“Elements of mathematical statistics” - Confidence interval. The science. Classification of hypotheses. Parts are made on different machines. Verification rules. Correlation dependence. Addiction. A set of criterion values. Find the confidence interval. Calculation of confidence intervals for unknown variance. Normal distribution.

“Probability and mathematical statistics” - Accuracy of the obtained values. Code for the safe. Descriptive statistics. Apple. Let's consider the events. Multiplication rule. Two shooters. Comparison of training programs. Caramel. Examples of bar charts. Math grades. Multiplication rule for three. White and red roses. 9 different books. The winter vacation.

“Fundamentals of mathematical statistics” - Conditional probability. Table of standardized values. Properties of the Student distribution. Confidence interval of mathematical expectation. Sample mean. Distribution. One test can be considered as a series of one test. Quantile – to the left should be the number of values ​​corresponding to the quantile index.

“Probability theory and statistics” - Interval boundaries. Critical areas. Probability multiplication theorem. Distribution of a normal random variable. Derivation of Bernoulli's formula. Laws of distribution of random variables. The formulation of the LBC. The meaning and formulation of the central limit theorem. Connection of nominal features. Stochastic dependence of two random variables.

"Statistical Research" - Relevance. Statistical characteristics and research. Plan. Range is the difference between the largest and smallest values ​​of a data series. Types of statistical observation. Do you enjoy studying mathematics? Let's look at a series of numbers. Who helps you understand a difficult topic in mathematics? Do you need mathematics in your future profession?

“Basic statistical characteristics” - Basic statistical characteristics. Find the arithmetic mean. Petronius. Scope. Fashion series. The arithmetic mean of a series of numbers. Row range. Median of the series. Statistics. Median. School notebooks.

There are a total of 17 presentations in the topic

Slide 2

Pn(k)=Cknpk(1-p)n-k If the Probability p of the occurrence of event A in each trial is constant, then the probability Pn(k) that event A will occur k times in n independent trials is equal to: T Statement of the theorem Bernoulli formula - a formula in probability theory that allows you to find the probability of the occurrence of event A during independent trials. Bernoulli's formula allows you to get rid of a large number of calculations - adding and multiplying probabilities - with a sufficiently large number of tests.

Slide 3

Historical information JACOB BERNOULLI (1654–1705) Date of birth: December 27, 1654 Place of birth: Basel Date of death: August 16, 1705 Place of death: Basel Citizenship: Switzerland Scientific field: Mathematician Place of work: University of Basel Scientific. director: Leibniz Jacob Bernoulli (German Jakob Bernoulli, December 27, 1654, Basel, - August 16, 1705, ibid.) - Swiss mathematician, brother of Johann Bernoulli; professor of mathematics at the University of Basel (since 1687). Jacob Bernoulli has significant achievements in series theory, differential calculus of variations, probability theory and number theory, where numbers with certain properties are named after him. Jacob Bernoulli also wrote works on physics, arithmetic, algebra and geometry.

Slide 4

An example of using the Bernoulli formula Every day, shares of ABC Corporation rise in price or fall in price by one point with probabilities of 0.75 and 0.25, respectively. Find the probability that the stock will return to its original price after six days. Accept the condition that changes in the stock price up and down are independent events. SOLUTION: In order for the shares to return to their original price in 6 days, they need to rise in price 3 times and fall in price three times during this time. The desired probability is calculated using the Bernoulli formula P6(3) =C36(3/4)3(1/4)3=0.13

Slide 5

Test yourself There are 20 white and 10 black balls in an urn. 4 balls are taken out in a row, and each removed ball is returned to the urn before the next one is taken out and the balls in the urn are mixed. What is the probability that out of four balls drawn, two will be white? ANSWER: SOLUTION: ANSWER: ANSWER: SOLUTION: SOLUTION: The auditor detects financial irregularities in the audited company with a probability of 0.9. Find the probability that among 4 violating firms more than half will be identified. The dice is rolled 3 times. What is the probability that 6 points will appear exactly 2 times in this series of tests? 0.01389 8/27 0.9477

Slide 6

Test yourself The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 2 times. ANSWER: SOLUTION: ANSWER: SOLUTION: Let the germination rate of wheat seeds be 90%. What is the probability that out of 7 seeds sown, 5 will sprout? 0.124 0.344

Slide 7

The probability of retrieving the white ball p=20/30=2/3 can be considered the same in all trials; 1-p=1/3 Using Bernoulli's formula, we get P4(2) = C42·p2·(1-p)2=(12/2)·(2/3)2·(1/3)2 = 8/ 27 BACK SOLUTION TO PROBLEM 1

Slide 8

BACK SOLUTION TO PROBLEM 2 The event is that out of 4 violating firms, three or four will be identified, i.e. P(A)=P4(3)+P4(4) P(A)= C340.93∙0.1+C44 0.94 = 0.93 (0.4+0.9)=0.9477

Transnistrian State University named after T.G. Shevchenko

DEPARTMENT OF APPLIED MATHEMATICS AND ECONOMIC-MATHEMATICAL METHODS

COURSE WORK

on the topic: "Repeated and independent tests. Bernoulli's theorem on probability frequency"

Completed:

student of group 303

Rudnitsky Alexander

Petrovich

Checked by: manager department

philosophy

Granevsky V.V.

Tiraspol, 2009

1. Introduction

2. Bernoulli formula

3. Local Moivre-Laplace formula

4. Poisson formula

5. Bernoulli's theorem on probability frequency

Bibliography

Applications

1. Introduction

In the practical application of probability theory, one often encounters problems in which the same test is repeated several times. As a result of each trial, some event A may or may not appear, and we are not interested in the result of each individual trial, but in the total number of occurrences of event A as a result of a series of experiments. For example, if a group of shots is fired at the same target, we are usually not interested in the result of each shot, but in the total number of hits. Such problems require the ability to determine the probability of any given number of occurrences of an event as a result of a series of experiments. Such problems will be considered. They can be solved quite simply when the tests are independent.

Definition. Tests are called independent if the probability of one or another outcome of each test does not depend on what outcomes the other tests had.

For example, multiple coin tosses constitute independent trials.

2. Bernoulli formula

Let two tests be performed (n=2). As a result, one of the following events may occur:

The corresponding probabilities of these events are: .

Or - the occurrence of an event in only one trial.

The probability of an event occurring twice.

The probability of an event occurring only once.

The probability of an event occurring is zero times.

Let now n=3. Then one of the following events is possible:

The corresponding probabilities are equal.

It is obvious that the results obtained for n=2 and n=3 are elements

Now let's say n tests are performed. Event A can occur n times, 0 times, n-1 times, etc. Let us write an event consisting of the occurrence of event A m times

It is necessary to find the number of trials in which event A occurs m times. To do this, you need to find the number of combinations of n elements in which A is repeated m times, and n-m times.

The probability of event A occurring.

The last formula is called Bernoulli's formula and represents the general term of the expansion:

From formula (1) it is clear that it is convenient to use when the number of tests is not too large.

Examples

№1 . A coin is tossed 7 times. Find the probability of heads coming three times.

Solution.

№2. Every day, ABC Corporation shares rise or fall in price by one point with probabilities of 0.75 and 0.25, respectively. Find the probability that the stock will return to its original price after six days. Accept the condition that changes in the stock price up and down are independent events.

Solution. In order for the shares to return to their original price in 6 days, they need to rise in price 3 times and fall in price three times during this time. The required probability is calculated using the Bernoulli formula

№3. The engines of a multi-engine aircraft fail during flight independently of one another with probability p. A multi-engine aircraft continues to fly if at least half of its engines are working. At what values ​​of p is a two-engine airplane more reliable than a four-engine airplane?

Solution. A twin-engine airplane crashes if both engines fail. This happens with probability p2. A four-engine airplane crashes if all 4 engines fail, which happens with probability p4, or if three engines out of 4 fail. The probability of the last event is calculated using Bernoulli's formula: . For a twin-engine aircraft to be more reliable than a four-engine aircraft, the following inequality must be satisfied:

p2<р4+4p3(1–p)

This inequality reduces to the inequality (3р–1)(р–1)<0. Второй сомножитель в левой части этого неравенства всегда отрицателен (по условию задачи). Следовательно, величина 3р–1 должна быть положительной, откуда следует, что должно выполняться условие р>1/3. It should be noted that if the probability of an airplane engine failure exceeded one third, the very idea of ​​using aviation for passenger transportation would be very doubtful.

№4. A team of ten people goes to lunch. There are two identical canteens, and each member of the team, independently of one another, goes to dine in any of these canteens. If by chance more visitors come to one of the canteens than there are seats available, a queue will form. What is the minimum number of seats that should be in each of the canteens so that the probability of a queue occurring is less than 0.15?

Solution. The solution to the problem will have to be sought by searching through possible options. First, note that if there are 10 seats in each dining room, then a queue is impossible. If each dining room has 9 seats, then a queue will only appear if all 10 visitors end up in one dining room. From the conditions of the problem it follows that each member of the team chooses this canteen with probability 1/2. This means that everyone will gather in one dining room with probability 2(1/2)10=1/512. This number is much less than 0.15 and should be calculated for eight-seat dining rooms. If each canteen has 8 seats, then a queue will occur if all members of the team come to one canteen, the probability of this event has already been calculated, or 9 people will go to one canteen, and 1 person will choose another canteen. The probability of this event is calculated using Bernoulli's formula. Thus, if there are 8 seats in the dining rooms, then a queue occurs with a probability of 11/512, which is still less than 0.15. Now let there be 7 seats in each of the canteens. In addition to the two options considered, in this case a queue will arise if 8 people come to one of the canteens, and 2 people to the other. This can happen with probability. This means that in this case a queue occurs with a probability of 56/512=0.109375<0,15. Действуя аналогичным образом, вычисляем, что если в каждой столовой 6 мест, то очередь возникает с вероятностью 56/512+120/512=176/512=0,34375. Отсюда получаем, что наименьшее число мест в каждой столовой должно равняться семи.

№5. There are 20 white and 10 black balls in an urn. 4 balls are taken out, and each removed ball is returned to the urn before the next one is taken out and the balls in the urn are mixed. Find the probability that out of four drawn balls there will be 2 white ones.

Solution. Event A- took out a white ball. Then the probabilities

According to Bernoulli's formula, the required probability is equal to

№6. Determine the probability that a family with 5 children will have no more than three girls. The probabilities of having a boy and a girl are assumed to be the same.

Solution. Probability of having a girl

Let's find the probabilities that there are no girls in the family, one, two or three girls were born:

Therefore, the required probability

№7. Among the parts processed by a worker, on average 4% are non-standard. Find the probability that among 30 parts taken for testing, two will be non-standard.

Solution. Here the experience consists of checking each of the 30 parts for quality. Event A is “the appearance of a non-standard part”, its probability is then . From here, using Bernoulli’s formula, we find

№8. With each individual shot from a gun, the probability of hitting the target is 0.9. Find the probability that out of 20 shots the number of successful shots will be at least 16 and not more than 19.

Solution. We calculate using Bernoulli's formula:

№9. Independent testing continues until the event A will not happen k once. Find the probability that it will be required n tests (n i k), if in each of them .

Solution. Event IN– exactly n tests before k- occurrence of an event A– is the product of the following two events:

D – in n-th test A happened;

C - first (n–1) -th tests A appeared (k-1) once.

The multiplication theorem and Bernoulli's formula give the required probability:

№10. Out of n batteries, k fail during a year of storage. m batteries are selected at random. Determine the probability that among them l are serviceable. n = 100, k = 7, m = 5, l = 3.

Solution: We have a Bernoulli circuit with parameters p=7/100=0.07 (the probability that the battery will fail), n = 5 (the number of tests), k = 5-3 =2 (the number of “successes”, faulty batteries). We will use Bernoulli's formula (the probability that an event will occur k times in n trials).

We get

№11. A device consisting of five independently operating elements is turned on in time T. The probability of failure of each of them during this time is 0.2. Find the probability that: a) three elements will be rejected; b) at least four elements; c) at least one element.

Solution: We have a Bernoulli scheme with parameters p = 0.2 (the probability that an element will fail), n = 5 (the number of tests, that is, the number of elements), k (the number of “successes”, failed elements). We will use Bernoulli's formula (the probability that for n elements failure will occur in k elements): . We get a) - the probability that exactly three out of five elements will fail. b) - the probability that at least four elements out of five will fail (that is, either four or five). c) - the probability that at least one element will fail (found through the probability of the opposite event - not a single element will fail).

№12. How many games of chess should be played with a probability of winning in one game equal to 1/3, so that the most probable number of victories is 5?

Solution: The most probable number of victories k is determined from the formula Here p = 1/3 (probability of victory), q = 2/3 (probability of loss), n is the unknown number of games. Substituting these values, we get:

We get that n = 15, 16 or 17.

3. Local Moivre-Laplace formula

It is easy to see that using Bernoulli's formula for large values ​​of n is quite difficult, since the formula requires operations on huge numbers. Naturally, the question arises: is it possible to calculate the probability of interest to us without resorting to Bernoulli’s formula.

In 1730, another solution method for p=1/2 was found by Moivre; in 1783, Laplace generalized Moivre's formula for arbitrary p, different from 0 and 1.

This formula is used for an unlimited increase in the number of trials, when the probability of an event occurring is not too close to zero or one. Therefore, the theorem in question is called the Moivre-Laplace theorem.

De Moivre-Laplace theorem. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then the probability that event A will appear in n trials exactly k times is approximately equal (the more accurate, the larger n) to the value of the function

There are tables containing the function values

corresponding to positive values ​​of the argument x (see Appendix 1). For negative values ​​of the argument, the same tables are used, since the function is even, i.e. .

So, the probability that event A will appear in n independent trials exactly k times is approximately equal to

№13. Find the probability that event A will occur exactly 80 times in 400 trials if the probability of this event occurring in each trial is 0.2.

Solution. According to the condition n=400; k=80; p=0.2; q=0.8. Let's use Laplace's formula:

Required probability

№14. The probability of a shooter hitting a target with one shot p=0.75.

Find the probability that with 10 shots the shooter will hit the target 8 times.

Solution. By condition n=10; k=8; p=0.75; q=0.25.

Let's use Laplace's formula:

Let us calculate the value x determined by the task data:

Using the table in Appendix 1 we find

Required probability

№15. Find the probability that event A will occur exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.

Solution. By condition n=243; k=70; p=0.25; q=0.75. Let's use Laplace's formula:

Let's find the value of x:

Using the table in Appendix 1 we find

Required probability

№16. Find the probability that event A will occur 1400 times in 2400 trials if the probability of this event occurring in each trial is 0.6.

Solution. According to the condition n=2400; k=1400; p=0.6; q=0.4. As in the previous example, we will use Laplace’s formula:

Let's calculate x:

Using the table in Appendix 1 we find

Required probability

4. Poisson formula

This formula is applied when the number of trials increases indefinitely, when the probability of an event occurring is sufficiently close to 0 or 1.

Proof.

Thus we got the formula:

Examples

№17. The probability of producing an unusable part is 0.0002. Find the probability that among 10,000 parts only 2 parts will be unusable.

Solution. n=10000; k=2; p=0.0002.

Required probability

.

№18. The probability of producing a defective part is 0.0004. Find the probability that among 1000 parts, only 5 parts will be defective.

Solution. n=1000; k=5; p=0.0004.

Required probability

№19. The probability of winning the lottery is 0.0001. Find the probability that out of 5000 attempts you will win 3 times.

Solution. n=5000; k=3; p=0.0001.

Required probability

.

5. Bernoulli's theorem on probability frequency

Theorem. The probability that in n independent trials, in each of which the probability of the occurrence of an event is equal to p, the absolute value of the deviation of the relative frequency of the occurrence of an event from the probability of the occurrence of an event will not exceed a positive number, is approximately equal to twice the Laplace function at:

Proof. We assume that n independent trials are performed, in each of which the probability of occurrence of event A is constant and equal to p. Let us set ourselves the task of finding the probability that the deviation of the relative frequency from the constant probability p in absolute value does not exceed a given number. In other words, we find the probability of the inequality

Let us replace inequality (*) with equivalent ones:

Multiplying these inequalities by a positive factor , we obtain inequalities equivalent to the original one:

Then we find the probability as follows:

The value of the function is found in the table (see Appendix 2).

Examples

№20. The probability that the part is not standard is p=0.1. Find the probability that among randomly selected 400 parts, the relative frequency of occurrence of non-standard parts will deviate from the probability p=0.1 in absolute value by no more than 0.03.

Solution. n=400; p=0.1; q=0.9; =0.03. You need to find the probability. Using the formula

Using the table in Appendix 2 we find . Hence, . So, the desired probability is 0.9544.

№21. The probability that the part is not standard is p=0.1. Find how many parts need to be selected so that, with a probability equal to 0.9544, it can be stated that the relative frequency of occurrence of non-standard parts (among those selected) will deviate from the constant probability p in absolute value by no more than 0.03.

Solution. According to the condition, p=0.1; q=0.9; =0.03; . You need to find n. Let's use the formula

Due to the conditions

Hence,

Using the table in Appendix 2 we find. To find the number n we get the equation. Hence the required number of parts n=400.

№22. The probability of an event occurring in each of the independent trials is 0.2. Find what deviation of the relative frequency of occurrence of an event from its probability can be expected with a probability of 0.9128 with 5000 trials.

Solution. Let's use the same formula from which it follows:

Literature

1. Gmurman E.V. "Probability theory and mathematical statistics", Moscow, "Higher School" 2003.

2. Gmurman E.V. "Guide to solving problems in probability theory and mathematical statistics", Moscow "Higher School" 2004.

3. Gnedenko B.V. "Course on Probability Theory", Moscow, "Science" 1988.

4. Kolemaev V.A., Kalinina V.N., Solovyov V.I., Malykhin V.I., Kurochkin A.P. "Probability theory in examples and problems", Moscow, 2001.

5. Ventzel E.S. "Theory of Probability", Moscow, "Higher School" 1998.

Applications

Annex 1

Function value table

Appendix 2

Function value table

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Repetition of tests. Bernoulli circuit

Guidelines for practical exercises

in the discipline "Higher Mathematics"

Compiled by: Egorova Yu.B.

Mamonov I.M.

Moscow 2006 introduction

The guidelines are intended for full-time and evening students of Faculty No. 14, specialties 150601, 160301, 230102. The guidelines highlight the basic concepts of the topic and determine the sequence of studying the material. A large number of examples discussed help in the practical development of the topic. Methodical instructions serve as a methodological basis for practical exercises and individual assignments.

BERNOULLI SCHEME. BERNOULLI FORMULA

Bernoulli scheme- a scheme of repeated independent tests in which some event A can be repeated many times with constant probability R (A)= R .

Examples of tests carried out according to the Bernoulli scheme: repeated tossing of a coin or dice, manufacturing a batch of parts, shooting at a target, etc.

Theorem. If the probability of an event occurring A in each test is constant and equal R, then the probability that the event A will come m once every n tests (no matter in what sequence), can be determined by Bernoulli’s formula:

Where q = 1 – p.

EXAMPLE 1. The probability that electricity consumption during one day will not exceed the established norm is equal to p= 0,75. Find the probability that in the next 6 days, electricity consumption for 4 days will not exceed the norm.

SOLUTION. The probability of normal electricity consumption for each of 6 days is constant and equal to R= 0.75. Consequently, the probability of excessive energy consumption every day is also constant and equal to q = 1R = 1  0,75 = 0,25.

The required probability according to Bernoulli’s formula is equal to:

EXAMPLE 2. The shooter fires three shots at the target. The probability of hitting the target with each shot is equal to p= 0,3. Find the probability that: a) one target is hit; b) all three targets; c) not a single target; d) at least one target; e) less than two targets.

SOLUTION. The probability of hitting the target with each shot is constant and equal to R=0.75. Therefore, the probability of a miss is equal to q = 1 R= 1  0.3= 0.7. Total number of experiments performed n=3.

a) The probability of hitting one target with three shots is equal to:

b) The probability of hitting all three targets with three shots is equal to:

c) The probability of three misses with three shots is equal to:

d) The probability of hitting at least one target with three shots is equal to:

e) The probability of hitting less than two targets, that is, either one target or none:

1. Local and integral theorems of Moivre-Laplace

If a large number of tests are performed, then calculating probabilities using Bernoulli's formula becomes technically difficult, since the formula requires operations with huge numbers. Therefore, there are simpler approximate formulas for calculating probabilities at large n. These formulas are called asymptotic and are determined by Poisson's theorem, local and integral theorem of Laplace.

Local theorem of Moivre-Laplace. A A will happen m once every n n (n →∞ ), is approximately equal to:

where is the function
and the argument

The more n, the more accurate the calculation of probabilities. Therefore, it is advisable to apply the Moivre-Laplace theorem when npq  20.

f ( x ) special tables have been compiled (see Appendix 1). When using the table you need to keep in mind function properties f(x) :

Function f(x) is even f(  x)=f(x) .

At X ∞ function f(x) 0. In practice, we can assume that already at X>4 function f(x) ≈0.

EXAMPLE 3. Find the probability that the event A will occur 80 times in 400 trials if the probability of the event occurring A in each trial is equal p= 0,2.

SOLUTION. By condition n=400, m=80, p=0,2, q=0.8. Hence:

Using the table, we determine the value of the function f (0)=0,3989.

Integral theorem of Moivre-Laplace. If the probability of an event occurring A in each trial is constant and different from 0 and 1, then the probability that the event A comes from m 1 before m 2 once every n tests with a sufficiently large number n (n →∞ ), is approximately equal to:

Where
 integral or Laplace function,

To find the values ​​of a function F( x ) Special tables have been compiled (for example, see Appendix 2). When using the table you need to keep in mind properties of the Laplace function Ф(x) :

Function Ф(x) is odd F(  x)=  Ф(x) .

At X ∞ function Ф(x) 0.5. In practice, we can assume that already at X>5 function Ф(x) ≈0,5.

F (0)=0.

EXAMPLE 4. The probability that the part did not pass the quality control inspection is 0.2. Find the probability that among 400 parts there will be from 70 to 100 parts untested.

SOLUTION. By condition n=400, m 1 =70, m 2 =100, p=0,2, q=0.8. Hence:

Using the table that shows the values ​​of the Laplace function, we determine:

Ф(x 1 ) = F(  1,25 )=  F( 1,25 )=  0,3944; Ф(x 2 ) = F( 2,5 )= 0,4938.

Slide 1

Bernoulli's theorem
17.03.2017

Slide 2

A series of n independent tests is carried out. Each test has 2 outcomes: A - “success” and - “failure”. The probability of “success” in each trial is the same and is equal to P(A) = p. Accordingly, the probability of “failure” also does not change from experiment to experiment and is equal.
Bernoulli scheme
What is the probability that in a series of n experiments there will be success k times? Find Pn(k) .

Slide 3

The coin is tossed n times. A card is drawn from the deck n times, and each time the card is returned, the deck is shuffled. n products of a certain production, selected at random, are examined for quality. The shooter shoots at the target n times.
Examples

Slide 4

Explain why the following questions fit Bernoulli's scheme. State what “success” is and what n and k are equal to. a) What is the probability of getting a “2” three times when throwing a die ten times? b) What is the probability that in one hundred coin tosses, heads will appear 73 times? c) A pair of dice were thrown twenty times in a row. What is the probability that the sum of the points never equals ten? d) Three cards were drawn from a deck of 36 cards, the result was recorded and returned to the deck, then the cards were shuffled. This was repeated 4 times. What is the probability that each time the queen of spades was among the cards drawn?

Slide 5

For the number of combinations from n to k, the following formula is valid:
For example:

Slide 6

Bernoulli's theorem
The probability Pn(k) of the occurrence of exactly k successes in n independent repetitions of the same trial is found by the formula, where p is the probability of “success”, q = 1- p is the probability of “failure” in a separate experiment.

Slide 7

The coin is tossed 6 times. What is the probability of getting the coat of arms 0, 1, ...6 times? Solution. Number of experiments n=6. Event A – “success” – loss of the coat of arms. According to Bernoulli's formula, the required probability is equal to
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Slide 8

The coin is tossed 6 times. What is the probability of getting the coat of arms 0, 1, ...6 times? Solution. Number of experiments n=6. Event A – “success” – loss of the coat of arms.
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Slide 9

The coin is tossed 10 times. What is the probability of the coat of arms appearing twice? Solution. Number of experiments n=10, m=2. Event A – “success” – loss of the coat of arms. According to Bernoulli's formula, the required probability is equal to
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Slide 10

There are 20 white and 10 black balls in an urn. 4 balls are taken out, and each removed ball is returned to the urn before the next one is taken out and the balls in the urn are mixed. Find the probability that out of four drawn balls there will be 2 white ones. Solution. Event A – a white ball is taken out. Then the probabilities According to Bernoulli's formula, the required probability is equal to

Slide 11

Determine the probability that a family with 5 children has no girls. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 12

Determine the probability that a family with 5 children will have one girl. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 13

Determine the probability that a family with 5 children will have two girls. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 14

Determine the probability that a family with 5 children will have three girls. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 15

Determine the probability that in a family with 5 children there will be no more than three girls. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy The required probability is equal to
.

Slide 16

Among the parts processed by a worker, on average 4% are non-standard. Find the probability that among 30 parts taken for testing, two will be non-standard. Solution. Here the experience consists of checking each of the 30 parts for quality. Event A - “appearance of a non-standard part”,