the random drawing of 5 balls. Elementary

6 b events are all possible combinations of 5 out of 14 balls. Their number is

a) A 1 - among the drawn balls 3 are white. So, among the drawn balls 3 are white and 2 are black. Using the multiplication rule, we get

P (A 1) \u003d 560/2002 \u003d 280/1001.

b) A 2 - among the drawn balls there are less than 3 white ones. This event consists of three incompatible events:

In 1 - among the drawn balls there are only 2 white and 3 black balls,

B 2 - among the drawn balls only one white and 4 black balls

In 3 - there is not a single white ball among the drawn balls, all 5 balls are black:

In 2 In 3.

Since events B 1 , B 2 and B 3 are incompatible, you can use the formula:

P (A 2) \u003d P (B 1) + P (B 2) + P (B 3);

P (A 2) \u003d 840/2002 + 70/2002 + 56/2002 \u003d 483/1001.

- there are no white balls among the drawn balls. In this case:

P(A 3) = 1 - P(

) = 1 - 28/1001 = 973/1001.

Answer: P (A 1) \u003d 280/1001, P (A 2) \u003d 483/1001, P (A 3) \u003d 973/1001.

Problem 1.6. The first urn contains 5 white and 7 black balls, and the second urn contains 6 white and 4 black balls. 2 balls are drawn randomly from the first urn and 2 balls from the second urn. Find the probability that among the drawn balls:

a) all balls of the same color;

b) only three white balls;

c) at least one white ball.

Urn 1 Urn 2 Balls were drawn from both urns independently. Trials

5 b 6 b are drawing two balls from the first urn and two balls

7h 4h from the second urn. Elementary events will be combinations

2 or 2 out of 12 or 10 balls respectively.

2 2 a) A 1 - all drawn balls of the same color, i.e. they are all white

or all black.

We define for each urn all possible events:

In 1 - 2 white balls are taken out from the first urn;

B 2 - 1 white and 1 black ball are drawn from the first urn;

In 3 - 2 black balls are drawn from the first urn;

C 1 - 2 white balls are drawn from the second urn;

C 2 - 1 white and 1 black ball are drawn from the second urn;

C 3 - 2 black balls are drawn from the second urn.

So A 1 =

, whence, taking into account the independence and incompatibility of events, we obtain

P (A 1) \u003d P (B 1) * P (C 1) + P (B 3) * P (C 3).

Let's find the number of elementary events n 1 and n 2 for the first and second urns, respectively. We have:

Find the number of each element of events that determine the following events:

C 1: m 21 = C 2: m 22 = C 3: m 23 =

Hence,

P (A 1) \u003d 10/66 * 15/45 + 21 * 6/45 \u003d 5/99 + 7/165 \u003d 46/495.

b) A 2 - among the drawn balls only 3 are white. In this case

C 2 (B 2 C 1);

P (A 2) \u003d P (B 1) * P (C 1) + P (B 2) * P (C 2)

P (A 2) \u003d 10/66 * 6/45 + 35/66 * 24/45 \u003d 33/99 \u003d 1/3.

c) A 3 - among the drawn balls there is at least one white one.

- there is not a single white ball among the extracted balls. Then ) \u003d P (B 3) * P (C 3) \u003d 21/66 * 6/45 \u003d 7/165;

P(A 3) = 1 - P(

) = 1 - 7/165 = 158/165.

Answer: P (A 1) \u003d 46/495, P (A 2) \u003d 1/3, P (A 3) \u003d 158/165.

Problem 1.7. An urn contains 5 black and white balls, 4 white balls are added to them. After that, 3 balls are randomly drawn from the urn. Find the probability that all the drawn balls are white, assuming that all possible proposals for the original contents of the urn are equally likely.

There are two types of tests here: first, the initial contents of the urn are given and then the 3rd ball is randomly drawn, and the result of the second test depends on the result of the first. Therefore, the total probability formula is used.

event A - 3 white balls are randomly drawn. The probability of this event depends on how original composition balls in the urn.

Consider the events:

In 1 - there were 5 white balls in the urn;

In 2 - there were 4 white and 1 black balls in the urn;

In 3 - there were 3 white and 2 black balls in the urn;

In 4 - there were 2 white and 3 black balls in the urn;

At 5 - there were 1 white and 4 black balls in the urn.

At 6 - there were 5 black balls in the urn;

Total number of elementary outcomes

Let us find the conditional probabilities of the event A under various conditions.

P (A / B 1) \u003d 1. P (A / B 2) \u003d 56/84 \u003d 2/3. P (A / B 3) \u003d 35/84 \u003d 5/12. P (A / B 4) \u003d 5/21. P (A / B 5) \u003d 5/42. P (A / B 6) \u003d 1/21.

P(A) = 1 * 1/6 + 2/3 * 1/6 + 5/12 * 1/6 + 5/21 * 1/6 + 5/42 * 1/6 + 1/21 * 1/6 = 209/504.

Problem 1.10. In the assembly shop, an electric motor is connected to the device. Electric motors are supplied by three manufacturers. There are electric motors of these factories in the warehouse, respectively, in the amount of M 1 =13, M 2 =12, and M 3 = 17 pieces, which can work without failure until the end of the warranty period with probabilities of 0.91, 0.82, and 0.77, respectively. The worker randomly takes one electric motor and mounts it to the device. Find the probability that an electric motor installed and operating without fail until the end of the warranty period was supplied by the first, second or third manufacturer, respectively.