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| Formulas for calculating the probability of events. Probability theory: formulas and examples of problem solving How to determine the probability of a random event |
1.3.1. Sequence of independent tests (Bernoulli circuit)Suppose that some experiment can be carried out repeatedly under the same conditions. Let this experience be made n times, i.e., a sequence of n tests. Definition. Subsequence n tests are called mutually independent , if any event related to a given test is independent of any events related to other tests. Let's assume that some event A likely to happen p as a result of one test or not likely to happen q= 1- p. Definition . Sequence of n tests forms a Bernoulli scheme if the following conditions are met: subsequence n tests are mutually independent, 2) probability of an event A does not change from trial to trial and does not depend on the result in other trials. Event A is called the “success” of the test, and the opposite event is called “failure.” Consider the event
To calculate the probability of this event, the Bernoulli formula is valid p( Where
Example 1.16. The die is tossed three times. Find: a) the probability that 6 points will appear twice; b) the probability that the number of sixes will not appear more than twice. Solution . We will consider the “success” of the test to be when the side with the image of 6 points appears on the die. a) Total number of tests – n=3, number of “successes” – m
= 2. Probability of “success” - p=
b) Let us denote by A an event that means that a side with a score of 6 will appear no more than twice. Then the event can be represented as the sum of three incompatible events A=
Where IN 3 0 – an event when the edge of interest never appears, IN 3 1 - event when the edge of interest appears once, IN 3 2 - event when the edge of interest appears twice. Using the Bernoulli formula (1.6) we find p(A)
= p ( =
1.3.2. Conditional probability of an eventConditional probability reflects the influence of one event on the probability of another. Changing the conditions under which the experiment is carried out also affects on the probability of occurrence of the event of interest. Definition. Let A And B– some events, and probability p(B)> 0. Conditional probability events A provided that the “event Balready happened” is the ratio of the probability of the occurrence of these events to the probability of an event that occurred earlier than the event whose probability is required to be found. Conditional probability is denoted as p(A B). Then by definition p
(A
B)
=
Example 1.17. Two dice are tossed. The space of elementary events consists of ordered pairs of numbers (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6). In Example 1.16 it was determined that the event A=(number of points on the first die > 4) and event C=(sum of points is 8) dependent. Let's make a relation
This relationship can be interpreted as follows. Let us assume that the result of the first roll is known to be that the number of points on the first die is > 4. It follows that throwing the second die can lead to one of the 12 outcomes that make up the event A: (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) . At this event C only two of them can match (5,3) (6,2). In this case, the probability of the event C
will be equal Probability of events happening Multiplication theorem Probability of events happeningA 1 A 2 A n is determined by the formula p(A 1 A 2 A n)= p(A 1)p(A 2 A 1)) p(A n A 1 A 2 A n- 1). (1.8) For the product of two events it follows that p(AB)= p(A B)p{B)= p(B A)p{A). (1.9) Example 1.18. In a batch of 25 products, 5 products are defective. 3 items are selected at random in succession. Determine the probability that all selected products are defective. Solution. Let's denote the events: A 1 = (first product is defective), A 2 = (second product is defective), A 3 = (third product is defective), A = (all products are defective). Event A is the product of three events A = A 1 A 2 A 3 . From the multiplication theorem (1.6) we get p(A)= p( A 1 A 2 A 3 ) = p(A 1) p(A 2 A 1))p(A 3 A 1 A 2). The classical definition of probability allows us to find p(A 1) is the ratio of the number of defective products to the total number of products: p(A 1)=
p(A 2) – This the ratio of the number of defective products remaining after the removal of one to the total number of remaining products: p(A 2
A 1))=
p(A 3) – this is the ratio of the number of defective products remaining after the removal of two defective ones to the total number of remaining products: p(A 3
A 1 A 2)= Then the probability of the event A will be equal p(A)
= |
TOPIC 1 . Classic formula for calculating probability. Basic definitions and formulas: An experiment whose outcome cannot be predicted is called random experiment(SE). An event that may or may not occur in a given SE is called random event. Elementary outcomes events that meet the requirements are called: 1.with any implementation of SE, one and only one elementary outcome occurs; 2. every event is a certain combination, a certain set of elementary outcomes. The set of all possible elementary outcomes completely describes the SE. Such a set is usually called space of elementary outcomes(PEI). The choice of PEI to describe a given SE is ambiguous and depends on the problem being solved. P(A) = n(A)/n, where n – total number equally possible outcomes, n (A) – the number of outcomes that make up event A, as they also say, favorable to event A. The words “at random”, “at random”, “randomly” guarantee the equal possibility of elementary outcomes. Solving typical examples Example 1. From an urn containing 5 red, 3 black and 2 white balls, 3 balls are drawn at random. Find the probabilities of events: A– “all the drawn balls are red”; IN– “all drawn balls are of the same color”; WITH– “among those extracted there are exactly 2 black ones.” Solution: The elementary outcome of this SE is a triple (disordered!) of balls. Therefore, the total number of outcomes is the number of combinations: n == 120 (10 = 5 + 3 + 2). Event A consists only of those triplets that were drawn from five red balls, i.e. n(A)==10. Event IN In addition to 10 red threes, black threes are also favorable, the number of which is = 1. Therefore: n (B)=10+1=11. Event WITH Those threes of balls that contain 2 black and one non-black are favored. Each method of selecting two black balls can be combined with selecting one non-black ball (out of seven). Therefore: n (C) = = 3 * 7 = 21. So: P(A) = 10/120; P(B) = 11/120; R(S) = 21/120.
Example 2. In the conditions of the previous problem, we will assume that the balls of each color have their own numbering, starting from 1. Find the probabilities of events: D– “the maximum extracted number is 4”; E– “The maximum number extracted is 3.” Solution: To calculate n(D), we can assume that the urn has one ball with number 4, one ball with a higher number and 8 balls (3k+3h+2b) with lower numbers. Event D Those threes of balls that necessarily contain a ball with number 4 and 2 balls with lower numbers are favored. Therefore: n(D) = P(D) = 28/120. To calculate n (E) we consider: there are two balls in the urn with number 3, two with large numbers and six balls with lower numbers (2k+2h+2b). Event E consists of triplets of two types: 1. one ball with number 3 and two with lower numbers; 2.two balls with number 3 and one with a lower number. Therefore: n(E)= P(E) = 36/120. Example 3. Each of M different particles is thrown at random into one of N cells. Find the probabilities of events: A– all particles fell into the second cell; IN– all particles fell into one cell; WITH– each cell contains no more than one particle (M £ N); D– all cells are occupied (M =N +1); E– the second cell contains exactly To particles. Solution: For each particle there are N ways to get into a particular cell. According to the basic principle of combinatorics for M particles we have N *N *N *…*N (M times). So, the total number of outcomes in this SE n = N M . For each particle we have one opportunity to get into the second cell, therefore n (A) = 1*1*…*1= 1 M = 1, and P(A) = 1/ N M. Getting into one cell (for all particles) means getting everyone into the first, or everyone into the second, or etc. everyone in Nth. But each of these N options can be implemented in one way. Therefore n (B)=1+1+…+1(N -times)=N and Р(В)=N/N M. Event C means that each particle has one less number of placement options than the previous particle, and the first one can fall into any of N cells. That's why: n (C) = N *(N -1)*…*(N +M -1) and Р(С) = In the particular case with M =N: Р(С)= Event D means that one of the cells contains two particles, and each of the (N -1) remaining cells contains one particle. To find n (D) we reason like this: choose a cell in which there will be two particles, this can be done in =N ways; then we will select two particles for this cell, there are ways to do this. After this, we distribute the remaining (N -1) particles one at a time into the remaining (N -1) cells, for this there are (N -1)! ways. So n(D) =
The number n(E) can be calculated as follows: To particles for the second cell can be done in ways; the remaining (M – K) particles are distributed randomly over the (N -1) cell (N -1) in M-K ways. That's why: |
|||||||||
=( in n tests happened exactly m“success”).
)
=
, m
= 1, 2, …, n
, (1.6)
- number of combinations of n elements by m
:
=
=
.
,
and the probability of “failure” is q= 1 - =. Then, according to Bernoulli's formula, the probability that, as a result of throwing a die three times, the side with six points will appear twice, will be equal to
.
,
)
=
p(
)=
+
+
=
.
.
(1.7)
.
. Thus, information about the occurrence of an event A influenced the likelihood of an event C.
;
;
.
=
.
.Probability in mathematics
In the modern mathematical approach, classical (that is, not quantum) probability is given by the Kolmogorov axiomatics. Probability is a measure P, which is defined on the set X, called probability space. This measure must have the following properties:
From specified conditions it follows that the probability measure P also has the property additivity: if sets A 1 and A 2 do not intersect, then . To prove you need to put everything A 3 , A 4 , ... equal to the empty set and apply the property of countable additivity.
The probability measure may not be defined for all subsets of the set X. It is enough to define it on a sigma algebra, consisting of some subsets of the set X. In this case, random events are defined as measurable subsets of space X, that is, as elements of sigma algebra.
Probability sense
When we find that the reasons for some possible fact actually occurring outweigh the contrary reasons, we consider that fact probable, otherwise - incredible. This preponderance of positive bases over negative ones, and vice versa, can represent an indefinite set of degrees, as a result of which probability(And improbability) It happens more or less .
Complex individual facts do not allow for an exact calculation of the degrees of their probability, but even here it is important to establish some large subdivisions. So, for example, in the legal field, when a personal fact subject to trial is established on the basis of testimony, it always remains, strictly speaking, only probable, and it is necessary to know how significant this probability is; in Roman law, a quadruple division was adopted here: probatio plena(where the probability practically turns into reliability), Further - probatio minus plena, then - probatio semiplena major and finally probatio semiplena minor .
In addition to the question of the likelihood of the case, the question may arise, both in the field of law and in the moral field (with a certain ethical point of view), of how likely it is that a given particular fact constitutes a violation common law. This question, which served as the main motive in the religious jurisprudence of the Talmud, also gave rise to Roman Catholic moral theology (especially with late XVI centuries) very complex systematic constructions and a huge literature, dogmatic and polemical (see Probabilism).
The concept of probability allows for a certain numerical expression when applied only to such facts that are part of certain homogeneous series. So (in the simplest example), when someone throws a coin a hundred times in a row, we find here one general or large series (the sum of all falls of the coin), consisting of two private or smaller, in this case numerically equal, series (falls " heads" and falls "tails"); The probability that this time the coin will land heads, that is, that this new member of the general series will belong to this of the two smaller series, is equal to the fraction expressing the numerical relationship between this small series and the larger one, namely 1/2, that is, the same probability belongs to one or the other of two particular series. In less simple examples the conclusion cannot be deduced directly from the data of the problem itself, but requires preliminary induction. So, for example, the question is: what is the probability for a given newborn to live to be 80 years old? Here there must be a general, or large, series of a certain number of people born in similar conditions and dying at different ages (this number must be large enough to eliminate random deviations, and small enough to maintain the homogeneity of the series, for for a person, born, for example, in St. Petersburg into a wealthy, cultured family, the entire million-strong population of the city, a significant part of which consists of people from various groups who can die prematurely - soldiers, journalists, workers in dangerous professions - represents a group too heterogeneous for a real determination of probability) ; let this general row consist of ten thousand human lives; it includes smaller series representing the number of people surviving to a particular age; one of these smaller series represents the number of people living to age 80. But it is impossible to determine the number of this smaller series (like all others) a priori; this is done purely inductively, through statistics. Let's say statistical research found that out of 10,000 middle-class St. Petersburg residents, only 45 live to be 80; Thus, this smaller series is related to the larger one as 45 is to 10,000, and the probability for a given person to belong to this smaller series, that is, to live to be 80 years old, is expressed as a fraction of 0.0045. The study of probability from a mathematical point of view constitutes a special discipline - probability theory.
see also
Notes
Literature
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See what “Probability” is in other dictionaries:
General scientific and philosophical. a category denoting the quantitative degree of possibility of the occurrence of mass random events under fixed observation conditions, characterizing the stability of their relative frequencies. In logic, semantic degree... ... Philosophical Encyclopedia
PROBABILITY, a number in the range from zero to one inclusive, representing the possibility of a given event occurring. The probability of an event is defined as the ratio of the number of chances that an event can occur to the total number of possible... ... Scientific and technical encyclopedic dictionary
In all likelihood.. Dictionary of Russian synonyms and similar expressions. under. ed. N. Abramova, M.: Russian Dictionaries, 1999. probability possibility, likelihood, chance, objective possibility, maza, admissibility, risk. Ant. impossibility... ... Synonym dictionary
probability- A measure that an event is likely to occur. Note The mathematical definition of probability is: “a real number between 0 and 1 that is associated with a random event.” The number may reflect the relative frequency in a series of observations... ... Technical Translator's Guide
Probability- “a mathematical, numerical characteristic of the degree of possibility of the occurrence of any event in certain specific conditions that can be repeated an unlimited number of times.” Based on this classic... ... Economic and mathematical dictionary
- (probability) The possibility of the occurrence of an event or a certain result. It can be presented in the form of a scale with divisions from 0 to 1. If the probability of an event is zero, its occurrence is impossible. With a probability equal to 1, the onset of... Dictionary of business terms
When a coin is tossed, we can say that it will land heads up, or probability this is 1/2. Of course, this does not mean that if a coin is tossed 10 times, it will necessarily land on heads 5 times. If the coin is "fair" and if it is tossed many times, then heads will land very close half the time. Thus, there are two types of probabilities: experimental And theoretical .
Experimental and theoretical probability
If you toss a coin a large number of times - say 1000 - and count the number of times heads are thrown, we can determine the probability of heads being thrown. If heads are thrown 503 times, we can calculate the probability of it landing:
503/1000, or 0.503.
This experimental determination of probability. This definition of probability comes from observation and study of data and is quite common and very useful. Here, for example, are some probabilities that were determined experimentally:
1. The probability that a woman will develop breast cancer is 1/11.
2. If you kiss someone who has a cold, then the probability that you will also get a cold is 0.07.
3. A person who has just been released from prison has an 80% chance of returning to prison.
If we consider tossing a coin and taking into account that it is just as likely that it will come up heads or tails, we can calculate the probability of getting heads: 1/2. This is a theoretical definition of probability. Here are some other probabilities that have been determined theoretically using mathematics:
1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding year) is 0.706.
2. During a trip, you meet someone, and during the conversation you discover that you have a mutual friend. Typical reaction: “This can’t be!” In fact, this phrase is not suitable, because the probability of such an event is quite high - just over 22%.
Thus, experimental probabilities are determined through observation and data collection. Theoretical probabilities are determined through mathematical reasoning. Examples of experimental and theoretical probabilities, such as those discussed above, and especially those that we do not expect, lead us to the importance of studying probability. You may ask, "What is true probability?" In fact, there is no such thing. Probabilities within certain limits can be determined experimentally. They may or may not coincide with the probabilities that we obtain theoretically. There are situations in which it is much easier to determine one type of probability than another. For example, it would be sufficient to find the probability of catching a cold using theoretical probability.
Calculation of experimental probabilities
Let us first consider the experimental definition of probability. The basic principle we use to calculate such probabilities is as follows.
Principle P (experimental)
If in an experiment in which n observations are made, a situation or event E occurs m times in n observations, then the experimental probability of the event is said to be P (E) = m/n.
Example 1 Sociological survey. An experimental study was conducted to determine the number of left-handed people, right-handed people and people whose both hands are equally developed. The results are shown in the graph.
a) Determine the probability that the person is right-handed.
b) Determine the probability that the person is left-handed.
c) Determine the probability that a person is equally fluent in both hands.
d) Most Professional Bowling Association tournaments are limited to 120 players. Based on the data from this experiment, how many players could be left-handed?
Solution
a)The number of people who are right-handed is 82, the number of left-handers is 17, and the number of those who are equally fluent in both hands is 1. The total number of observations is 100. Thus, the probability that a person is right-handed is P
P = 82/100, or 0.82, or 82%.
b) The probability that a person is left-handed is P, where
P = 17/100, or 0.17, or 17%.
c) The probability that a person is equally fluent in both hands is P, where
P = 1/100, or 0.01, or 1%.
d) 120 bowlers, and from (b) we can expect that 17% are left-handed. From here
17% of 120 = 0.17.120 = 20.4,
that is, we can expect about 20 players to be left-handed.
Example 2 Quality control
. It is very important for a manufacturer to maintain the quality of its products at high level. In fact, companies hire quality control inspectors to ensure this process. The goal is to produce the minimum possible number of defective products. But since the company produces thousands of products every day, it cannot afford to test every product to determine whether it is defective or not. To find out what percentage of products are defective, the company tests far fewer products.
Ministry Agriculture The US requires that 80% of the seeds sold by growers must germinate. To determine the quality of the seeds that an agricultural company produces, 500 seeds from those that were produced are planted. After this, it was calculated that 417 seeds sprouted.
a) What is the probability that the seed will germinate?
b) Do the seeds meet government standards?
Solution a) We know that out of 500 seeds that were planted, 417 sprouted. Probability of seed germination P, and
P = 417/500 = 0.834, or 83.4%.
b) Since the percentage of seeds germinated has exceeded 80% as required, the seeds meet government standards.
Example 3 Television ratings. According to statistics, there are 105,500,000 households with televisions in the United States. Every week, information about viewing programs is collected and processed. In one week, 7,815,000 households tuned in to the hit comedy series "Everybody Loves Raymond" on CBS and 8,302,000 households tuned in to the hit series "Law & Order" on NBC (Source: Nielsen Media Research). What is the probability that one household's TV is tuned to "Everybody Loves Raymond" during a given week? to "Law & Order"?
Solution The probability that the TV in one household is tuned to "Everybody Loves Raymond" is P, and
P = 7,815,000/105,500,000 ≈ 0.074 ≈ 7.4%.
The chance that the household's TV was tuned to Law & Order is P, and
P = 8,302,000/105,500,000 ≈ 0.079 ≈ 7.9%.
These percentages are called ratings.
Theoretical probability
Suppose we are conducting an experiment, such as throwing a coin or darts, drawing a card from a deck, or testing products for quality on an assembly line. Each possible result of such an experiment is called Exodus . The set of all possible outcomes is called outcome space . Event it is a set of outcomes, that is, a subset of the space of outcomes.
Example 4 Throwing darts. Suppose that in a dart throwing experiment, a dart hits a target. Find each of the following:
b) Outcome space
Solution
a) The outcomes are: hitting black (B), hitting red (R) and hitting white (B).
b) The space of outcomes is (hitting black, hitting red, hitting white), which can be written simply as (H, K, B).
Example 5 Throwing dice.
A die is a cube with six sides, each with one to six dots on it. 
Suppose we are throwing a die. Find
a) Outcomes
b) Outcome space
Solution
a) Outcomes: 1, 2, 3, 4, 5, 6.
b) Outcome space (1, 2, 3, 4, 5, 6).
We denote the probability that an event E occurs as P(E). For example, “the coin will land on heads” can be denoted by H. Then P(H) represents the probability that the coin will land on heads. When all outcomes of an experiment have the same probability of occurring, they are said to be equally likely. To see the differences between events that are equally likely and events that are not, consider the target shown below. 
For target A, the events of hitting black, red and white are equally probable, since the black, red and white sectors are the same. However, for target B, the zones with these colors are not the same, that is, hitting them is not equally probable.
Principle P (Theoretical)
If an event E can happen in m ways out of n possible equally probable outcomes from the outcome space S, then theoretical probability
events, P(E) is
P(E) = m/n.
Example 6 What is the probability of rolling a die to get a 3?
Solution There are 6 equally probable outcomes on a dice and there is only one possibility of rolling the number 3. Then the probability P will be P(3) = 1/6.
Example 7 What is the probability of rolling an even number on a die?
Solution The event is the throwing of an even number. This can happen in 3 ways (if you roll a 2, 4 or 6). The number of equally probable outcomes is 6. Then the probability P(even) = 3/6, or 1/2.
We will use a number of examples involving a standard 52 card deck. This deck consists of the cards shown in the figure below. 
Example 8 What is the probability of drawing an Ace from a well-shuffled deck of cards?
Solution There are 52 outcomes (the number of cards in the deck), they are equally likely (if the deck is well shuffled), and there are 4 ways to draw an Ace, so according to the P principle, the probability
P(draw an ace) = 4/52, or 1/13.
Example 9 Suppose we choose, without looking, one ball from a bag with 3 red balls and 4 green balls. What is the probability of choosing a red ball?
Solution There are 7 equally probable outcomes of drawing any ball, and since the number of ways to draw a red ball is 3, we get
P(red ball selection) = 3/7.
The following statements are results from Principle P.
Properties of Probability
a) If event E cannot happen, then P(E) = 0.
b) If event E is certain to happen then P(E) = 1.
c) The probability that event E will occur is a number from 0 to 1: 0 ≤ P(E) ≤ 1.
For example, in a coin toss, the event that the coin lands on its edge has zero probability. The probability that a coin is either heads or tails has a probability of 1.
Example 10 Let's assume that 2 cards are drawn from a 52-card deck. What is the probability that both of them are peaks?
Solution The number n of ways to draw 2 cards from a well-shuffled deck of 52 cards is 52 C 2 . Since 13 of the 52 cards are spades, the number of ways m to draw 2 spades is 13 C 2 . Then,
P(pulling 2 peaks) = m/n = 13 C 2 / 52 C 2 = 78/1326 = 1/17.
Example 11 Suppose 3 people are randomly selected from a group of 6 men and 4 women. What is the probability that 1 man and 2 women will be selected?
Solution The number of ways to select three people from a group of 10 people is 10 C 3. One man can be chosen in 6 C 1 ways, and 2 women can be chosen in 4 C 2 ways. According to the fundamental principle of counting, the number of ways to choose 1 man and 2 women is 6 C 1. 4 C 2 . Then, the probability that 1 man and 2 women will be selected is
P = 6 C 1 . 4 C 2 / 10 C 3 = 3/10.
Example 12 Throwing dice. What is the probability of rolling a total of 8 on two dice?
Solution Each dice has 6 possible outcomes. The outcomes are doubled, meaning there are 6.6 or 36 possible ways in which the numbers on the two dice can appear. (It’s better if the cubes are different, say one is red and the other is blue - this will help visualize the result.) 
The pairs of numbers that add up to 8 are shown in the figure below. There are 5 possible ways receiving a sum equal to 8, hence the probability is 5/36.
Brought to date in open jar Unified State Examination problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is the classical definition of probability.
The easiest way to understand the formula is with examples.
Example 1. There are 9 red balls and 3 blue balls in the basket. The balls differ only in color. We take out one of them at random (without looking). What is the probability that the ball chosen in this way will be blue?
A comment. In probability problems, something happens (in this case, our action of drawing the ball) that can have different result- outcome. It should be noted that the result can be looked at in different ways. “We pulled out some kind of ball” is also a result. “We pulled out the blue ball” - the result. “We pulled out exactly this ball from all possible balls” - this least generalized view of the result is called an elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability.
Solution. Now let's calculate the probability of choosing the blue ball.
Event A: “the selected ball turned out to be blue”
Total number of all possible outcomes: 9+3=12 (the number of all balls that we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25
For the same problem, let's calculate the probability of choosing a red ball.
The total number of possible outcomes will remain the same, 12. Number of favorable outcomes: 9. Probability sought: 9/12=3/4=0.75
The probability of any event always lies between 0 and 1.
Sometimes in everyday speech (but not in probability theory!) the probability of events is estimated as a percentage. The transition between math and conversational scores is accomplished by multiplying (or dividing) by 100%.
So,
Moreover, the probability is zero for events that cannot happen - incredible. For example, in our example this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0, if calculated using the formula)
Probability 1 has events that are absolutely certain to happen, without options. For example, the probability that “the selected ball will be either red or blue” is for our task. (Number of favorable outcomes: 12, P(A)=12/12=1)
We looked at a classic example illustrating the definition of probability. All similar problems of the Unified State Exam in probability theory are solved by using this formula.
In place of the red and blue balls there may be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on some topic (prototypes,), defective and high-quality bags or garden pumps (prototypes,) - the principle remains the same.
They differ slightly in the formulation of the problem of the probability theory of the Unified State Examination, where you need to calculate the probability of some event occurring on a certain day. ( , ) As in previous problems, you need to determine what is the elementary outcome, and then apply the same formula.
Example 2. The conference lasts three days. On the first and second days there are 15 speakers, on the third day - 20. What is the probability that Professor M.’s report will fall on the third day if the order of reports is determined by drawing lots?
What is the elementary outcome here? – Assigning the professor’s report one of all possible serial numbers for the speech. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report may receive one of 50 issues. This means there are only 50 elementary outcomes.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4
The drawing of lots here represents the establishment of a random correspondence between people and ordered places. In example 2, the establishment of correspondence was considered from the point of view of which of the places could be taken special person. You can approach the same situation from the other side: which of the people with what probability could get to a specific place (prototypes , , , ):
Example 3. The draw includes 5 Germans, 8 French and 3 Estonians. What is the probability that the first (/second/seventh/last – it doesn’t matter) will be a Frenchman.
Number of elementary outcomes – number of all possible people, who could get to this place by drawing lots. 5+8+3=16 people.
Favorable outcomes - French. 8 people.
Required probability: 8/16=1/2=0.5
Answer: 0.5
The prototype is slightly different. There are still problems about coins () and dice(), somewhat more creative. The solution to these problems can be found on the prototype pages.
Here are a few examples of tossing a coin or dice.
Example 4. When we toss a coin, what is the probability of landing on heads?
There are 2 outcomes – heads or tails. (it is believed that the coin never lands on its edge) A favorable outcome is tails, 1.
Probability 1/2=0.5
Answer: 0.5.
Example 5. What if we toss a coin twice? What is the probability of getting heads both times?
The main thing is to determine what elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP – both times it came up heads
2) PO – first time heads, second time heads
3) OP – heads the first time, tails the second time
4) OO – heads came up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one, 1, is favorable.
Probability: 1/4=0.25
Answer: 0.25
What is the probability that two coin tosses will result in tails?
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5
In such problems, another formula may be useful.
If during one toss of a coin possible options we have 2 results, then for two throws the results will be 2 2 = 2 2 = 4 (as in example 5), for three throws 2 2 2 = 2 3 = 8, for four: 2 2 2 2 =2 4 =16, ... for N throws the possible results will be 2·2·...·2=2 N .
So, you can find the probability of getting 5 heads out of 5 coin tosses.
Total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR – heads all 5 times)
Probability: 1/32=0.03125
The same is true for dice. With one throw, there are 6 possible results. So, for two throws: 6 6 = 36, for three 6 6 6 = 216, etc.
Example 6. We throw the dice. What is the probability that an even number will be rolled?
Total outcomes: 6, according to the number of sides.
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5
Example 7. We throw two dice. What is the probability that the total will be 10? (round to the nearest hundredth)
For one die there are 6 possible outcomes. This means that for two, according to the above rule, 6·6=36.
What outcomes will be favorable for the total to roll 10?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. This means that the following options are possible for the cubes:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
Total, 3 options. Required probability: 3/36=1/12=0.08
Answer: 0.08
Other types of B6 problems will be discussed in a future How to Solve article.
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