What is a probability?

Faced with this term for the first time, I would not understand what it is. So I'll try to explain in an understandable way.

Probability is the chance that the desired event will occur.

For example, you decided to visit a friend, remember the entrance and even the floor on which he lives. But I forgot the number and location of the apartment. And now you are standing on the stairwell, and in front of you are the doors to choose from.

What is the chance (probability) that if you ring the first doorbell, your friend will open it for you? Whole apartment, and a friend lives only behind one of them. With equal chance, we can choose any door.

But what is this chance?

Doors, the right door. Probability of guessing by ringing the first door: . That is, one time out of three you will guess for sure.

We want to know by calling once, how often will we guess the door? Let's look at all the options:

1. you called to 1st door
2. you called to 2nd door
3. you called to 3rd door

And now consider all the options where a friend can be:

A. Behind 1st door
b. Behind 2nd door
V. Behind 3rd door

Let's compare all the options in the form of a table. A tick indicates the options when your choice matches the location of a friend, a cross - when it does not match.

How do you see everything Maybe options friend's location and your choice of which door to ring.

A favorable outcomes of all . That is, you will guess the times from by ringing the door once, i.e. .

This is the probability - the ratio of a favorable outcome (when your choice coincided with the location of a friend) to the number of possible events.

The definition is the formula. Probability is usually denoted p, so:

It is not very convenient to write such a formula, so let's take for - the number of favorable outcomes, and for - the total number of outcomes.

The probability can be written as a percentage, for this you need to multiply the resulting result by:

Probably, the word “outcomes” caught your eye. Since mathematicians call various actions (for us, such an action is a doorbell) experiments, it is customary to call the result of such experiments an outcome.

Well, the outcomes are favorable and unfavorable.

Let's go back to our example. Suppose we rang at one of the doors, but it was opened to us stranger. We didn't guess. What is the probability that if we ring one of the remaining doors, our friend will open it for us?

If you thought that, then this is a mistake. Let's figure it out.

We have two doors left. So we have possible steps:

1) Call to 1st door
2) Call 2nd door

A friend, with all this, is definitely behind one of them (after all, he was not behind the one we called):

a) a friend 1st door
b) a friend for 2nd door

Let's draw the table again:

As you can see, there are all options, of which - favorable. That is, the probability is equal.

Why not?

The situation we have considered is example of dependent events. The first event is the first doorbell, the second event is the second doorbell.

And they are called dependent because they affect the following actions. After all, if a friend opened the door after the first ring, what would be the probability that he was behind one of the other two? Right, .

But if there are dependent events, then there must be independent? True, there are.

A textbook example is tossing a coin.

1. We toss a coin. What is the probability that, for example, heads will come up? That's right - because the options for everything (either heads or tails, we will neglect the probability of a coin to stand on edge), but only suits us.
2. But the tails fell out. Okay, let's do it again. What is the probability of coming up heads now? Nothing has changed, everything is the same. How many options? Two. How much are we satisfied with? One.

And let tails fall out at least a thousand times in a row. The probability of falling heads at once will be the same. There are always options, but favorable ones.

Distinguishing dependent events from independent events is easy:

1. If the experiment is carried out once (once a coin is tossed, the doorbell rings once, etc.), then the events are always independent.
2. If the experiment is carried out several times (a coin is tossed once, the doorbell is rung several times), then the first event is always independent. And then, if the number of favorable or the number of all outcomes changes, then the events are dependent, and if not, they are independent.

Let's practice a little to determine the probability.

Example 1

The coin is tossed twice. What is the probability of getting heads up twice in a row?

Solution:

Consider all possible options:

1. eagle eagle
2. tails eagle
3. tails-eagle
4. Tails-tails

As you can see, all options. Of these, we are satisfied only. That is the probability:

If the condition asks simply to find the probability, then the answer must be given in the form decimal fraction. If it were indicated that the answer must be given as a percentage, then we would multiply by.

Answer:

Example 2

In a box of chocolates, all candies are packed in the same wrapper. However, from sweets - with nuts, cognac, cherries, caramel and nougat.

What is the probability of taking one candy and getting a candy with nuts. Give your answer in percentage.

Solution:

How many possible outcomes are there? .

That is, taking one candy, it will be one of those in the box.

And how many favorable outcomes?

Because the box contains only chocolates with nuts.

Answer:

Example 3

In a box of balls. of which are white and black.

1. What is the probability of drawing a white ball?
2. We added more black balls to the box. What is the probability of drawing a white ball now?

Solution:

a) There are only balls in the box. of which are white.

The probability is:

b) Now there are balls in the box. And there are just as many whites left.

Answer:

Full Probability

The probability of all possible events is ().

For example, in a box of red and green balls. What is the probability of drawing a red ball? Green ball? Red or green ball?

Probability of drawing a red ball

Green ball:

Red or green ball:

As you can see, the sum of all possible events is equal to (). Understanding this point will help you solve many problems.

Example 4

There are felt-tip pens in the box: green, red, blue, yellow, black.

What is the probability of drawing NOT a red marker?

Solution:

Let's count the number favorable outcomes.

NOT a red marker, that means green, blue, yellow, or black.

Probability of all events. And the probability of events that we consider unfavorable (when we pull out a red felt-tip pen) is .

Thus, the probability of drawing NOT a red felt-tip pen is -.

Answer:

The probability that an event will not occur is minus the probability that the event will occur.

Rule for multiplying the probabilities of independent events

You already know what independent events are.

And if you need to find the probability that two (or more) independent events will occur in a row?

Let's say we want to know what is the probability that by tossing a coin once, we will see an eagle twice?

We have already considered - .

What if we toss a coin? What is the probability of seeing an eagle twice in a row?

Total possible options:

1. Eagle-eagle-eagle
2. Eagle-head-tails
3. Head-tails-eagle
4. Head-tails-tails
5. tails-eagle-eagle
6. Tails-heads-tails
7. Tails-tails-heads
8. Tails-tails-tails

I don't know about you, but I made this list wrong once. Wow! And only option (the first) suits us.

For 5 rolls, you can make a list of possible outcomes yourself. But mathematicians are not as industrious as you.

Therefore, they first noticed, and then proved, that the probability of a certain sequence of independent events decreases each time by the probability of one event.

In other words,

Consider the example of the same, ill-fated, coin.

Probability of coming up heads in a trial? . Now we are tossing a coin.

What is the probability of getting tails in a row?

This rule does not only work if we are asked to find the probability that the same event will occur several times in a row.

If we wanted to find the TAILS-EAGLE-TAILS sequence on consecutive flips, we would do the same.

The probability of getting tails - , heads - .

The probability of getting the sequence TAILS-EAGLE-TAILS-TAILS:

You can check it yourself by making a table.

The rule for adding the probabilities of incompatible events.

So stop! New definition.

Let's figure it out. Let's take our worn out coin and flip it once.
Possible options:

1. Eagle-eagle-eagle
2. Eagle-head-tails
3. Head-tails-eagle
4. Head-tails-tails
5. tails-eagle-eagle
6. Tails-heads-tails
7. Tails-tails-heads
8. Tails-tails-tails

So here are incompatible events, this is a certain, given sequence of events. are incompatible events.

If we want to determine what is the probability of two (or more) incompatible events, then we add the probabilities of these events.

You need to understand that the loss of an eagle or tails is two independent events.

If we want to determine what is the probability of a sequence) (or any other) falling out, then we use the rule of multiplying probabilities.
What is the probability of getting heads on the first toss and tails on the second and third?

But if we want to know what is the probability of getting one of several sequences, for example, when heads come up exactly once, i.e. options and, then we must add the probabilities of these sequences.

Total options suits us.

We can get the same thing by adding up the probabilities of occurrence of each sequence:

Thus, we add probabilities when we want to determine the probability of some, incompatible, sequences of events.

There is a great rule to help you not get confused when to multiply and when to add:

Let's go back to the example where we tossed a coin times and want to know the probability of seeing heads once.
What is going to happen?

Should drop:
(heads AND tails AND tails) OR (tails AND heads AND tails) OR (tails AND tails AND heads).
And so it turns out:

Let's look at a few examples.

Example 5

There are pencils in the box. red, green, orange and yellow and black. What is the probability of drawing red or green pencils?

Solution:

What is going to happen? We have to pull out (red OR green).

Now it’s clear, we add up the probabilities of these events:

Answer:

Example 6

A die is thrown twice, what is the probability that a total of 8 will come up?

Solution.

How can we get points?

(and) or (and) or (and) or (and) or (and).

The probability of falling out of one (any) face is .

We calculate the probability:

Answer:

Training.

I think now it has become clear to you when you need to how to count the probabilities, when to add them, and when to multiply them. Is not it? Let's get some exercise.

Tasks:

Let's take a deck of cards in which the cards are spades, hearts, 13 clubs and 13 tambourines. From to Ace of each suit.

1. What is the probability of drawing clubs in a row (we put the first card drawn back into the deck and shuffle)?
2. What is the probability of drawing a black card (spades or clubs)?
3. What is the probability of drawing a picture (jack, queen, king or ace)?
4. What is the probability of drawing two pictures in a row (we remove the first card drawn from the deck)?
5. What is the probability, taking two cards, to collect a combination - (Jack, Queen or King) and Ace The sequence in which the cards will be drawn does not matter.

Answers:

1. In a deck of cards of each value, it means:
2. The events are dependent, since after the first card drawn, the number of cards in the deck has decreased (as well as the number of "pictures"). Total jacks, queens, kings and aces in the deck initially, which means the probability of drawing the “picture” with the first card:

   Since we are removing the first card from the deck, it means that there is already a card left in the deck, of which there are pictures. Probability of drawing a picture with the second card:

   Since we are interested in the situation when we get from the deck: “picture” AND “picture”, then we need to multiply the probabilities:

   Answer:

3. After the first card is drawn, the number of cards in the deck will decrease. Thus, we have two options:
   1) With the first card we take out Ace, the second - jack, queen or king
   2) With the first card we take out a jack, queen or king, the second - an ace. (ace and (jack or queen or king)) or ((jack or queen or king) and ace). Don't forget about reducing the number of cards in the deck!

If you were able to solve all the problems yourself, then you are a great fellow! Now tasks on the theory of probability in the exam you will click like nuts!

PROBABILITY THEORY. AVERAGE LEVEL

Consider an example. Let's say we throw a die. What kind of bone is this, do you know? This is the name of a cube with numbers on the faces. How many faces, so many numbers: from to how many? Before.

So we roll a die and want it to come up with an or. And we fall out.

In probability theory they say what happened favorable event(not to be confused with good).

If it fell out, the event would also be auspicious. In total, only two favorable events can occur.

How many bad ones? Since all possible events, then the unfavorable of them are events (this is if it falls out or).

Definition:

Probability is the ratio of the number of favorable events to the number of all possible events.. That is, the probability shows what proportion of all possible events are favorable.

The probability is denoted by a Latin letter (apparently, from English word probability - probability).

It is customary to measure the probability as a percentage (see topics and). To do this, the probability value must be multiplied by. In the dice example, probability.

And in percentage: .

Examples (decide for yourself):

1. What is the probability that the toss of a coin will land on heads? And what is the probability of a tails?
2. What is the probability that an even number will come up when a dice is thrown? And with what - odd?
3. In a drawer of plain, blue and red pencils. We randomly draw one pencil. What is the probability of pulling out a simple one?

Solutions:

1. How many options are there? Heads and tails - only two. And how many of them are favorable? Only one is an eagle. So the probability

   Same with tails: .

2. Total options: (how many sides a cube has, so many different options). Favorable ones: (these are all even numbers :).
   Probability. With odd, of course, the same thing.
3. Total: . Favorable: . Probability: .

Full Probability

All pencils in the drawer are green. What is the probability of drawing a red pencil? There are no chances: probability (after all, favorable events -).

Such an event is called impossible.

What is the probability of drawing a green pencil? There are exactly as many favorable events as there are total events (all events are favorable). So the probability is or.

Such an event is called certain.

If there are green and red pencils in the box, what is the probability of drawing a green or a red one? Yet again. Note the following thing: the probability of drawing green is equal, and red is .

In sum, these probabilities are exactly equal. That is, the sum of the probabilities of all possible events is equal to or.

Example:

In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of not drawing green?

Solution:

Remember that all probabilities add up. And the probability of drawing green is equal. This means that the probability of not drawing green is equal.

Remember this trick: The probability that an event will not occur is minus the probability that the event will occur.

Independent events and the multiplication rule

You flip a coin twice and you want it to come up heads both times. What is the probability of this?

Let's go through all the possible options and determine how many there are:

Eagle-Eagle, Tails-Eagle, Eagle-Tails, Tails-Tails. What else?

The whole variant. Of these, only one suits us: Eagle-Eagle. So, the probability is equal.

Fine. Now let's flip a coin. Count yourself. Happened? (answer).

You may have noticed that with the addition of each next throw, the probability decreases by a factor. The general rule is called multiplication rule:

The probabilities of independent events change.

What are independent events? Everything is logical: these are those that do not depend on each other. For example, when we toss a coin several times, each time a new toss is made, the result of which does not depend on all previous tosses. With the same success, we can throw two different coins at the same time.

More examples:

1. A die is thrown twice. What is the probability that it will come up both times?
2. A coin is tossed times. What is the probability of getting heads first and then tails twice?
3. The player rolls two dice. What is the probability that the sum of the numbers on them will be equal?

Answers:

1. The events are independent, which means that the multiplication rule works: .
2. The probability of an eagle is equal. Tails probability too. We multiply:
3. 12 can only be obtained if two -ki fall out: .

Incompatible events and the addition rule

Incompatible events are events that complement each other to full probability. As the name implies, they cannot happen at the same time. For example, if we toss a coin, either heads or tails can fall out.

Example.

In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of drawing green or red?

Solution .

The probability of drawing a green pencil is equal. Red - .

Auspicious events of all: green + red. So the probability of drawing green or red is equal.

The same probability can be represented in the following form: .

This is the addition rule: the probabilities of incompatible events add up.

Mixed tasks

Example.

The coin is tossed twice. What is the probability that the result of the rolls will be different?

Solution .

This means that if heads come up first, tails should be second, and vice versa. It turns out that there are two pairs of independent events here, and these pairs are incompatible with each other. How not to get confused about where to multiply and where to add.

There is a simple rule for such situations. Try to describe what should happen by connecting the events with the unions "AND" or "OR". For example, in this case:

Must roll (heads and tails) or (tails and heads).

Where there is a union "and", there will be multiplication, and where "or" is addition:

Try it yourself:

1. What is the probability that two coin tosses come up with the same side both times?
2. A die is thrown twice. What is the probability that the sum will drop points?

Solutions:

1. (Heads up and heads up) or (tails up and tails up): .
2. What are the options? And. Then:
   Rolled (and) or (and) or (and): .

Another example:

We toss a coin once. What is the probability that heads will come up at least once?

Solution:

Oh, how I don’t want to sort through the options ... Head-tails-tails, Eagle-heads-tails, ... But you don’t have to! Let's talk about full probability. Remembered? What is the probability that the eagle will never drop? It's simple: tails fly all the time, that means.

PROBABILITY THEORY. BRIEFLY ABOUT THE MAIN

Probability is the ratio of the number of favorable events to the number of all possible events.

Independent events

Two events are independent if the occurrence of one does not change the probability of the other occurring.

Full Probability

The probability of all possible events is ().

The probability that an event will not occur is minus the probability that the event will occur.

Rule for multiplying the probabilities of independent events

The probability of a certain sequence of independent events is equal to the product of the probabilities of each of the events

Incompatible events

Incompatible events are those events that cannot possibly occur simultaneously as a result of an experiment. A number of incompatible events form a complete group of events.

The probabilities of incompatible events add up.

Having described what should happen, using the unions "AND" or "OR", instead of "AND" we put the sign of multiplication, and instead of "OR" - addition.

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Evening gradually enveloped the majestic castle of Zmiulan. Gradually, torches were lit in the corridors, the students hurried to go to their rooms. And now, when the corridors were already empty, a man came out from around the corner: an expensive black suit ideally sat on his taut figure, blond hair was combed back, pistachio-colored eyes looked only forward with an indifferent look. Norton Ognev, and it was him, approached the office of the Great Spirit Ostala. After knocking and receiving permission, the man entered the room. - So, why did you come, Norton? - the owner of the castle himself stood with his back to Vasilisa's father, looking out the window. The indifference did not disappear from Ognev's face, but he tensed inwardly. “Mr. Astragor, I need to go to Chernovod for a few days,” the head of the Dragocievs turned around. - As I understand, you will not go alone? - Norton Sr. slowly nodded: - Yes, Mr. Astragor. If you don't mind, I'll take my daughter, Fasha, and Zaharra with me. -And why do you, Norton, take my nephews with you? - with a certain interest, the head Dragotsiev looked at Ognev. Vasilisa asked, - as if reluctantly answered Norton Sr. Astragoras stared thoughtfully at the flames in the fireplace. Ognev patiently waited for an answer... *** The night shrouded the majestic castle in a starry canvas. A light breeze rustled the foliage of the garden. Vasilisa was already getting ready for bed in the Green Room. “Oh, how long have I been here…” the girl drawled, looking around the room. She did not even remember the last time she was here, but she saw that everything was in its place. Suddenly, a guy flew in through the open window. Ogneva looked at the unexpected guest in surprise. Having hidden the black wings, the dark-haired one smiled at the hostess of the room: -Hi owls! -You scared me! - the girl exclaimed, irritably looking at the guy. “Oh, come on,” the guest chuckled. I think you will always be afraid of me. -Do not be silly! I will be afraid of such an arrogant guy like you,” Vasilisa said irritably. - By the way, Flash, why did you come, especially so late? Can't sleep again? "Yeah," Dragotius nodded. - I decided to arrange a tour of Chernovod for myself ... But walking alone is not very fun, and dangerous. In an unfamiliar castle after all, - Flash slyly flashed his eyes. - Would you like me to give you a tour? Vasilisa looked at her friend in bewilderment. -Why not? Do you know everything here? The brunette raised an eyebrow questioningly. "Almost," the redhead replied evasively. - Well, that's good, - Dragotsy went to the door. The fireman had nothing left to do but follow him. The guys walked along the dark corridors, lighting the lamps for hours. Vasilisa told Fash what she remembers in this castle. He listened attentively to her, sometimes interrupting or snorting maliciously at this or that proposal. Soon he got tired of just walking and listening to chatter, and he, remembering something, asked the question: -By the way, what kind of tower is there that we saw when we were driving in a carriage? -What do you mean? Ogneva asked thoughtfully. “It seems to be Western,” drawled Dragotsy. “Ah, this one,” the red-haired woman immediately realized. - We call her Lonely, there once were prisoners. Let's take a look there, shall we? Excitement flashed in the brunette's icy blue eyes. “Well, I don’t know…” Vasilisa drawled uncertainly. - Are you afraid? Dragotius chuckled. As Flash expected, they managed to take her weakly: the girl's face flared up, and she clenched her fists: -Let's go, - and Vasilisa led the rather smiling brunette to this tower. Having opened the door without obstacles, the guys entered the room. The door soon slammed shut. Flash went to the open window and jumped on the windowsill, inhaling the invigorating smell of the sea: - Oh, well ... - then turned to the redhead. - Come on, sit down, - and hit his palm on the spot next to him. The girl immediately settled down next to him. Full moon shone in the heights, and below the sea was agitated. Wave after wave rolled, crashing against the rocks. “What a bright moon,” Vasilisa looked up at the sky again. -I have a song about the moon. I've been composing for a long time, - Flash said suddenly. - So you can sing? - the red-haired Dragotia looked in surprise. He silently nodded. -What, you don't believe me? - the brunette approached Ognevoy's face, looking into the eyes of the interlocutor with a grin. I noticed that her cheeks turned pink, and her smile became wider. “No, it’s just…” Vasilisa, blushing, stammered, looking away from her icy blue eyes, which reflected the light of the moon. - It was simply not possible to confirm your words, - she again looked into those eyes. Flash began to slowly lean towards the redhead. She walked towards him. Only a few millimeters remain between their faces. Ogneva already felt a light breeze of exhalations on her lips. Their lips almost touched, and… -Oh, how sweet! - Vasilisa immediately pulled away from Dragotius and blushed even worse than before. Flash turned. Before his clear eyes appeared ... -Zakharra?! exclaimed two doves in surprise. -What are you doing here? The brunet looked at his sister in annoyance. Yes, I saw you flying somewhere, I decided to find out. I went out, I look, you are walking, chatting. The main thing is that you don't notice me. Well, I followed you and went, - laid out everything bobtailed. - Connecting native blood ... - Flash muttered, got down from the windowsill and went to his room. Vasilisa followed suit. Zakharra instantly slipped into the corridor behind Ogneva and also returned to her room ...

To build a probability tree, first of all, you need to draw the tree itself, then write down all the information known for this problem in the figure, and, finally, use the basic rules to calculate the missing numbers and complete the tree.

1. The probabilities are indicated at each of the endpoints and circled. At each level of the tree, the sum of these probabilities must equal 1 (or 100%). So, for example, in Fig. 6.5.1 The sum of probabilities at the first level is 0.20 + 0.80 = 1.00 and at the second level - 0.03 + 0.17 + 0.56 + 0.24 = 1.00. This rule helps to fill one empty circle in a column if the values ​​of all other probabilities of this level are known.

Rice. 6.5.1

2. Conditional probabilities are indicated next to each of the branches (except,
possibly first-level branches). For each of the groups of branches emerging from one point, the sum of these probabilities is also equal to 1 (or 100%).
For example, in fig. 6.5.1 for the first group of branches we get 0.15 + 0.85 =
1.00 and for the second group - 0.70 + 0.30 = 1.00. This rule allows
calculate one unknown value of conditional probability in a group of branches emanating from one point.

3. The circled probability at the beginning of the branch multiplied by the conditional
the probability next to this branch gives the probability written in a circle in
end of the branch. For example, in fig. 6.5.1 for the upper branch leading to the right
we have 0.20 x 0.15 = 0.03, for the next branch - 0.20 x 0.85 = 0.17; similar relationships hold for the other two branches. This rule can be used to compute a single unknown value
probabilities of three corresponding to some branch.

4. The value of the probability written in a circle is equal to the sum of the circled probabilities at the ends of all branches emerging from this circle
to the right. So, for example, for Fig. 6.5.1 exit the circle with a value of 0.20
two branches, at the ends of which are circled probabilities, the sum of which is equal to this value: 0.03 + 0.17 = 0.20. This rule allows you to find one unknown probability value in a group,
including this probability and all probabilities at the ends of the branches of the tree,
coming out of the corresponding circle.

Using these rules, it is possible, knowing all but one probability value for some branch or at some level, to find this unknown value.

37. What sample is called representative? How can a representative sample be taken?

Representativeness is the ability of the sample to represent the population under study. The more accurately the composition of the sample represents the population on the issues under study, the higher its representativeness.

A representative sample is one of the key concepts of data analysis. A representative sample is a sample from a population with a distribution F(x) representing the main features of the general population. For example, if there are 100,000 people in a city, half of which are men and half are women, then a sample of 1,000 people of which 10 are men and 990 are women will certainly not be representative. A public opinion poll built on its basis will, of course, contain a bias in estimates and lead to falsified results.

A necessary condition for constructing a representative sample is an equal probability of including each element of the general population in it.

The sample (empirical) distribution function gives, with a large sample size, a fairly good idea of ​​the distribution function F(x) of the original general population.

The leading principle underlying such a procedure is the principle of randomization, randomness. A sample is said to be random (sometimes we'll say simple random or pure random) if two conditions are met. First, the sample must be designed in such a way that any person or object within the population has an equal opportunity to be selected for analysis. Second, the sample must be designed so that any combination of n items (where n is simply the number of items, or cases, in the sample) has an equal chance of being selected for analysis.

When examining populations that are too large to run a real lottery, simple random samples are often used. Writing down the names of several hundred thousand objects, putting them into a drum and selecting a few thousand is still not an easy job. In such cases, a different, but equally reliable method is used. Each object in the collection is assigned a number. The sequence of numbers in such tables is usually given by a computer program called a random number generator, which, in essence, places a large number of numbers, randomly pulls them out and prints them out in the order they were received. In other words, the same process that is characteristic of the lottery takes place, but the computer, using numbers rather than names, makes a universal choice. This choice can be used by simply assigning a number to each of our objects.

Random number table like toy, can be used by multiple different ways, and in each case three Decisions must be made. First, it is necessary to decide how many digits We will use; secondly, it is necessary to develop a decision rule for their use; thirdly, you need to choose the starting point and the method of passing through the table.

Once this is done, we must develop a rule that links the numbers in the table to our object numbers. There are two possibilities here. The easiest way (although not necessarily the most correct) is to use only those numbers that fall into the number of numbers assigned to our objects. So if we have a population of 250 features (and thus use three digit numbers) and decide to start at the top left corner of the table and move down the columns, we will include feature numbers 100, 084, and 128 in our sample, and let's skip the numbers 375 and 990, which do not correspond to our objects. This process will continue until the number of objects needed for our sample is determined.

A more time-consuming, but methodologically more correct procedure is based on the premise that in order to preserve the randomness characteristic of the table, every number of a given dimension (for example, every three-digit number) must be used. Following this logic, and again dealing with a collection of 250 objects, we must divide the region of three-digit numbers from 000 to 999 into 250 equal intervals. Since there are 1000 such numbers, we divide 1000 by 250 and find that each part contains four numbers. So table numbers from 000 to 003 will correspond to object 004 to 007 - object 2, and so on. Now, in order to determine which object number corresponds to the number in the table, you should divide the three-digit number from the table and round to the nearest whole number.

And finally, we must choose in the table the starting point and the method of passage. The starting point can be the top left corner (as in the previous example), the bottom right corner, the left edge of the second line, or anywhere else. This choice is completely arbitrary. However, when working with the table, we must act systematically. We could take the first three digits of each five-digit sequence, the middle three digits, the last three digits, or even the first, second, and fourth digits. (From the first five-digit sequence, these various procedures produce the numbers 100, 009, 097, and 109, respectively.) We could apply these procedures from right to left, getting 790, 900, 001, and 791. We could go along the rows , considering each next digit in turn and ignoring the division into fives (for the first row, the numbers 100, 973, 253, 376 and 520 will be obtained). We could only deal with every third group of digits (eg 10097, 99019, 04805, 99970). There are many different possibilities, and each next one is no worse than the previous one. However, once we have made a decision about one way or another, we must systematically follow it in order to respect the randomness of the elements in the table as much as possible.

38. What interval do we call confidence interval?

The confidence interval is the allowable deviation of the observed values ​​from the true values. The size of this assumption is determined by the researcher, taking into account the requirements for the accuracy of the information. If the margin of error increases, the sample size decreases even if the confidence level remains at 95%.

The confidence interval shows in what range the results of sample observations (surveys) will be located. If we conduct 100 identical surveys in identical samples from a single population (for example, 100 samples of 1000 people each in a city with a population of 5 million), then at a 95% confidence level, 95 out of 100 results will fall within the confidence interval (for example, from 28% to 32% with a true value of 30%).

For example, the true number of city residents who smoke is 30%. If we select 1000 people 100 times in a row and in these samples we ask the question "do you smoke?", in 95 of these 100 samples at a 2% confidence interval, the value will be from 28% to 32%.

39 What is called the level of confidence (confidence level)?

The confidence level reflects the amount of data needed by the evaluator in order to assert that the program being examined has the desired effect. The social sciences traditionally use the 95% confidence level. However, for most community programs, 95% is overkill. A confidence level in the range of 80-90% is sufficient for an adequate assessment of the program. In this way, the size of the representative group can be reduced, thereby reducing the cost of the evaluation.

The statistical evaluation process tests the null hypothesis that the program did not have the intended effect. If the results obtained differ significantly from the initial assumptions about the correctness of the null hypothesis, then the latter is rejected.

40. Which of the two confidence intervals is larger: two-tailed 99% or two-tailed 95%? Explain.

The 2-sided 99% confidence interval is larger than the 95% because more values ​​fall into it. Doc-in:

Using z-scores, you can more accurately estimate the confidence interval and determine the overall shape of the confidence interval. The exact formulation of the confidence interval for the sample mean is as follows:

Thus, for a random sample of 25 observations satisfying normal distribution, with the confidence interval of the sample mean has the following form:

Thus, you can be 95% sure that the value lies within ±1.568 units of the sample mean. Using the same method, it can be determined that the 99% confidence interval lies within ±2.0608 units of the sample mean

value Thus, we have and hence , Similarly, we obtain lower limit, which is equal to

According to the many-worlds interpretation of quantum physics, we live in an infinite network of alternate universes. This is a serious statement that has certain and extremely serious scientific, philosophical and existential implications. Let's look at ten of them.

According to the hypothesis of the creator of quantum mechanics Hugh Everett, we live in the Universe, more precisely in the multiverse, in which many successive worlds are constantly born and branch off, each of which has a different version of you.

Quantum physicists have used the many-worlds interpretation to eliminate the nasty shortcoming of the Copenhagen interpretation, namely the claim that an unobservable phenomenon can exist in two states. That is, instead of claiming that it is both alive and dead, the many-world interpretation says that the cat simply "branched" into different worlds: in one he is alive, in the other he is dead.

60 years after its introduction, the many-worlds interpretation remains quite controversial issue. In a 2013 survey among quantum physicists, only a fifth indicated that they welcomed the many-worlds interpretation (by comparison, 42% of physicists adhere to the Copenhagen interpretation). Nevertheless, among the supporters of the multiverse there are very eminent scientists from the field of quantum physics - David Deutsch, Scott Aaronson, Sean Carroll.

No matter what state this theory is in, it is extremely interesting to speculate on its implications.

We live in a gigantic multiverse

Cosmologists take the fact that the world we observe is one as a matter of course. Speculation about a multiple universe has long been considered scientific heresy, but the likelihood that this is true is growing more and more. Physicists and metaphysicians, cosmologists, anthropologists, quantum fanatics - everyone is starting to think about it.

The main claim of the many-worlds interpretation is that everything that exists is made up of a quantum superposition of an unimaginably large—or infinite—number of universes. If this interpretation is correct, there must be an absolutely astonishing number of alternate worlds.

The wholeness of your life is an illusion

MMI also violates our concept of personality. We all perceive our life as a single and integral journey through space and time. In reality, we are an exponentially growing set of events that branch out from moment to moment. As a result, we must think of ourselves not as a person, but as a fraction.

The reason for this illusion is that multiple experiences are impossible, so we are left with the knowledge that we are one person. But this does not mean that our experience of reality is authentic or real. We must recognize - through MMI - that our lives are not exactly what they seem.

There are many versions of you

If MMI is correct, there are (or an infinite) number of versions of you, each of which perceives the world as a separate person and is unaware of the existence of other versions. Consequently, the sheer volume of alternative life paths is extremely large. Since birth, you - or what you think you are - have branched into different worlds. The complete set of you is a massive root system that grows exponentially, and each root represents new life.

Because MWI is about constant change, dependence on probabilities, each new instance of you must be different, seeing the world in which an alternative outcome of your life events has occurred. Therefore, there are worlds in which you still live with your ex, are more or less successful, have already died or experienced the death of loved ones who are alive in the present world. There may even be evil versions of you where you are terrorists or assassins. The possibilities are almost limitless as long as the basics of physics are not violated.

Do you still have free will

Considering that all possible decisions will be made by different versions of you, it is quite difficult for MMI to explain the issue of free will. If all choices have already been made in alternate worlds, why then go through all the trouble, weighing the pros and cons, making decisions? The collective fate of your alter egos is already predetermined, the choice is made for you.

MMI expert Michael Clive-Price points out that although all decisions have already been made, some are being made more often than others. In other words, each branch of the decision has its own "weight" that affects the usual laws of quantum statistics.

In addition, MMI would mean a certain non-determinism of being, albeit in a non-intuitive way. Whenever we ask the question: “Could I have made a different decision or act differently?”, MMI answers that yes, of course. And not only you, but also alternative version could you too. But why you chose this option, achieved certain results, it all comes down to the effect of quantum events on classical objects - including the reflections in your head.

Somewhere out there may exist extremely strange worlds

MMI necessarily leads to very strange possibilities. Again, all branch points are possible exactly as long as you don't break the laws of physics. It is important to note, however, that given the full range of possible worlds, it is more likely that you will find yourself in the most possible and rational of the worlds, since they occur with high frequency.

But there are also worlds in which extremely strange things happen. For example, someone flips a coin 1,000 times, and with that, a world arises in which he flips heads 1,000 times in a row.

There are also worlds in which someone will guess absolutely all the predictions of sports matches. Worlds in which a person without a musical education, seeing the piano for the first time, will play Rachmaninov's 3rd Piano Concerto, as the maestro himself would have played. The chances, however, of such an event are negligible and go beyond the limits of astronomical probabilities, although, of course, there are among the infinitely possible options.

However, it is this point that skeptics single out as the most acute, reducing the rationality of the MMI to a minimum.

You are somehow immortal

This thought experiment is called "quantum suicide". Imagine a situation in which a person is playing Russian roulette, in which half the barrel of a revolver is filled with bullets. In such a superposition, each turn of the drum will reset the chances of a person's suicide to 50/50. But MMI tells us that there must be a world where a man will never shoot himself even after 50 turns of the drum. Although the chances of this happening are close to zero, it must happen somewhere.

Curiously, physicist Max Tegmark says that this experiment could serve as proof of MMI, only it would require the death of many people before one lucky person gets to the finish line.

Another view of quantum immortality argues that a version of ourselves must always exist in order to observe the universe. Paul Halpern, author of Schrödinger's Cat, put it this way:

“What is human survival? We are all a collection of particles, set by quantum rules on deepest level. If every time a quantum transition occurs, our bodies and minds split, there will be copies that experience every possible outcome, including the one that determines whether we live or die. Suppose that in one case, a particular set of quantum transitions leads to an abnormal distribution of cells and causes a deadly form of cancer. For every transition there will always be an alternative that does not lead to cancer. It turns out that there will always be branches with survivors. Add to this the assumption that our consciousness will always reside only in living copies, and we can survive any number of potentially dangerous events associated with quantum transitions.

Communication between parallel worlds may be possible

In 1995, the quantum physicist Rainer Plaga proposed to experimentally test MMI, describing the procedure for the "inter-world" exchange of information and energy through "weak coupling".

Using standard quantum optical equipment, a single ion can be isolated from its environment in an ion trap. A quantum mechanical measurement can then be made with two separate results on another system, thus creating two parallel worlds. Depending on the result, the ion will only be excited in one of these parallel worlds before the ion decoheres during the interaction. environment. Plaga claims that we could detect this excitation in another parallel world, which would provide MMI with evidence - and would provide possible way send a message to a parallel reality.

No paradoxes of time travel

It's simple: the presence of alternative worlds will mean the absence of a single time scale on which you can navigate.

If one travels back in time, it would mean moving into completely new temporal paradigms. Accordingly, in MMI, paradoxes like returning to the past and killing grandfather simply do not find a place.

Everything has happened before and will happen again.

The most interesting consequence of an infinite number of worlds is that everything has already happened. Moreover, it will happen an infinite number of times.

Based on materialsIO9

1. Ω = (11,12,13,14,15,16, 21, 22,..., 66),

2. Ω = (2,3,4,5,6, 7,8,9,10,11,12)

3. ● A = (16,61,34, 43, 25, 52);

● B = (11.12, 21.13,31.14, 41.15, 51.16, 61)

● C = (12, 21.36,63.45, 54.33.15, 51, 24.42.66).

● D= (SUM POINTS IS 2 OR 3 );

● E= (TOTAL POINTS IS 10).

Describe the event: WITH= (CIRCUIT CLOSED) for each case.

Solution. Let's introduce the notation: event A- contact 1 is closed; event IN- contact 2 is closed; event WITH- the circuit is closed, the light is on.

1. For a parallel connection, the circuit is closed when at least one of the contacts is closed, so C = A + B;

2. For a series connection, the circuit is closed when both contacts are closed, so C \u003d A B.

Task. 1.1.4 Two electrical circuits have been drawn up:

Event A - the circuit is closed, event A i - I-th contact is closed. For which of them is the ratio

A1 (A2 + A3 A4) A5 = A?

Solution. For the first circuit, A = A1 (A2 A3 + A4 A5), since the sum of events corresponds to a parallel connection, and serial connection- production of events. For the second scheme A = A1 (A2+A3 A4 A5). Therefore, this relation is valid for the second scheme.

Task. 1.1.5 Simplify the expression (A + B)(B + C)(C + A).

Solution. Let us use the properties of operations of addition and multiplication of events.

(A+ B)(B + C)(A + C) =

(AB+ AC + B B + BC)(A + C) =

= (AB+ AC + B + BC)(A + C) =

(AB + AC + B)(A + C) = (B + AC)(A + C) =

= BA + BC + ACA + ACC = B A + BC + AC.

Task. 1.1.6Prove that the events A, AB and A+B form a complete group.

Solution. When solving the problem, we will use the properties of operations on events. First, we show that these events are pairwise incompatible.

Let us now show that the sum of these events gives the space of elementary events.

Task. 1.1.7Using the Euler–Venn scheme, check the de Morgan rule:

A) Event AB is shaded.

B) Event A - vertical hatching; event B - horizontal hatching. Event

(A+B) - shaded area.

From a comparison of figures a) and c) it follows:

Task. 1.2.1In how many ways can 8 people be seated?

1. In one row?

2. Behind round table?

Solution.

1. The desired number of ways is equal to the number of permutations out of 8, i.e.

P8 = 8! = 1 2 3 4 5 6 7 8 = 40320

2. Since the choice of the first person at the round table does not affect the alternation of elements, then anyone can be taken first, and the remaining ones will be ordered relative to the chosen one. This action can be done in 8!/8 = 5040 ways.

Task. 1.2.2The course covers 5 subjects. In how many ways can you make a schedule for Saturday if there are to be two different couples on that day?

Solution. The desired number of ways is the number of placements

From 5 to 2, since you need to take into account the order of the pairs:

Task. 1.2.3How many examining committees, consisting of 7 people, can be made up of 15 teachers?

Solution. The desired number of commissions (without regard to order) is the number of combinations of 15 to 7:

Task. 1.2.4 From a basket containing twenty numbered balls, 5 balls are chosen for good luck. Determine the number of elements in the space of elementary events of this experience if:

The balls are selected sequentially one after the other with a return after each extraction;

The balls are chosen one by one without returning;

5 balls are selected at once.

Solution.

The number of ways to extract the first ball from the basket is 20. Since the extracted ball is returned to the basket, the number of ways to extract the second ball is also 20, and so on. Then the number of ways to extract 5 balls in this case is 20 20 20 20 20 = 3200000.

The number of ways to extract the first ball from the basket is 20. Since the extracted ball did not return to the basket after extraction, the number of ways to extract the second ball became 19, etc. Then the number of ways to extract 5 balls without replacement is 20 19 18 17 16 = A52 0

The number of ways to extract 5 balls from the basket at once is equal to the number of combinations of 20 by 5:

Task. 1.2.5 Two dice. Find the probability of event A that at least one 1 will be rolled.

Solution. Any number of points from 1 to 6 can fall on each die. Therefore, the space of elementary events contains 36 equally possible outcomes. Event A is favored by 11 outcomes: (1.1), (1.2), (2.1), (1.3), (3.1), (1.4), (4.1), (1 .5), (5.1), (1.6), (6.1), so

Task. 1.2.6 The letters y, i, i, k, c, f, n are written on red cards, the letters a, a, o, t, t, s, h are written on blue cards. After thorough mixing, which is more likely: from the first time from the letters to use the red cards to make the word "function" or the letters on the blue cards to make the word "frequency"?

Solution. Let event A be the word "function" randomly composed of 7 letters, event B - the word "frequency" randomly composed of 7 letters. Since two sets of 7 letters are ordered, the number of all outcomes for events A and B is n = 7!. Event A is favored by one outcome m = 1, since all the letters on the red cards are different. Event B is favored by m = 2! · 2! outcomes, since the letters "a" and "t" occur twice. Then P(A) = 1/7! , P(B) = 2! 2! /7! , P(B) > P(A).

Task. 1.2.7 At the exam, the student is offered 30 tickets; Each ticket has two questions. Of the 60 questions included in the tickets, the student knows only 40. Find the probability that the ticket taken by the student will consist of

1. from the issues known to him;

2. from questions unknown to him;

3. from one known and one unknown question.

Solution. Let A be the event that the student knows the answer to both questions; B - does not know the answer to both questions; C - he knows the answer to one question, he does not know the answer to another. The choice of two questions out of 60 can be done in n = C260 = 60 2 59 = 1770 ways.

1. There are m = C240 ​​= 40 2 39 = 780 choices of questions known to the student. Then P(A) = M N = 17 78 70 0 = 0.44

2. The choice of two unknown questions from 20 can be done in m = C220 = 20 2 19 = 190 ways. In this case

P(B) = M N = 11 79 70 0 = 0.11

3. There are m = C14 0 C21 0 = 40 20 = 800 ways to choose a ticket with one known and one unknown question. Then P(C) = 18 70 70 0 = 0.45.

Task. 1.2.8Some information has been sent through three channels. Channels operate independently of each other. Find the probability that the information will reach the goal

1. Only on one channel;

2. At least one channel.

Solution. Let A be an event consisting in the fact that information reaches the goal through only one channel; B - at least one channel. Experience is the transmission of information through three channels. The outcome of the experience - the information has reached the goal. Denote Ai - information reaches the target through the i-th channel. The space of elementary events has the form:

Event B is favored by 7 outcomes: all outcomes except Then n = 8; mA = 3; mB = 7; P(A) = 3 8 ; P(B) = 7 8.

Task. 1.2.9A point randomly appears on a segment of unit length. Find the probability that the distance from the point to the ends of the segment is greater than 1/8.

Solution. According to the condition of the problem, the desired event is satisfied by all points that appear on the interval (a; b).

Since its length is s = 1 - 1 8 + 1 8 = 3 4, and the length of the entire segment is S = 1, the required probability is P = s/S = 3/14 = 0.75.

Task. 1.2.10In a batch ofNproductsKproducts are defective. For control, m products are selected. Find the probability that from M Products L They turn out to be defective (event A).

Solution. The choice of m products from n can be done in ways, and the choice L defective out of k defective - in ways. After selection L defective products will remain (m - L) fit, located among (n - k) products. Then the number of outcomes favoring the event A is

And the desired probability

Task. 1.3.1BAn urn contains 30 balls: 15 red, 10 blue and 5 white. Find the probability that a randomly drawn ball is colored.

Solution. Let event A - a red ball is drawn, event B - a blue ball is drawn. Then events (A + B) - a colored ball is drawn. We have P(A) = 1 3 5 0 = 1 2 , P(B) = 1 3 0 0 = 1 3. Since

Events A and B are incompatible, then P(A + B) = P(A) + P(B) = 1 2 + 1 3 = 5 6 = 0.83.

Task. 1.3.2The probability that it will snow (an event A ), is equal to 0.6, And the fact that it will rain (event B ), is equal to 0.45. Find the probability of bad weather if the probability of rain and snow (event AB ) is equal to 0.25.

Solution. Events A and B are joint, so P(A + B) = P(A) + P(B) - P(AB) = 0.6 + 0.45 - 0.25 = 0.8

Task. 1.3.3BThe first box contains 2 white and 10 black balls, the second - 3 white and 9 black balls, and the third - 6 white and 6 black balls. A ball was taken from each box. Find the probability that all the balls drawn are white.

Solution. Event A - a white ball is drawn from the first box, B - from the second box, C - from the third. Then P(A) = 12 2 = 1 6; P(B) = 13 2 = 1 4; P(C) = 16 2 = 1 2. Event ABC - all taken out

Balls are white. Events A, B, C are independent, therefore

P(ABC) = P(A) P(B) P(C) = 1 6 1 4 1 2 = 41 8 = 0.02

Task. 1.3.4Belectrical circuit connected in series 5 Elements that work independently of each other. The probability of failures of the first, second, third, fourth, fifth elements, respectively, are 0.1; 0.2; 0.3; 0.2; 0.1. Find the probability that there will be no current in the circuit (event A ).

Solution. Since the elements are connected in series, there will be no current in the circuit if at least one element fails. Event Ai(i =1...5) - will fail I-th element. Events

Task. 1.3.5The circuit consists of independent blocks connected in a system with one input and one output.

Failure in time T of various circuit elements are independent events with the following probabilitiesP 1 = 0.1; P 2 = 0.2; P 3 = 0.3; P 4 = 0.4. The failure of any of the elements leads to an interruption of the signal in the branch of the circuit where this element is located. Find the reliability of the system.

Solution. If event A - (SYSTEM IS RELIABLE), Ai - (i - th UNIT WORKS FAULTY), then A = (A1 + A2)(A3 + A4). Events A1+A2, A3+A4 are independent, events A1 and A2, A3 and A4 are joint. According to the formulas for multiplication and addition of probabilities

Task. 1.3.6The worker serves 3 machines. The probability that within an hour the machine does not require the attention of a worker is 0.9 for the first machine, 0.8 for the second machine, and 0.7 for the third machine.

Find the probability that during some hour

1. The second machine will require attention;

2. Two machines will require attention;

3. At least two machines will need attention.

Solution. Let Ai - the i-th machine require the attention of the worker, - the i-th machine will not require the attention of the worker. Then

Space of elementary events:

1. Event A - will require the attention of the second machine: Then

Since the events are incompatible and independent. P(A) = 0.9 0.8 0.7 + 0.1 0.8 0.7 + 0.9 0.8 0.3 + 0.1 0.8 0.3 = 0.8

2. Event B - two machines will require attention:

3. Event C - at least two stuns will require attention
cov:

Task. 1.3.7Bmachine "Examiner" introduced 50 questions. The student is offered 5 Questions and an “excellent” mark is given if all questions are answered correctly. Find the probability of getting "excellent" if the student prepared only 40 questions.

Solution. A - (RECEIVED "EXCELLENT"), Ai - (ANSWERED TO i - th QUESTION). Then A = A1A2A3A4A5, we have:

Or, in another way - using the classical probability formula: AND

Task. 1.3.8The probabilities that the part needed by the assembler is inI, II, III, IVbox, respectively, are equal 0.6; 0.7; 0.8; 0.9. Find the probability that the collector will have to check all 4 boxes (eventA).

Solution. Let Ai - (The part required by the assembler is in i-th box.) Then

Since the events are incompatible and independent, then

Task. 1.4.1 A group of 10,000 people over the age of 60 was examined. It turned out that 4000 people are permanent smokers. 1800 smokers showed serious changes in the lungs. Among non-smokers, 1500 people had changes in the lungs. What is the probability that a randomly examined person with lung changes is a smoker?

Solution. Let's introduce the hypotheses: H1 - the examined is a permanent smoker, H2 - is a non-smoker. Then by the condition of the problem

P(H1)= -------=0.4, P(H2)=---------=0.6

Denote by A the event that the examined person has changes in the lungs. Then by the condition of the problem

By formula (1.15) we find

The desired probability that the examined person is a smoker, according to the Bayes formula, is equal to

Task. 1.4.2Televisions from three factories go on sale: 30% from the first factory, 20% from the second, 50% from the third. The products of the first factory contain 20% of TVs with a hidden defect, the second - 10%, the third - 5%. What is the probability of getting a working TV?

Solution. Let's consider the following events: A - a serviceable TV was purchased; hypotheses H1, H2, H3 - the TV went on sale from the first, second, third factory, respectively. According to the task

By formula (1.15) we find

Task. 1.4.3There are three identical boxes. The first has 20 white balls, the second has 10 white and 10 black balls, and the third has 20 black balls. A white ball is drawn from a randomly selected box. Find the probability that this ball is from the second box.

Solution. Let the event A - a white ball is taken out, hypotheses H1, H2, H3 - the ball is taken out from the first, second, third boxes respectively. From the condition of the problem we find

Then
By formula (1.15) we find

By formula (1.16) we find

Task. 1.4.4A telegraph message consists of the dot and dash signals. The statistical properties of interference are such that they are distorted on average 2/5 Dot messages and 1/3 Dash messages. It is known that among the transmitted signals "dot" and "dash" occur in the ratio 5: 3. Determine the probability that a transmitted signal is received if:

A) a "point" signal is received;

B)dash signal received.

Solution. Let the event A - the "dot" signal is received, and the event B - the "dash" signal is received.

Two hypotheses can be made: H1 - the "dot" signal is transmitted, H2 - the "dash" signal is transmitted. By condition P(H1) : P(H2) =5: 3. In addition, P(H1 ) + P(H2)= 1. Therefore P( H1 ) = 5/8, P(H2 ) = 3/8. It is known that

Event probabilities A AND B We find by the formula of total probability:

The desired probabilities will be:

Task. 1.4.5Of the 10 radio channels, 6 channels are protected from interference. Probability that a secure channel over timeTwill not fail is 0.95, for an unprotected channel - 0.8. Find the probability that two randomly selected channels will not fail in timeT, and both channels are not protected from interference.

Solution. Let the event A - both channels will not fail during the time t, the event A1- Secure channel selected A2- An unsecured channel is selected.

Let's write the space of elementary events for the experiment - (two channels are selected):

Ω = (A1A1, A1A2, A2A1, A2A2)

Hypotheses:

H1 - both channels are protected from interference;

H2 - the first selected channel is protected, the second selected channel is not protected from interference;

H3 - the first selected channel is not protected, the second selected channel is protected from interference;

H4 - both selected channels are not protected from interference. Then

AND

Task. 1.5.1Transmitted over the communication channel 6 Messages. Each of the messages can be distorted by noise with a probability 0.2 Regardless of others. Find the probability that

1. 4 messages out of 6 are not distorted;

2. At least 3 out of 6 were transmitted distorted;

3. At least one message out of 6 is garbled;

4. No more than 2 out of 6 are not distorted;

5. All messages are transmitted without distortion.

Solution. Since the probability of distortion is 0.2, the probability of transmitting a message without interference is 0.8.

1. Using the Bernoulli formula (1.17), we find the probability
transmission rate of 4 out of 6 messages without interference:

2. at least 3 out of 6 are transmitted distorted:

3. at least one message out of 6 is garbled:

4. at least one message out of 6 is garbled:

5. all messages are transmitted without distortion:

Task. 1.5.2The probability that the day will be clear in summer is 0.42; the probability of an overcast day is 0.36 and partly cloudy is 0.22. How many days out of 59 can be expected to be clear and overcast?

Solution. It can be seen from the condition of the problem that it is necessary to look for the most probable number of clear and cloudy days.

For clear days P= 0.42, N= 59. We compose inequalities (1.20):

59 0.42 + 0.42 - 1 < m0 < 59 0.42 + 0.42.

24.2 ≤ Mo≤ 25.2 → Mo= 25.

For cloudy days P= 0.36, N= 59 and

0.36 59 + 0.36 - 1 ≤ M0 ≤ 0.36 59 + 0.36;

Hence 20.16 ≤ M0 ≤ 21.60; → M0 = 21.

Thus, the most probable number of clear days Mo= 25, cloudy days - M0 = 21. Then in summer we can expect Mo+ M0 =46 clear and cloudy days.

Task. 1.5.3There are 110 students of the course at the lecture on probability theory. Find the probability that

1. k students (k = 0,1,2) of those present were born on the first of September;

2. at least one student of the course was born on the first of September.

P=1/365 is very small, so we use the Poisson formula (1.22). Let's find the Poisson parameter. Because

N= 110, then λ = np = 110 1 /365 = 0.3.

Then by the Poisson formula

Task. 1.5.4The probability that a part is not standard is 0.1. How many details need to be selected so that with probability P = 0.964228 It could be argued that the relative frequency of occurrence of non-standard parts deviates from the constant probability p = 0.1 In absolute terms, no more than 0.01 ?

Solution.

Required number N Let us find by formula (1.25). We have:

P = 1.1; q = 0.9; P= 0.96428. Substitute the data in the formula:

Where do we find

According to the table of values ​​of the function Φ( X) we find that

Task. 1.5.5The probability of failure in time T of one capacitor is 0.2. Determine the probability that in time T out of 100 capacitors will fail.

1. Exactly 10 capacitors;

2. At least 20 capacitors;

3. Less than 28 capacitors;

4. From 14 to 26 capacitors.

Solution. We have P = 100, P= 0.2, Q = 1 - P= 0.8.

1. Exactly 10 capacitors.

Because P Veliko, let's use the local de Moivre-Laplace theorem:

Compute

Since the function φ(x)- even, then φ (-2.5) = φ (2.50) = 0.0175 (we find from the table of function values φ(x). Desired probability

2. At least 20 capacitors;

The requirement that at least 20 out of 100 capacitors fail means that either 20, or 21, ..., or 100 will fail. Thus, T1 = 20, T 2=100. Then

According to the table of function values Φ(x) Let us find Φ(x1) = Φ(0) = 0, Φ(x2) = Φ(20) = 0.5. Required probability:

3. Less than 28 capacitors;

(here it was taken into account that the Laplace function Ф(x) is odd).

4. From 14 to 26 capacitors. By condition M1= 14, m2 = 26.
Calculate x 1,x2:

Task. 1.5.6The probability of occurrence of some event in one experiment is equal to 0.6. What is the probability that this event will occur in most of the 60 trials?

Solution. Quantity M The occurrence of an event in a series of tests is in the interval. "In most experiments" means that M Belongs to interval By condition N= 60, P= 0.6, Q = 0.4, M1 = 30, m2 = 60. Calculate x1 and x2:

Random variables and their distributions

Task. 2.1.1There is a table where the top row shows possible values random variable X , and at the bottom - their probabilities.

Can this table be a distribution series X ?

Answer: Yes, since p1 + p2 + p3 + p4 + p5 = 1

Task. 2.1.2Released 500 Lottery tickets, and 40 Tickets will bring their owners a prize for 10000 Rub., 20 Tickets - by 50000 Rub., 10 Tickets - by 100000 Rub., 5 Tickets - by 200000 Rub., 1 Ticket - 500000 Rub., the rest - without a win. Find the winning distribution law for the owner of one ticket.

Solution.

Possible values ​​of X: x5 = 10000, x4 = 50000, x3 = 100000, x2 = 200000, x1 = 500000, x6 = 0. The probabilities of these possible values ​​are:

The desired distribution law:

Task. 2.1.3shooter, having 5 Cartridges, shoots until the first hit on the target. The probability of hitting each shot is 0.7. Construct the law of distribution of the number of cartridges used, find the distribution functionF(X) and plot its graph, find P(2< x < 5).

Solution.

Space of elementary events of experience

Ω = {1, 01, 001, 0001, 00001, 11111},

Where event (1) - hit the target, event (0) - did not hit the target. Elementary outcomes correspond to the following values ​​of the random value of the number of cartridges used: 1, 2, 3, 4, 5. Since the result of each next shot does not depend on the previous one, the probabilities of possible values ​​are:

P1 = P(x1= 1) = P(1)= 0.7; P2 = P(x2= 2) = P(01)= 0.3 0.7 = 0.21;

P3 = P(x3= 3) = P(001) = 0.32 0.7 = 0.063;

P4 = P(x4= 4) = P(0001) = 0.33 0.7 = 0.0189;

P5 = P(x5= 5) = P(00001 + 00000) = 0.34 0.7 + 0.35 = 0.0081.

The desired distribution law:

Find the distribution function F(X), Using formula (2.5)

X≤1, F(x)= P(X< x) = 0

1 < x ≤2, F(x)= P(X< x) = P1(X1 = 1) = 0.7

2 < x ≤ 3, F(x) = P1(X= 1) + P2(x = 2) = 0.91

3 < x ≤ 4, F(x) = P1 (x = 1) + P2(x = 2) + P3(x = 3) =

= 0.7 + 0.21 + 0.063 = 0.973

4 < x ≤ 5, F(x) = P1(x = 1) + P2(x = 2) + P3(x = 3) +

+ P4(x = 4) = 0.973 + 0.0189 = 0.9919

X >5, F(x) = 1

Find P(2< x < 5). Применим формулу (2.4): P(2 < X< 5) = F(5) - F(2) = 0.9919 - 0.91 = 0.0819

Task. 2.1.4DanaF(X) of some random variable:

Write down the distribution series for X.

Solution.

From properties F(X) It follows that the possible values ​​of the random variable X - Function break points F(X), And the corresponding probabilities are jumps of the function F(X). Find the possible values ​​of the random variable X=(0,1,2,3,4).

Task. 2.1.5Set which function

Is a distribution function of some random variable.

If the answer is yes, find the probability that the corresponding random value takes values ​​on[-3,2].

Solution. Let's plot the functions F1(x) and F2(x):

The function F2(x) is not a distribution function, since it is not non-decreasing. The function F1(x) is

The distribution function of some random variable, since it is non-decreasing and satisfies condition (2.3). Let's find the probability of hitting the interval:

Task. 2.1.6Given the probability density of a continuous random variable X :

Find:

1. Coefficient C ;

2. distribution function F(x) ;

3. The probability of a random variable falling into the interval(1, 3).

Solution. From the normalization condition (2.9) we find

Hence,

By formula (2.10) we find:

Thus,

By formula (2.4) we find

Task. 2.1.7Random downtime of electronic equipment in some cases has a probability density

Where M = lge = 0.4343...

Find distribution function F(x) .

Solution. By formula (2.10) we find

Where

Task. 2.2.1A distribution series of a discrete random variable is given X :

Find the mathematical expectation, variance, standard deviation, M, D[-3X + 2].

Solution.

According to the formula (2.12) we find the mathematical expectation:

M[X] = x1p1 + x2p2 + x3p3 + x4p4 = 10 0.2 + 20 0.15 + 30 0.25 + 40 0.4 = 28.5

M = 2M[X] + M = 2M[X] + 5 = 2 28.5 + 5 = 62. Using formula (2.19), we find the dispersion:

Task. 2.2.2Find the mathematical expectation, variance and standard deviation of a continuous random variable X , whose distribution function

.

Solution. Find the probability density:

The mathematical expectation is found by the formula (2.13):

We find the dispersion by the formula (2.19):

Let us first find the mathematical expectation of the square of the random variable:

Standard deviation

Task. 2.2.3Xhas a number of distributions:

Find the mathematical expectation and variance of a random variableY = EX .

Solution. M[ Y] = M[ EX ] = e-- 1 0.2 + e0 0.3 + e1 0.4 + e2 0.1 =

0.2 0.3679 + 1 0.3 + 2.71828 0.4 + 7.389 0.1 = 2.2.

D[Y] = D = M[(eX)2 - M2[E X] =

[(e-1)2 0.2 + (e0)2 0.3 + (e1)2 0.4 + (e2)2 0.1] - (2.2)2 =

= (e--2 0.2 + 0.3 + e2 0.4 + e4 0.1) - 4.84 = 8.741 - 4.84 = 3.9.

Task. 2.2.4Discrete random variable X Can only take two values X1 AND X2 , and X1< x2. Known Probability P1 = 0.2 Possible value X1 , expected value M[X] = 3.8 And dispersion D[X] = 0.16. Find the law of distribution of a random variable.

Solution. Since the random variable X takes only two values ​​x1 and x2, then the probability p2 = P(X = x2) = 1 - p1 = 1 - 0.2 = 0.8.

By the condition of the problem, we have:

M[X] = x1p1 + x2p2 = 0.2x1 + 0.8x2 = 3.8;

D[X] = (x21p1 + x22p2) - M2[X] = (0.2x21 + 0.8x22) - (0.38)2 = 0.16.

Thus, we got the system of equations:

Condition x1

Task. 2.2.5The random variable X is subject to the distribution law, the density graph of which has the form:

Find the mathematical expectation, variance and standard deviation.

Solution. Let us find the differential distribution function f(x). Outside the interval (0, 3) f(x) = 0. On the interval (0, 3) the density graph is a straight line with slope k = 2/9 passing through the origin. Thus,

Expected value:

Find the variance and standard deviation:

Task. 2.2.6Find the mathematical expectation and variance of the sum of points on four dice in one roll.

Solution. Let's denote A - the number of points on one die in one throw, B - the number of points on the second die, C - on the third die, D - on the fourth die. For random variables A, B, C, D, the distribution law one.

Then M[A] = M[B] = M[C] = M[D] = (1+2+3+4+5+6) / 6 = 3.5

Task. 2.3.1The probability that a particle emitted from a radioactive source will be registered by a counter is equal to 0.0001. During the observation period, 30000 particles. Find the probability that the counter registered:

1. Exactly 3 particles;

2. Not a single particle;

3. At least 10 particles.

Solution. By condition P= 30000, P= 0.0001. The events consisting in the fact that particles emitted from a radioactive source are registered are independent; number P Great, but the probability P Small, so we use the Poisson distribution: Let's find λ: λ = n P = 30000 0.0001 = 3 = M[X]. Desired probabilities:

Task. 2.3.2There are 5% non-standard parts in the lot. 5 items were randomly selected. Write the distribution law of a discrete random variable X - the number of non-standard parts among the five selected; find the mathematical expectation and variance.

Solution. The discrete random variable X - the number of non-standard parts - has a binomial distribution and can take the following values: x1 = 0, x2 = 1, x3 = 2, x4 = 3, x5 = 4, x6 = 5. Probability of a non-standard part in a batch p = 5 /100 = 0.05. Let's find the probabilities of these possible values:

Let's write the desired distribution law:

Let's find numerical characteristics:

0 0.7737809 + 1 0.2036267 + 2 0.0214343+

3 0.0011281 + 4 0.0000297 + 5 0.0000003 = 0.2499999 ≈ 0.250

M[X] = Np= 5 0.05 = 0.25.

D[X] = M– M2 [X]= 02 0.7737809 + 12 0.2036267+

22 0.0214343 + 32 0.0011281 + 42 0.0000297 + 52 0.0000003- 0.0625 =

0.2999995 - 0.0625 = 0.2374995 ≈ 0.2375

Or D[ X] = np (1 - P) = 5 0.05 0.95 = 0.2375.

Task. 2.3.3The radar target detection time is distributed according to the exponential law

Where1/ λ = 10 Sec. - average target detection time. Find the probability that the target will be found within the time5 Before15 Sec. after the start of the search.

Solution. Probability of hitting a random variable X In interval (5, 15) Let us find by formula (2.8):

At We get

0.6065(1 - 0.3679) = 0.6065 0.6321 = 0.3834

Task. 2.3.4Random measurement errors are subject to the normal law with parameters a = 0, σ = 20 Mm. Write differential distribution functionF(X) and find the probability that the measurement made an error in the interval from 5 Before 10 Mm.

Solution. Let us substitute the values ​​of the parameters a and σ into the differential distribution function (2.35):

Using formula (2.42), we find the probability of hitting a random variable X In the interval , i.e. A= 0, B= 0.1. Then the differential distribution function F(x) Will look like