Example 1. 2 cards are removed sequentially from a deck of 32 cards without returning. Find the probability that both of them are aces.

Solution. Since the first card can be drawn from the deck in 32 ways, and the second - 31 (since there are 31 cards left in the deck), then the number of possible outcomes of the experiment is . Let's determine the number of favorable outcomes. The first ace can be chosen from the four in the deck, the second - from the remaining three. This means that the number of favorable outcomes and the desired probability is equal to

Example 2. Five cakes were taken out of a box containing five eclairs and seven Napoleons. Find the probability that among them there are two “éclairs” and three “Napoleons”.

Solution. The number of possible outcomes of the experiment is the number of combinations of 12 by 5:

The number of favorable outcomes is the product of the number of ways in which two “éclairs” can be selected from the five available, and the number of sets of three “Napoleons” from the seven:

Therefore, the required probability is equal to

Example 3. A dot is thrown into the circle at random. Find the probability that it will not fall into a regular triangle inscribed in this circle.

Solution. In this case, the measure of the set of possible outcomes is the area of ​​the circle: and the measure of the set of favorable outcomes is the difference between the areas of the circle and the triangle: . Therefore, the probability of a given event is equal to

Example 4. Two shooters each fire one shot at a target. The probabilities of their hit are 0.6 and 0.9, respectively. Find the probabilities of the following events:

Both hit the target;

At least one hit the target.

Solution. Let us call the hitting of the target by the first and second shooter respectively the events and note that and are joint but independent events (in other words, both shooters can hit the target, and the probability of hitting each does not depend on the result of the other). An event is a product of events and therefore

An event is a sum and to determine its probability we use the general form of the addition theorem:

Example 5. Three identical urns contain balls: the first contains 5 white and 3 black, the second contains 2 white and 6 black, the third contains 3 white and 1 black. A ball is drawn from a randomly selected urn. Find the probability that he is white.

The conditional probability of an event, that is, drawing a white ball from the urn, is determined by the classical definition of probability (the number of favorable outcomes is the number of white balls, and the number of possible outcomes is the total number of balls in the urn). That's why

Using the total probability formula, we get:

Example 6. There are 20 students in the student group. Of these, 5 are excellent students who know all the exam questions, 8 students know the answers to 70% of the questions and 7 know the answers to 50%. The first student called answered the first question on the exam paper. Find the probability that he is an excellent student.

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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 this case there are 6 possible outcomes and one event, the probability of which 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%.

Answer: 0.7157

2.

3.

4. the number is not divisible by 5

Solution: P(A) = m/n; m=1/

It is equal to 90 and subtract from these numbers those that are divisible by 5 (10,15,20,25...90,95). Their number is 18 => n=90-18=72

Answer: 1/72

Solution: P(A)=m/n

a) P(A)=6/36 =1/6

Solution: C m n = n! /m!(n-m)!

m = C 3 7 = 7! / 3!*4! = 35

P (A1) = m/n = 35/220 = 7/44

b) you can get 3 reds out of 7 in 7 ways, and 3 blacks out of 5 =>

With 3 5 ways.

P(A2) = m/n = 45/220 = 9/44

Answer:

Solution:

Answer: 0.3.

Solution:

A – exit from the maze.

P(A/H3) =0.2 – from the 3rd labyrinth

P(A/H4) = 0.1 – from 4 labyrinths

Answer: 1/3; 2/5

9.

10.

11. .

Solution:

Solution:

P(A/H3)=8/10=4/5;

P(A)=1/3(1/2+5/6+4/5) = 62/45

13.

Solution:

Let B have no hits

P(C)= 1 - 0.216 = 0.784

Answer: 0.784

Solution:

H1=1/3; H2=1/3; H3=1/3

Answer: 15/48 = 0.3125

16.

Solution:

17.

Solution:

P(H2/A)=0.7/1.6=0.42

Solution:

Answer: P(A) = 0.925

A student visits 3 libraries in search of a book. The probability that they are in the library is 0.4; 0.5; 0.1; and the fact that they were issued or not are equally probable events. What is the probability that the book you need is found?

Solution: A-book is in the library, B – book is not issued.

P(B) = P(B -) = ½

P(A1) = 0.4 P(A2) = 0.5 P(A3) = 0.1

Let's determine the probability that the required book is found:

P = P(A1)* P(B) + P(A2)*P(B) + P(A3)*P(B) = P(B)(P(A1) + P(A2) + P(A3 ) = 1/2 * (0.4 + 0.5 +0.1) = 1/2 * 1 = ½

Answer: 1/2

23. Find the probabilities that the birthdays of 12 people will fall in different months of the year.

Solution: P(A)= m/n

n = --- A 12 = 12 12

P = 12! / 12 12 = 11! / 12 11 = (11*10*9*8*7*6*5*4*3*2*1) / (12*12*12*12*12 7) = (11*5*7*5* 1) / 12 7 = 7*8*25 / 12 7 = 1925 / 12 7

Answer: 1925/12 7

24. An urn contains 10 white, 5 black and 15 red balls. 2 balls are drawn sequentially. Two events are considered: A - at least one of the two drawn balls is red, B - at least one drawn ball is white. Find the probability of event C = A + B.

25. The randomly dialed number consists of 5 digits. Determine the probability that all the numbers in it are different.

26. The knitwear store received socks, 60% of which came from one factory, 25% from another and 15% from a third. Find the probability that the socks purchased by the buyer are made in the second or third factory.

Solution. A1-from 1 factory, P(A1) = 0.6;

A2 – from factory 2; P(A2) = 0.25

A3 – from 3 factories; P(A3) = 0.15

P(A2+A3) = 0.25 + 0.15 = 0.4

Answer: 0.4

A passenger can apply to one of the ticket offices to obtain a ticket. The probability of going to the 1st cash desk is 0.4; in the second 0.35; and 3rd 0.25. The probability that by the time the passenger arrives the tickets available at the ticket office will be sold is equal to 0.3 for the 1st ticket office; for the 2nd 0.4, for the 3rd 0.6. Find the probability that the passenger will buy a ticket.

P(A) – probability of not buying a ticket.

P(A) =0.4*0.3 + 0.35*0.4 + 0.25*0.6 =

0,12 + 0,14 + 0,15 = 0,41

P(A1) – probability of buying a ticket = 1-P(A) = 1 – 0.41 = 0.59.

Answer: P(A1) = 0.59.

28. 4 dice are thrown. Find the probability that: a) at least one of them will have 2 points, b) they will have the same number of points.

Solution:

29. From 9 tokens numbered with different single-digit numbers, 3 is selected. Find the probability that sequential recording of their numbers will show an increase in the values ​​of the digits.

Solution:

30. The probability of winning on a lottery ticket is 0.1. What is the probability that at least one ticket out of three purchased will win?

31. From a full deck of cards (52 sheets), 4 cards are taken out at once. Find the probability that all these cards will be of different suits.

Solution: The probability of drawing a specific suit is C 1 13

C 1 13 = 13 (number of possible ways).

Possibility to draw cards from 52 = C 4 52 = 52! / 4!* 48! = 48!*49*50*51* 52 / 2*3*4*48! = 270725
P(A) = C 1 13 * C 1 13 * C 1 13 * C 1 13 / C 4 52 = 28561 / 270725 = 0.1054982

Answer: P(A) = 0.1054982.

32. There are 3 urns. The first of them has 5 white and 6 black balls, the second has 4 white and 3 black balls, the third has 5 white and 3 black balls. Someone chooses one of the urns at random and draws a ball from it. This ball turned out to be white. Find the probability that this ball is drawn from the second urn.

Solution:

Answer: 0.9125

52. What is the probability of getting 1 ace, an ace and a king when dealing 6 cards from a deck of 52 cards?

The cars were delivered to the service station. Moreover, 5 of them had a chassis malfunction, 8 had engine malfunctions, and 10 were fully operational. What is the probability that a car with a faulty chassis also has a faulty motor?

Solution:

11111111 8 with faulty motor

5 with inappropriate moves part 11111 1111111111 10 are working

11111111111111111111 total 20

3 with faulty motor and stroke part 111

P = m/n m-number of cars with a faulty chassis and a faulty motor; m=3

n – number of vehicles with faulty chassis; n=5

P = 3/5 – probability that a car with a faulty chassis has a faulty motor.

Answer: 3/5

Answer: 21/625; 219/625; 247/625

67. In the first brigade of 8 tractors, 2 require repairs, in the second, out of 6-1. One tractor is selected at random from each brigade. Determine the probability that a) both are working, b) at least one is working, c) only one is working

a)P(A)=P(A1*A2) =3/4*5/6=5/8

b)P(A) = 1-P(--- A)=1-2/8*1/6=1-1/24=23/24

c) P(A)=3/4*1/6+5/6*1/4=1/8+5/24=8/24=1/3

68. The organization employs 12 men and 8 women. 3 prizes have been allocated for them. Determine the probability that the bonus will be received by: a) two men and one woman; b) only women; c) at least one man.

Solution: a) A-1 man

B- 2 men

S- 1 woman

P(A) = 12/20; P(B/A) = 11/19; P(C/AB) = 8/18

P(ABC) = P(A)*P(B/A)*P(C/AB) = 1056/6840 = 0.154

b) A-1 woman

B-2 women

S-3 women

P(A) = 8/20 ; P(B/A) = 7/19; P(C/AB) = 6/18

P(ABC) = P(A)*P(B/A)* P(C/AB) = 336/6840 = 0.049

c) A-at least 1 man

A all women

P(A)=1- P(---A)

P(---A) = 8/20 * 7/19 * 6/18 = 0.049

69. Out of 25 employees, 10 enterprises have a higher education: Determine the probability that out of three randomly selected people have a higher education; a) three people; b) one person; c) at least one person.

Solution:

70. The letters “K”, “A”, “P”, “T”, “O”, “Ch”, “K”, “A” are written on the cards. The cards are shuffled and placed in the order they are drawn. What is the probability that you get: a) the word “CARD”; b) the word “MAP”; c) the word “CURRENT”.

71. There are 15 high quality products in a box of 25 items. 3 items are drawn at random. Determine the probability that: a) one of them is of increased quality; b) all three products are of improved quality; c) at least one product of improved quality.

Solution:

72. Three dice are thrown. What is the probability that: a) at least one of them will have 5 points; b) everyone will get odd numbers; c) all dice will show the same numbers

73. The first box of 6 balls contains 4 red and 2 black, the second box of 7 balls contains 2 red and 5 black. One ball was transferred from the first box to the second, then one ball was transferred from the second to the first. Find the probability that the ball then drawn from the first box is black.

74. Two enterprises produce the same type of products. Moreover, the second produces 55% of the products of both enterprises. The probability of the first enterprise producing a non-standard product is 0.1, and the second is 0.15. a) Determine the probability that a product taken at random will turn out to be non-standard, b) The product taken will turn out to be non-standard. What is the probability that it was produced at the second plant.

Solution:

75. There are three urns. The first has 3 white and 2 black balls, the second and third have 4 white and 3 black balls. A ball is drawn from a randomly selected urn. He turned out to be white. What is the probability that the ball is drawn from the third urn?

Solution: P(H1) = 1/3; P(H2) =1/3; P(H3) = 1/3.

P(A) – probability of drawing a white ball.

If the 1st urn is chosen P(A/H1) = 3/5

2nd P(A/H2) = 4/7

3rd P(A/H3) = 4/7

P(A) = 1/3 * 3/5 + 1/3 * 4/7 + 1/3 * 4/7 = 12/21

P(H3/A) = (4/7 * 1/3) / (12/21) = 1/3

Answer: 1/3

76. Seeds for sowing are supplied to the farm from three seed farms. Moreover, the first and second farms each send 40% of all seeds. The germination rate of seeds from the first farm is 90%, the second is 85%, and the third is 95%. a) Determine the probability that a seed taken at random will not germinate, b) A seed taken at random will not germinate. What is the probability that it came from a second farm?

77. The exam program consists of 30 questions. Of the 20 students in the group, 8 people learned all the questions, 6 people learned 25 questions, 5 people learned 20 questions, and one person learned 10 questions. Determine the probability that a randomly called student will answer two questions on the ticket.

Solution: H1 is the choice of a student who has learned everything, H2 is the choice of a student who has learned 25 questions, H3 is the choice of a student who has learned 20 questions, H4 is the choice of a student who has learned 10 questions.

P(H1) = m/n = 8/20 = 2/5 m-those who have learned all the questions, n-all students.

P(H2) = 6/20 = 3/10

P(H3) = 5/20 = ¼

P(A/H1) = 1 – The probability that a student who has learned everything answered 2 questions on the ticket out of 25 questions he learned.

P(A/H2) = 25/30 = 5/6 – the probability that the student will answer 2 questions on the ticket out of 25 questions he has learned.

P(A/H3) = 20/30 = 2/3 – the probability that a student who has learned 20 questions will answer 2 questions on the ticket.

P(A/H4) = 10/30 = 1/3 – the probability that a student who has learned 10 questions will answer 2 questions on the ticket.

Using the total probability formula, we find the probability that a randomly called student will answer 2 questions on the ticket:

P(A) = ∑ P(H i) P(A/H i) = P(H1)P(A/H1) + P(H2)P(A/H2) + P(H3)P(A/H3) ) + P(H4) P(A/H4)

P(A) = 2/5*1 + 3/10*5/6 + 1/4*2/3 + 1/20*1/3 = 2/5 + 1/4+ 1/6 + 1/60 = 24/60 +15/60 +10/60 + 1/60 = 50/60 = 5/6

Answer: 5/6

78. Before sowing, 95% of seeds are treated with a special solution. Seed germination after treatment is 99%, untreated 85%. A) What is the probability that a randomly selected seed will germinate? B) The randomly taken seed sprouted. What is the probability that it came from treated seed?

Solution: H1-treated seeds, H2 – untreated seeds, A – sprouted seed.

95% + 5% = 100% => P(H1) = 0.95 ; P(H2) = 0.05

P(A/H1) = 0.99 – the probability that a randomly taken seed will germinate if it is processed.

P(A/H2) = 0.85 – The probability that a randomly selected seed will germinate if it is untreated.

A) using the total probability formula, we find the probability that a randomly taken seed will sprout:

P(A) = ∑ P(H i) P(A/H i) = ∑ P(H i)P(A/H i) = P(H1) P(A/H1) + P(H2)P( A/H2)

P(A) = 0.95*0.99 + 0.05*0.85 = 0.9405 +0.0425 = 0.983

Answer: 0.983

79. The store receives televisions from four factories. The probability that the TV will not have a malfunction during the year is: for the first plant 0.9, for the second 0.8, for the third 0.8 and for the fourth 0.99. A randomly selected TV failed within a year. What is the probability that it was manufactured in the first plant?

80. A buyer is equally likely to visit each of the three stores. The probability that a customer will buy a product in the first store is 0.4, the second is 0.6 and the third is 0.8. Determine the probability that a customer will buy a product in a particular store. The buyer bought the product. Find the probability that he bought it in the second store.

Answer: 0.7157

2. A worker operates 3 machines. The probability of failure-free operation of the first of them is 0.75, the second is 0.85,
third 0.95. Find the probability that a) two machines will fail, b) all three machines will work without failure, c) at least one machine will fail.

3. From a deck containing 52 cards, 3 is drawn at random. Find the probability that it is a three, a seven, and an ace.

4. Find the probability that a subscriber will dial the correct two-digit number if he knows that the given the number is not divisible by 5

Solution: P(A) = m/n; m=1/

Let's count the total number of two-digit numbers. It is equal to 90 and subtract from these numbers those that are divisible by 5 (10,15,20,25...90,95). Their number is 18 => n=90-18=72

Answer: 1/72

5. A die is tossed 2 times: a) Find the probability that the sum of points on the upper faces will be 7. b) find the probability that at least 2 points will appear during one toss.

Solution: P(A)=m/n

a) P(A)=6/36 =1/6

b) P(B)=1-5/6*5/6=1-25/36 =11/36

6. There are 5 black and 7 red balls in the urn. Three balls are drawn sequentially (without returning). Find the probability that a) all three balls will be red, b) three balls will be red or black.

Solution: C m n = n! /m!(n-m)!

C 3 12 = 220 - options for drawing three balls.

a) You can get 3 reds out of 7 in 7 ways.

m = C 3 7 = 7! / 3!*4! = 35

P (A1) = m/n = 35/220 = 7/44

b) you can get 3 reds out of 7 in 7 ways, and 3 blacks out of 5 =>

With 3 5 ways.

m = C 3 7 + C 3 5 = 35 + 5! / 3!*2! = 35 + 10 = 45

P(A2) = m/n = 45/220 = 9/44

Answer: a) P(A) = 7/44; b) P(A2) = 9/44

In a group of 15 people, 6 people play sports. Find the probability that out of 7 randomly selected people, 5 people go in for sports.

Solution: P(A) = C 5 6 * C 2 9 / C 7 15 = ((6!/(5!*1!))*(9!/(2!*7!)) / (15! / (7 !*8!) = (5*36) / (15* 14* 13* 12* 11* 10* 9* 8!) / (1*2*3*4*5*6*7*8) = ( 5*36*12) / (15*13*11*3) = 4/143 =0.03

Answer: 0.3.

The mouse can choose one of 5 mazes at random. It is known that the probability of her exiting various labyrinths in 3 minutes is 0.5; 0.6; 0.2; 0.1; 0.1. Let it turn out that the mouse got out of the maze in 3 minutes. What is the probability that she chose the first maze? Second labyrinth?

Solution: Initially, the probabilities of choosing a maze with the mouse are equal to:

P(H1) = P(H2) = P(H3) = P(H4) = P(H5) = 1/5 – probability of choosing the 1,2,3,4,5 maze, respectively.

A – exit from the maze.

P(A/H1) = 0.5 – Probability of a mouse exiting 1 maze

P(A/H2) = 0.6 – from 2 labyrinths.

P(A/H3) =0.2 – from the 3rd labyrinth

P(A/H4) = 0.1 – from 4 labyrinths

P(A/H5) = 0.1 – from 5 maze

According to the total probability formula:

P(A) = ∑ P(H i)P(A/H i) = P(H1)P(A/H1) + P(H2)P(A/H2) + P(H3)P(A/H3 ) +P(H4)P(A/H4) +P(H5)P(A/H5)

P(A) = 1/5*0.5 + 1/5*0.6 + 1/5*0.2 + 1/5*0.1 +1/5*0.1 = 1/5 (0 .5+0.6+0.2+0.1+0.1)=1/5*1.5=1.5*3/2 = 3/10 – probability of a mouse exiting the maze in 3 minutes.

A) Find the probability that the mouse chose the first maze (using Bayes’ formula):

P(H1/A) = P(H1)P(A/H1) / P(A) = (0.5*1/5)/(3/10) = (1/2*1/5) /( 3/10) = 1/10*10/3 = 1/3

B) Find the probability that the mouse chose the second maze (using Bayes’ formula)

P(H2/A) = P(H2)P(A/H2) / P(A) = (1/5*0.6) / 3/10 = (1/5*3/5) / 3/10 = 3/25* 10/3 = 10/25 = 2/5

Answer: 1/3; 2/5

9. Out of 10 tickets, 2 are winning. Find the probability that out of 5 tickets, one is winning.

10. In September the probability of a rainy day is 0.3. Team "Statistician" wins on a clear day with a probability of 0.8, and on a rainy day this probability is 0.3. It is known that in September they won a certain game. What is the probability that on that day: a) it rained; b) it was a clear day.

11. The probability of the first shooter hitting the target is 0.7, the second - 0.5, and the third -0.4. Find the probability that at least one shooter will hit the target .

Solution:

The first box contains 20 parts, of which 10 are standard, the second box contains 30 parts, of which 25 are standard, the third box contains 10 parts, of which 8 are standard. One part was taken at random from a randomly selected box, which turned out to be standard. Find the probability that it was taken from the second box.

Solution: P(H i) = 1/3; P(A/H1)=10/20=1/2; P(A/H2)=25/30=5/6;

P(A/H3)=8/10=4/5;

P(A)=1/3(1/2+5/6+4/5) = 62/45

P(H2/A) = (P(H2)*P(A/H2)) / P(A) = (1/3*5/6) /62/45 = 0.39

13. Each of five identical cards contains one of the following letters: A, E, N, C, T. Cards
mixed. Determine the probability that from the cards taken out and placed in a row a) it is possible to make
the word “WALL”, b) from three cards you can make the word “NO”.

To hit the target, at least one projectile is enough to hit it. Two salvoes were fired from two guns. Find the probability of hitting a target if the probability of hitting the target with one shot from the first gun is 0.46, the second is 0.6.

Solution:

Let B have no hits

A1 – hits on the 1st shot.

A2 – hit on the 2nd shot.

P(B) = -- A1 - A2 = 0.54* 0.4 = 0.216

Then C - at least one hit.

P(C)= 1 - 0.216 = 0.784

Answer: 0.784

There are 3 urns. The first urn contains 6 blacks and 4 whites, the second contains 5 whites and 5 blacks, the third contains 7 whites and 3 blacks. An urn is randomly selected and a ball is drawn from it, which turns out to be white. Find the probability that the second urn is chosen.

Solution:

H1=1/3; H2=1/3; H3=1/3

P(H/H1) = 4/10; P(H/H2) = 1/2; P(H/H3) = 7/10

P(H) = 1/3*4/10 + 1/3*1/2 + 1/3*1/7 = 16/30

P(H2/H) = (1/2*1/3)/ (8/15) = 1/6* 15/8 = 15/48

Answer: 15/48 = 0.3125

16. The coin is tossed 3 times. Find the probability that the coat of arms will appear: a) all 3 times, b) only once, c) at least once

Solution:

17. The numbers 0, 1,2, 3, 4, 5, 6, 7, 8, 9 are written on individual cards. All cards are mixed, after which 5 cards are taken at random and laid out in a row. Determine the probability that the number 1 2 0 3 5 will be obtained. (Solve the problem using the definition of the probability of an event and theorems of probability theory)

Three famous economists simultaneously proposed their theories, which were considered equally probable. After observing the state of the economy, it turned out that the probability of the development that it actually received in accordance with the first theory is 0.5; from the second – 0.7; from the third – 0.4. How will this change the probabilities of correctness of the three theories.

Solution:

P(A/H1)=0.5; P(A/H2)=0.7; P(A/H3)=0.4

P(A)=P(H1)*P(A/H1)+…=1/3*0.5+1/3*0.7+

1/3*0,4=1/3(0,5+0,7+0,4)=1,6/3=0,533

P(H1/A)=(1/3*0.5)/(1/3*1.6)=0.5/1.6=0.32.

P(H2/A)=0.7/1.6=0.42

The Store sells 4 tape recorders. The probability that they will withstand the warranty period is respectively equal to: 0.91; 0.9; 0.95; 0.94. Find the probability that a randomly purchased tape recorder will survive the warranty period.

Solution: Probability of buying 1 tape recorder –1/4; 2 – 1/4; 3 – 1/4; 4 –1/4.

P(A) = 1/4 * 0.91 + ¼ * 0.9 + ¼ * 0.95 + ¼ * 0.94 = 0.2275 + 0.225 + 0.2375 + 0.235 = 0.925

Answer: P(A) = 0.925

Problem No. 1.26

The car number contains four digits, each of which can equally take values ​​from 0 to 9 (the number 0000 is possible). Determine the probability that the second digit of the number is four.

Let's find the number of all possible combinations of the car number:

The 2nd digit of the number is 4 if its combination is a set of the form: X 4 XX, where X is any digit from 0 to 9.

Therefore, the number of such numbers is equal to:

The probability that the second digit of the number is four.

Answer:

Problem No. 2.11

A diagram of the connection of elements forming a circuit with one input and one output is given (Figure 1). It is assumed that element failures are collectively independent events. Failure of any of the elements leads to an interruption of the signal in the branch of the circuit where this element is located. The failure probabilities of elements 1, 2, 3, 4, 5 are respectively equal to q1=0.1; q2=0.2; q3=0.3; q4=0.4; q5=0.5. Find the probability that the signal will pass from input to output.

Picture 1

According to Figure 1, elements 1, 2, 3 are connected in parallel to each other and in series with element 4.

Let's enter the events: A­ 1 – element 1 is OK, A­ 2 – element 2 is OK, A­ 3 – element 3 is OK, A­ 4 – element 4 is OK, B– the signal passes from the point a to the point b, C– the signal passes from the point a to the point c(from entrance to exit).

Event B will happen if either element 1, or element 2, or element 3 works:

B :

Event C will happen if the event occurs B and event A 4 :

Probability of an event occurring C :

Answer:

Problem No. 3.28

Devices of the same name are manufactured at three factories. The first plant supplies 45% of all products entering production, the second - 30% and the third - 25%. The probability of failure-free operation of a device manufactured at the first plant is 0.8, at the second - 0.85 and at the third - 0.9. The device entered production turned out to be in good working order. Determine the probability that it was manufactured at the second plant.

Let us denote by A the event - the device received for production is in good working order.

Let's make a number of assumptions:

The device came from the 1st factory:

The device came from the 2nd factory:

The device came from the 3rd factory:

The corresponding conditional probabilities for each of the hypotheses are:

Using the total probability formula, we find the probability of an event A:

Let's calculate the probability that a working device came from the 2nd plant:

Answer:

Problem No. 4.26

A coin is tossed 100 times. What is the probability that it will never land with the coat of arms facing up?

Event - the coin never landed face up in 100 tosses.

Probability that the coin did not land face up p=0,5 and therefore, the probability that the coin fell with the coat of arms up q=0,5 :

Let's determine the probability of the event A according to Bernoulli's formula ( n = 100; k =100 )

Answer:

Problem No. 5.21

A discrete random variable X can take one of five fixed values ​​x1, x2, x3, x4, x5 with probabilities p1, p2, p3, p4, p5, respectively. Calculate the mathematical expectation and variance of the value X. Calculate and plot the distribution function.

Table 1 – Initial data

Mathematical expectation and dispersion of the X value:

Let's construct a series of distribution of SV X:

Table 2 – Distribution series SV X

Let's plot the distribution function (Figure 2):

Figure 2 - graph of the distribution function F(X i)

Problem No. 6.3

Random value X given by the probability density:

Define a constant WITH, mathematical expectation, dispersion, distribution function of the value X, as well as the probability of it falling into the interval.

Hence the constant:

Let us determine the mathematical expectation of SV X:

Let us determine the dispersion of SV X:

Let us define the distribution function of the value X:

Answer:

Problem No. 7.15

Random value X distributed uniformly over the interval [ a,b]. Plot a random variable Y= (X) and determine the probability density g(y).

there are no inverse functions

Figure 3 – function graph

Since the random variable X is distributed uniformly over the interval, then its probability density is equal to:

Let us determine the probability density of the quantity:

Problem No. 8.30

2D random vector ( X, Y) is uniformly distributed within the area B highlighted by bold straight lines in Figure 4. Two-dimensional probability density f(x,y) is the same for any point in this region B:

Calculate the correlation coefficient between the values ​​of X and Y.

Table 3 - Initial data

Figure 4

Let's build an area B according to the coordinates from Table 5 and Figure 4.

Figure 5

Let's analyze Figure 5: area B on the interval is bounded on the left by a straight line, on the right, on the interval is bounded on the left by a straight line, on the right -

Therefore, the joint probability density will take the form:

Thus:

Let's check the result obtained geometrically. Volume of a body limited by the distribution surface IN and the plane xOy is equal to 1, i.e.:

Therefore, the constant was calculated correctly.

Let's calculate the mathematical expectations:

Let's calculate the variances:

Let's calculate the correlation moment:

Let's calculate the correlation coefficient between the values ​​of X and Y:

Answer:

Problem No. 9

Based on a sample of a one-dimensional random variable:

Get a variation series;

Plot the empirical distribution function F * (x) ;

Construct a histogram using the equal-interval method;

Construct a histogram using an equal probability method;

Calculate point estimates of expectation and variance;

Calculate interval estimates of mathematical expectation and dispersion (γ = 0.95);

Put forward a hypothesis about the distribution law of a random variable and test it using the goodness-of-fit test  2 and Kolmogorov criterion (  = 0,05).

Univariate sampling:

Sample size

Solution

1. We obtain a variation series from the original one:

Let's build a histogram using the equal-interval method (Figure 7).

To construct a histogram, we will compose an interval statistical series, taking into account that the length of all intervals must be the same.

Number of intervals;

- interval width;

Frequency of SV X hitting the j-th interval;

Statistical density in the jth interval.

Table 4 – Interval statistical series

f * (x)

Figure 7

Let's build a histogram using an equal probability method (Figure 8).

To construct a histogram, we will compile an interval statistical series, taking into account that the frequency of SV X hitting in each j-th interval should be the same (Table 5).

Table 5 – Interval statistical series

f * (x)

Figure 8

Let us calculate point estimates of the mathematical expectation and variance:

Let us calculate the interval estimates of the mathematical expectation and dispersion (γ = 0.95):

H 0 – the value of X is distributed according to the exponential law:

H 1 – the value of X is not distributed according to an exponential law

Thus, we obtain a fully defined hypothetical distribution function:

Let's check the hypothesis about the normal law using the Pearson criterion. Let's calculate the value of the criterion based on an equal-interval statistical series:

We calculate the theoretical probabilities of falling into the intervals using the formula:

Table 6 – Calculation results

Let's check the correctness of the calculations:

Let's calculate the Pearson criterion:

Let's determine the number of degrees of freedom:

We select the critical value of the Pearson criterion from the table for the degree of freedom and the given significance level:

Since the condition is met, the hypothesis H 0 about the exponential distribution law is accepted (there is no reason to reject it).

8) Let's check the hypothesis using the Kolmogorov criterion. To do this, we will plot a hypothetical distribution function in the same coordinate system with the empirical function (Figure 6). We use 10 values ​​from Table 6 as reference points. Using the graph, we determine the maximum absolute deviation between the functions and :

Let us calculate the value of the Kolmogorov criterion:

From the Kolmogorov table, according to a given significance level, we select the critical value of the criterion:

Since the condition is satisfied, the hypothesis H 0 about the exponential distribution law is accepted (there is no reason to reject it).