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Conditional probability. Probability of several random events

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

The required probability is equal to

Total probability formula.

Let some event A happen together with one of incompatible events, constituting a complete group of events. Let the probabilities of these events and the conditional probabilities of the occurrence of event A be known when the event occurs H i .

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

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

Proof.

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

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

Wherein

Finally we get:

The theorem has been proven.

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

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

The probabilities that one of the shooters firing shots hits the target twice are equal to:

For the first shooter:

For the second shooter:

For the third shooter:

The required probability is:

LECTURE 2.

Bayes formula. (hypothesis formula)

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

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

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

This formula is called Bayes formula.

Proof.

Using the Probability Multiplication Theorem we get:

Then if .

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

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

Repetition of tests.

Bernoulli's formula.

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

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

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

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

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

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

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

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

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

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

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

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

In case of five hits out of five possible:

Four hits out of five shots:

Three hits out of five:

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

Random variables.

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

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

Random variables can be divided into two categories.

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

This set can be either finite or infinite.

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

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

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

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

Distribution law of a discrete random variable.

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

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

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

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

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

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

Similarly we find:

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

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

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

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

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

We get:

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

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

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

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

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

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

The probability that a white ball will be selected again is

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

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

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

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

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

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

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

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

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

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

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

Let's substitute these values into Bayes' formula:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Finally, we get: .

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

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

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

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

Finally we get:

Binomial distribution.

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

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

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

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

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

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

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

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

Let us find the probabilities that among the selected parts:

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

2) One is non-standard.

3) Two non-standard parts.

4) Three non-standard parts.

5) Four non-standard parts.

Let's construct a distribution polygon.

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

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

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

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

Individual tasks in mathematics

Problem 1

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

Solution

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

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

Since one of the white balls has already been drawn.

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

or two black balls:

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

Answer: 1)

2) 3) .

Problem 2

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

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

Solution

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

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

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

Answer: 1)

2) 3) .

Problem 3

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

Solution

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

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

Answer:

Problem 4

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

Solution

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

The probability that both parts taken will be first grade:

The probability that both parts taken will be second grade:

The probability that both parts taken will be third grade:

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

Answer:

Problem 5

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

Solution

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

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

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

Answer:

Problem 8

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

1) only one shooter will hit the target;

2) two shooters will hit the target;

3) at least one will hit the target.

Solution

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

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

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

, .

Hence, the desired probability:

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

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

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

What is probability?

The first time I encountered this term, I would not have understood what it was. Therefore, I will try to explain clearly.

Probability is the chance that the event we want will happen.

For example, you decided to go to a friend’s house, you 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 staircase, and in front of you there are doors to choose from.

What is the chance (probability) that if you ring the first doorbell, your friend will answer the door for you? There are only apartments, and a friend lives only behind one of them. With an equal chance we can choose any door.

But what is this chance?

The door, the right door. Probability of guessing by ringing the first doorbell: . That is, one time out of three you will accurately guess.

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

You called 1st door
You called 2nd door
You called 3rd door

Now let’s look at all the options where a friend could be:

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

Let's compare all the options in table form. A checkmark indicates options when your choice coincides with a friend's location, a cross - when it does not coincide.

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

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

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

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

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

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

The word “outcomes” probably caught your eye. Since mathematicians call various actions (in our case, such an action is a doorbell) experiments, the result of such experiments is usually called the outcome.

Well, there are favorable and unfavorable outcomes.

Let's go back to our example. Let's say we rang one of the doors, but a stranger opened it for us. We didn't guess right. 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 1st door
2) Call 2nd door

The friend, despite all this, is definitely behind one of them (after all, he wasn’t behind the one we called):

a) Friend for 1st the door
b) Friend for 2nd the door

Let's draw the table again:

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

Why not?

The situation we 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 influence the following actions. After all, if after the first ring the doorbell was answered by a friend, what would be the probability that he was behind one of the other two? Right, .

But if there are dependent events, then there must also be independent? That's right, they do happen.

A textbook example is tossing a coin.

Toss a coin once. What is the probability of getting heads, for example? That's right - because there are all the options (either heads or tails, we will neglect the probability of the coin landing on its edge), but it only suits us.
But it came up heads. Okay, let's throw it again. What is the probability of getting heads now? Nothing has changed, everything is the same. How many options? Two. How many are we happy with? One.

And let it come up heads at least a thousand times in a row. The probability of getting heads at once will be the same. There are always options, and favorable ones.

It is easy to distinguish dependent events from independent ones:

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

Let's practice determining probability a little.

Example 1.

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

Solution:

Let's consider all possible options:

Eagle-eagle
Heads-tails
Tails-Heads
Tails-tails

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

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

Answer:

Example 2.

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

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

Solution:

How many possible outcomes are there? .

That is, if you take one candy, it will be one of those available in the box.

How many favorable outcomes?

Because the box contains only chocolates with nuts.

Answer:

Example 3.

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

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

Solution:

a) There are only balls in the box. Of them are white.

The probability is:

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

Answer:

Total probability

The probability of all possible events is equal to ().

Let's say there are red and green balls in a box. 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 markers 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 take out a red marker) is .

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

Answer:

The probability that an event will not occur is equal to 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 happen in a row?

Let's say we want to know what is the probability that if we flip a coin once, we will see heads twice?

We have already considered - .

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

Total possible options:

Eagle-eagle-eagle
Heads-heads-tails
Heads-tails-heads
Heads-tails-tails
Tails-heads-heads
Tails-heads-tails
Tails-tails-heads
Tails-tails-tails

I don’t know about you, but I made mistakes several times when compiling this list. Wow! And only option (the first) suits us.

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

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

In other words,

Let's look at the example of the same ill-fated coin.

Probability of getting heads in a challenge? . Now we flip the coin once.

What is the probability of getting heads in a row?

This rule doesn't only work if we are asked to find the probability that the same event will happen several times in a row.

If we wanted to find the sequence TAILS-HEADS-TAILS for consecutive tosses, we would do the same.

The probability of getting tails is , heads - .

Probability of getting the sequence TAILS-HEADS-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 toss it once.
Possible options:

Eagle-eagle-eagle
Heads-heads-tails
Heads-tails-heads
Heads-tails-tails
Tails-heads-heads
Tails-heads-tails
Tails-tails-heads
Tails-tails-tails

So, incompatible events are a certain, given sequence of events. - these are incompatible events.

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

You need to understand that heads or tails are two independent events.

If we want to determine the probability of a sequence (or any other) occurring, 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 tosses?

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

Total options suit 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 certain, inconsistent, sequences of events.

There is a great rule to help you avoid getting confused when to multiply and when to add:

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

Should fall out:
(heads AND tails AND tails) OR (tails AND heads AND tails) OR (tails AND tails AND heads).
This is how 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 (red OR green).

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

Answer:

Example 6.

Dice roll twice, what is the probability of getting a total of 8 points?

Solution.

How can we get points?

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

The probability of getting one (any) face is .

We calculate the probability:

Answer:

Training.

I think now you understand when you need to calculate probabilities, when to add them, and when to multiply them. Is not it? Let's practice a little.

Tasks:

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

What is the probability of drawing clubs in a row (we put the first card pulled out back into the deck and shuffle it)?
What is the probability of drawing a black card (spades or clubs)?
What is the probability of drawing a picture (jack, queen, king or ace)?
What is the probability of drawing two pictures in a row (we remove the first card drawn from the deck)?
What is the probability, taking two cards, to collect a combination - (jack, queen or king) and an ace? The sequence in which the cards are drawn does not matter.

Answers:

In a deck of cards of each value, it means:
Events are dependent, since after the first card pulled out, the number of cards in the deck decreased (as did the number of “pictures”). There are total jacks, queens, kings and aces in the deck initially, which means the probability of drawing a “picture” with the first card:
Since we remove the first card from the deck, it means that there are already cards left in the deck, including pictures. Probability of drawing a picture with the second card:
Since we are interested in the situation when we take out a “picture” AND a “picture” from the deck, we need to multiply the probabilities:
Answer:
After the first card pulled out, the number of cards in the deck will decrease. Thus, two options suit us:
1) The first card is Ace, the second is Jack, Queen or King
2) We take out a jack, queen or king with the first card, and an ace with the second. (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 great! Now you will crack probability theory problems in the Unified State Exam like nuts!

PROBABILITY THEORY. AVERAGE LEVEL

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

So we roll the dice and we want it to come up or. And we get it.

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

If it happened, the event would also be favorable. In total, only two favorable events can happen.

How many are unfavorable? Since there are total possible events, it means that the unfavorable ones are events (this is if or falls out).

Definition:

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

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

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

And in percentage: .

Examples (decide for yourself):

What is the probability of getting heads when tossing a coin? What is the probability of landing heads?
What is the probability of getting an even number when throwing a die? Which one is odd?
In a box of simple, blue and red pencils. We draw one pencil at random. What is the probability of getting a simple one?

Solutions:

How many options are there? Heads and tails - just two. How many of them are favorable? Only one is an eagle. So the probability
It's the same with tails: .
Total options: (how many sides the cube has, so many different options). Favorable ones: (these are all even numbers:).
Probability. Of course, it’s the same with odd numbers.
Total: . Favorable: . Probability: .

Total probability

All pencils in the box 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 the same number of favorable events as there are total events (all events are favorable). So the probability is equal to or.

Such an event is called reliable.

If a box contains green and red pencils, what is the probability of drawing green or red? Yet again. Let's note this: the probability of pulling out green is equal, and red is equal.

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, plain, yellow, and the rest are orange. What is the probability of not drawing green?

Solution:

We remember that all probabilities add up. And the probability of getting 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 equal to minus the probability that the event will occur.

Independent events and the multiplication rule

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

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

Heads-Heads, Tails-Heads, Heads-Tails, Tails-Tails. What else?

Total options. Of these, only one suits us: Eagle-Eagle. In total, the probability is equal.

Fine. Now let's flip a coin once. Do the math yourself. Happened? (answer).

You may have noticed that with the addition of each subsequent throw, the probability decreases by half. General rule 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 throw a coin several times, each time a new throw is made, the result of which does not depend on all previous throws. We can just as easily throw two different coins at the same time.

More examples:

The dice are thrown twice. What is the probability of getting it both times?
The coin is tossed once. What is the probability that it will come up heads the first time, and then tails twice?
The player rolls two dice. What is the probability that the sum of the numbers on them will be equal?

Answers:

The events are independent, which means the multiplication rule works: .
The probability of heads is equal. The probability of tails is the same. Multiply:
12 can only be obtained if two -ki are rolled: .

Incompatible events and the addition rule

Events that complement each other to the point of full probability are called incompatible. As the name suggests, they cannot happen simultaneously. For example, if we flip a coin, it can come up either heads or tails.

Example.

In a box of pencils, among them are blue, red, green, plain, 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 - .

Favorable events in all: green + red. This means that the probability of drawing green or red is equal.

The same probability can be represented in this form: .

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

Mixed type problems

Example.

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

Solution .

This means that if the first result is heads, the second must be tails, and vice versa. It turns out that there are two pairs of independent events, 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 is going to happen using the conjunctions “AND” or “OR”. For example, in this case:

It should come up (heads and tails) or (tails and heads).

Where there is a conjunction “and” there will be multiplication, and where there is “or” there will be addition:

Try it yourself:

What is the probability that if a coin is tossed twice, the coin will land on the same side both times?
The dice are thrown twice. What is the probability of getting a total of points?

Solutions:

(Heads fell and tails fell) or (tails fell and tails fell): .
What are the options? And. Then:
Dropped (and) or (and) or (and): .

Another example:

Toss a coin once. What is the probability that heads will appear at least once?

Solution:

Oh, how I don’t want to go through the options... Heads-tails-tails, Eagle-heads-tails,... But there’s no need! Let's remember about total probability. Do you remember? What is the probability that the eagle will never fall out? It’s simple: heads fly all the time, that’s why.

PROBABILITY THEORY. BRIEFLY ABOUT THE MAIN THINGS

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.

Total probability

The probability of all possible events is equal to ().

The probability that an event will not occur is equal to 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 event

Incompatible events

Incompatible events are those 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 conjunctions “AND” or “OR”, instead of “AND” we put a multiplication sign, and instead of “OR” we put an addition sign.

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Probability theory

There are 12 boys and 8 girls in the group. 5 students were selected at random from the magazine. Find the probability that among the selected students there are exactly 3 girls.

Number of selected students according to the journal.

The probability of choosing a girl at random from the entire group.

The probability of not choosing a girl at random from the entire group (the probability of choosing a boy).

k = 3 - number of selected girls.

The probability that among the selected 5 students there are exactly 3 girls.

In a batch of 6 parts there are 4 standard ones. We took 3 parts at random. Find the probability that among the selected parts at least one is non-standard.

Number of parts in the batch.

Number of standard parts in a batch.

The probability of taking at random one non-standard part from the batch.

The probability of not taking at random one non-standard part from a batch (the probability of taking at random one standard part from a batch).

The probability of not taking two non-standard parts from a lot at random (the probability of taking two standard parts from a lot at random).

The probability of not taking three non-standard parts from the lot at random (the probability of taking three standard parts from the lot at random).

The probability that among the selected parts at least one is non-standard.

The machine consists of 3 independently working parts. The probability of failure of parts is correspondingly equal to 0.1; 0.2; 0.15. Find the probability of a machine failure if the failure of at least one part is sufficient for this to happen.

The probability that the 1st part will fail.

The probability that the 2nd part will fail.

The probability that the 3rd part will fail.

Probability that the 1st part will not fail.

Probability that the 2nd part will not fail.

Probability that the 3rd part will not fail.

The likelihood of a machine breakdown if the failure of at least one part is enough for this to happen.

Two shooters are shooting at a target. The probability of hitting the target with one shot is 0.5 for the first shooter, and 0.6 for the second. Find the probability that during one volley only one of the shooters will hit the target.

The probability that the first shooter will hit the target.

The probability that the second shooter will hit the target.

The probability that the first shooter will miss the target.

The probability that the second shooter will miss the target.

The probability that during one salvo only one of the shooters will hit the target.

There are 6 devices in the box, 4 of which are working. We took 3 pieces at random. Find the probability that all taken devices will be working.

The number of devices taken at random.

The probability of taking a working device from the box.

The probability of not taking a working device out of the box.

Let's use Bernoulli's formula:

k = 3 - the number of working devices taken at random.

The probability is that all the devices taken will be working.

The first urn contains 4 white and 1 black balls, the second urn contains 2 white and 5 black balls. 2 balls were transferred from the first to the second, then one ball was taken from the second urn. Find the probability that the ball chosen from the second urn is black.

Let's determine the possible outcomes of events when moving 2 balls from the 1st urn to the 2nd.

H1 - hypothesis that 2 white balls were drawn from the first urn.

H2 - hypothesis that 1 white and 1 black ball were drawn from the first urn.

Probability of drawing a black ball from the 1st urn.

Probability of drawing a white ball from the 1st urn.

Probability of hypothesis H1.

Probability of hypothesis H2.

Now consider the probability of the event when each of the hypotheses occurred.

The probability of drawing a black ball from the 2nd urn if hypothesis H1 occurs.

The probability of drawing a black ball from the 2nd urn if hypothesis H2 occurs.

The probability that the ball chosen from the second urn is black.

The likelihood is that the part manufactured at plant No. 1 is of excellent quality.

The likelihood is that the part manufactured at plant No. 2 is of excellent quality.

The likelihood is that the part manufactured at plant No. 3 is of excellent quality.

Possibility of pulling out of a box a part manufactured at factory No. 1.

Possibility of pulling out of a box a part manufactured at factory No. 2.

Possibility of pulling out of a box a part manufactured at factory No. 3.

According to the total probability formula:

The probability that a part removed at random will be of excellent quality.

There are three batches of products, each containing 25 products. The number of standard products is respectively 20, 21, 22. From a randomly selected batch, a product that turned out to be standard was extracted at random. Find the probability that it was taken from 1 batch.

The probability that a randomly selected part from the 1st batch is standard.

The probability that a randomly selected part from the 2nd batch is standard.

The probability that a randomly selected part from the 3rd batch is standard.

The probability of choosing one of three games at random.

According to Bayes' formula:

The probability that a randomly removed item was taken from 1 batch.

Two machines produce parts. The productivity of the second machine is twice that of the first. The first machine produces 80% of parts of excellent quality, and the second - 90%. The part taken at random turned out to be of excellent quality. Find the probability that this part was produced by 1 machine.

theory probability finding choice hit

The probability that the part produced by the 1st machine is of excellent quality.

The probability is that the part produced by the 2nd machine is of excellent quality.

Since the productivity of the second machine is twice that of the first, then out of 3 conditionally manufactured parts, two parts are from the 2nd machine and one is from the 1st machine.

The probability of randomly selecting a part produced by the 1st machine.

The probability of randomly selecting a part produced by the 2nd machine.

According to Bayes' formula:

It is likely that a randomly selected part of excellent quality turned out to be a part produced by the 1st machine.

The coin is tossed 9 times. Find the probability that the “coat of arms” will appear: a.) less than 4 times; b.) at least 4 times.

The probability that a "coat of arms" will appear.

The probability that the “coat of arms” will not appear.

Let's use Bernoulli's formula:

Number of coin tosses.

The probability of getting a coin with a coat of arms is less than 4 times.

k = 0, 1, 2, 3 - the number of times the “coat of arms” appears.

The probability of getting a coin with a coat of arms is 0 times out of 9.

The probability of getting a coin with a coat of arms is 1 time in 9.

The probability of getting a coin with a coat of arms is 2 times out of 9.

The probability of getting a coin with a coat of arms is 3 times out of 9.

The probability of the coin appearing as a coat of arms is at least 4 times.

k = 4, 5, 6, 7, 8, 9 - the number of times the “coat of arms” appears.

The probability of getting a coin with a coat of arms is 4 times out of 9.

The probability of getting a coin with a coat of arms is 5 times out of 9.

The probability of getting a coin with a coat of arms is 6 times out of 9.

The probability of getting a coin with a coat of arms is 7 times out of 9.

The probability of getting a coin with a coat of arms is 8 times out of 9.

The probability of getting a coin with a coat of arms is 9 times out of 9.

The probability of having a boy is 0.51. Find the probability that among 100 newborns there will be 50 boys.

Probability of having a boy.

Probability of not having a boy (probability of having a girl).

Number of newborns.

Number of boys born.

Let us use the local theorem of Moivre-Laplace, since

Tabulated even Gaussian function,

Using the table we find the value

The probability that among 100 newborns there will be 50 boys.

The probability of an event occurring in each of 100 independent trials is 0.8. Find the probability that the event will appear: a.) at least 75 times and not more than 90 times; b.) at least 90 times.

The probability of an event occurring.

The probability of an event not occurring.

Total number of trials.

Number of tests.

Using the table we find the value

The probability that the event will appear at least 75 times and at most 90 times.

Number of tests.

Let's take advantage integral theorem Moivre-Laplace because

Tabulated odd Laplace function,

Using the table we find the value

The probability that the event will appear at least 90 times.

A discrete random variable is specified by the distribution law:

a.) construct a distribution polygon and find the distribution function F(x);

b.) Find M(X), D(X), .

Expected value.

Dispersion.

Standard deviation.

The distribution density f(x) of a continuous random variable X is given.

a.) find A and the distribution function F(x);

b.) find M(x), D(x),

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Steps

Probability of a single random event

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.
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.
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%.
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.
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%.