Tasks for die probability are no less popular than coin toss problems. The condition of such a problem usually sounds like this: when throwing one or more dice (2 or 3), what is the probability that the sum of the points will be 10, or the number of points is 4, or the product of the number of points, or is the product of the number of points divided by 2 and etc.
The use of the classical probability formula is the main method for solving problems of this type.
One die, probability.
The situation is quite simple with one die. is determined by the formula: P = m / n, where m is the number of outcomes favorable to the event, and n is the number of all elementary equally possible outcomes of an experiment with a toss of a dice or dice.
Problem 1. A die is thrown once. What is the probability of getting an even number of points?
Since the dice is a cube (or it is also called the correct dice, the cube will fall on all faces with the same probability, since it is balanced), the cube has 6 faces (the number of points from 1 to 6, which are usually indicated by dots), this means what is in the problem total number outcomes: n = 6. The event is favored only by outcomes in which a face with even points 2,4 and 6 falls out, the cube has such faces: m = 3. Now we can determine the required probability of the die: P = 3/6 = 1/2 = 0.5.
Problem 2. A dice is thrown once. What is the probability that at least 5 points will be dropped?
This problem is solved by analogy with the example indicated above. When throwing a dice, the total number of equally possible outcomes is: n = 6, and the condition of the problem (at least 5 points dropped out, that is, 5 or 6 points were dropped) only 2 outcomes, which means m = 2. Next, we find the required probability: P = 2/6 = 1/3 = 0.333.
Two dice, probability.
When solving problems with throwing 2 dice, it is very convenient to use a special table of points dropping out. On it horizontally, the number of points dropped on the first die is deposited, and vertically - the number of points that fell on the second die. The workpiece looks like this:
But the question arises, what will be in the empty cells of the table? It depends on the problem that needs to be solved. If the task it comes about the sum of points, then the sum is written there, and if about the difference, then the difference is recorded and so on.
Problem 3. 2 dice are thrown at the same time. What is the probability of getting less than 5 points?
First, you need to figure out what the total number of outcomes of the experiment will be. Everything was obvious when throwing one die 6 sides of the dice - 6 outcomes of the experiment. But when there are already two dice, then the possible outcomes can be represented as ordered pairs of numbers of the form (x, y), where x shows how many points fell on the first dice (from 1 to 6), and y - how many points fell on the second dice (from 1 until 6). The total of such numerical pairs will be: n = 6 * 6 = 36 (36 cells correspond to them in the table of outcomes).
Now you can fill in the table, for this, the number of the sum of points that fell on the first and second dice is entered in each cell. The completed table looks like this:
Thanks to the table, we will determine the number of outcomes that favor the event "there will be less than 5 points in total". Let's count the number of cells, the value of the sum in which will be less number 5 (these are 2, 3 and 4). For convenience, paint over such cells, there will be m = 6:
Given the data in the table, die probability equals: P = 6/36 = 1/6.
Problem 4. Two dice were thrown. Determine the probability that the product of the number of points will be divisible by 3.
To solve the problem, let's make a table of the products of the points that fell on the first and second dice. In it, we immediately select numbers that are multiples of 3:
We write down the total number of outcomes of the experiment n = 36 (the reasoning is the same as in the previous task) and the number of favorable outcomes (the number of cells that are filled in the table) m = 20. The probability of the event is: P = 20/36 = 5/9.
Problem 5. The die is thrown twice. What is the probability that the difference in the number of points will be from 2 to 5 on the first and second dice?
To determine die probability Let's write down the table of score differences and select those cells in it, the value of the difference in which will be between 2 and 5:
The number of favorable outcomes (the number of cells filled in the table) is equal to m = 10, the total number of equally possible elementary outcomes will be n = 36. Determines the probability of the event: P = 10/36 = 5/18.
In the case of a simple event and when throwing 2 dice, you need to build a table, then select the necessary cells in it and divide their number by 36, this will be considered a probability.
Another popular problem in probability theory (along with the problem of tossing coins) is dice toss problem.
Usually the problem sounds like this: one or more dice are thrown (usually 2, less often 3). It is necessary to find the probability that the number of points is 4, or the sum of the points is 10, or the product of the number of points is divided by 2, or the numbers of points differ by 3, and so on.
The main method for solving such problems is using the classical probability formula, which we will analyze in the examples below.
After familiarizing yourself with the solution methods, you can download a super-useful one for throwing 2 dice (with tables and examples).
One dice
With one die, the matter is obscenely simple. Let me remind you that the probability is found by the formula $ P = m / n $, where $ n $ is the number of all equally possible elementary outcomes of an experiment with a toss of a dice or dice, and $ m $ is the number of those outcomes that favor the event.
Example 1. The dice is rolled once. What is the probability that an even number of points are scored?
Since the dice is a cube (they also say correct die, that is, the cube is balanced, so it falls on all faces with the same probability), the cube has 6 faces (with the number of points from 1 to 6, usually denoted by dots), then the total number of outcomes in the problem is $ n = 6 $. The event is favored only by such outcomes when a face with 2, 4 or 6 points (only even) falls out, such faces are $ m = 3 $. Then the required probability is equal to $ P = 3/6 = 1/2 = 0.5 $.
Example 2. The dice is thrown. Find the probability of dropping out at least 5 points.
We reason in the same way as in the previous example. The total number of equally possible outcomes when throwing a dice is $ n = 6 $, and the condition “at least 5 points fell out”, that is, “either 5 or 6 points fell out” satisfy 2 outcomes, $ m = 2 $. The required probability is $ P = 2/6 = 1/3 = $ 0.333.
I don't even see any reason to give more examples, let's move on to two dice, where everything is more interesting and more complicated.
Two dice
When it comes to problems with throwing 2 dice, it is very convenient to use points drop table... Horizontally we put off the number of points that fell on the first die, vertically - the number of points that fell on the second die. We will get such a blank (usually I make it in Excel, you can download the file):
And what about the cells of the table, you ask? And it depends on what problem we are going to solve. There will be a problem about the sum of points - we will write down the sum there, about the difference - we will write down the difference, and so on. Getting started?
Example 3. 2 dice are thrown at the same time. Find the probability that the total will be less than 5 points.
First, let's look at the total number of outcomes in the experiment. when we threw one die, everything was obvious, 6 sides - 6 outcomes. There are already two dice, so the outcomes can be represented as ordered pairs of numbers like $ (x, y) $, where $ x $ is how many points fell on the first dice (from 1 to 6), $ y $ is how many points fell on the second dice (from 1 to 6). Obviously, the total of such pairs of numbers will be $ n = 6 \ cdot 6 = 36 $ (and they correspond to just 36 cells in the outcome table).
Now it's time to fill out the table. In each cell we will enter the sum of the number of points dropped on the first and second dice and we will get the following picture:
Now this table will help us find the number of outcomes favorable to the event "in total less than 5 points will be drawn". To do this, count the number of cells in which the value of the sum will be less than 5 (that is, 2, 3 or 4). For clarity, we will fill in these cells, they will be $ m = 6 $:
Then the probability is: $ P = 6/36 = 1/6 $.
Example 4. Two dice are thrown. Find the probability that the product of the number of points is divisible by 3.
We make a table of the products of the points dropped on the first and second dice. We immediately select in it those numbers that are multiples of 3:
It only remains to note that the total number of outcomes is $ n = 36 $ (see the previous example, the reasoning is the same), and the number of favorable outcomes (the number of filled cells in the table above) is $ m = 20 $. Then the probability of the event will be equal to $ P = 20/36 = 5/9 $.
As you can see, this type of problem can be solved quickly and easily with proper preparation (analyze a couple of three more problems). Let's do one more task for a change with another table (all tables can be downloaded at the bottom of the page).
Example 5. The dice are rolled twice. Find the probability that the difference in the number of points on the first and second dice will be from 2 to 5.
Let's write down a table of score differences, select cells in it, in which the value of the difference will be between 2 and 5:
So, the total number of equally possible elementary outcomes is $ n = 36 $, and the number of favorable outcomes (the number of filled cells in the table above) is $ m = 10 $. Then the probability of the event will be equal to $ P = 10/36 = 5/18 $.
So, in the case when it comes to throwing 2 dice and a simple event, you need to build a table, select the necessary cells in it and divide their number by 36, this will be the probability. In addition to problems for the sum, product and difference of the number of points, there are also problems for the modulus of the difference, the lowest and highest dropped number of points (you can find suitable tables in).
Other problems about dice and cubes
Of course, the above two classes of dice-throwing problems are not limited to the case (it's just that they are most often encountered in problem books and manuals), there are others. For a change and understanding of an approximate solution, we will analyze three more typical examples: for throwing 3 dice, for conditional probability and for Bernoulli's formula.
Example 6. Throw 3 dice. Find the probability that the total is 15 points.
In the case of 3 dice, the tables are made up less often, since they will need as many as 6 pieces (and not one, as above), they get by with a simple enumeration of the necessary combinations.
Let's find the total number of outcomes of the experiment. Outcomes can be represented as ordered triplets of numbers of the form $ (x, y, z) $, where $ x $ is how many points fell on the first die (from 1 to 6), $ y $ is how many points fell on the second die (from 1 to 6), $ z $ - how many points fell on the third die (from 1 to 6). Obviously, the total of such triplets of numbers will be $ n = 6 \ cdot 6 \ cdot 6 = 216 $.
Now let's select such outcomes that add up to 15 points.
$$ (3,6,6), (6,3,6), (6,6,3),\\ (4,5,6), (4,6,5), (5,4,6), (6,5,4), (5,6,4), (6,4,5),\\ (5,5,5). $$
We got $ m = 3 + 6 + 1 = $ 10 outcomes. The desired probability is $ P = 10/216 = 0.046 $.
Example 7. Throw 2 dice. Find the probability that the first die has no more than 4 points, provided that the sum of the points is even.
The easiest way to solve this problem is to use the table again (everything will be clear), as before. We write out a table of the sums of points and select only cells with even values:
We get that according to the condition of the experiment, there are not 36, but $ n = 18 $ outcomes in total (when the sum of the points is even).
Now of these cells select only those that correspond to the event "no more than 4 points fell on the first die" - that is, in fact, the cells in the first 4 rows of the table (highlighted in orange), there will be $ m = 12 $.
The sought probability $ P = 12/18 = 2/3. $
The same task can be decide differently using the conditional probability formula. Let's introduce events:
A = The sum of the number of points is even
B = The first die has a maximum of 4 points
AB = The sum of the number of points is even and the first die has no more than 4 points
Then the formula for the required probability is: $$ P (B | A) = \ frac (P (AB)) (P (A)). $$ Find the probabilities. The total number of outcomes is $ n = 36 $, for event A the number of favorable outcomes (see tables above) $ m (A) = 18 $, and for event AB - $ m (AB) = 12 $. We get: $$ P (A) = \ frac (m (A)) (n) = \ frac (18) (36) = \ frac (1) (2); \ quad P (AB) = \ frac (m (AB)) (n) = \ frac (12) (36) = \ frac (1) (3); \\ P (B | A) = \ frac (P (AB)) (P (A)) = \ frac (1/3) (1/2) = \ frac (2) (3). $$ The answers matched.
Example 8. The dice is rolled 4 times. Find the probability that an even number of points will be drawn exactly 3 times.
In the case when the dice rushes several times, and the event is not about the amount, product, etc. integral characteristics, but only about number of drops of a certain type, we can use to calculate the probability
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Pedagogical technologies: Technology of explanatory-illustrated teaching, computer technology, student-centered approach to learning, health-preserving technologies.
Lesson type: lesson in gaining new knowledge.
Duration: 1 lesson.
Class: Grade 8.
Lesson objectives:
Educational:
- to repeat the skills of using the formula for finding the probability of an event and teach how to use it in problems with dice;
- conduct evidence-based reasoning when solving problems, evaluate the logical correctness of reasoning, recognize logically incorrect reasoning.
Developing:
- develop skills in searching, processing and presenting information;
- develop the ability to compare, analyze, draw conclusions;
- develop observation, as well as communication skills.
Educational:
- cultivate attentiveness, perseverance;
- to form an understanding of the importance of mathematics as a way of knowing the world around.
Lesson equipment: computer, multimedia, markers, mimio copier (or interactive whiteboard), envelope (it contains an assignment for practical work, homework, three cards: yellow, green, red), dice models.
Lesson plan
Organizing time.
In the previous lesson, we got acquainted with the formula for classical probability.
The probability P of the occurrence of a random event A is the ratio of m to n, where n is the number of all possible outcomes of the experiment, and m is the number of all favorable outcomes.
The formula is the so-called classical definition of the Laplace probability, which came from the domain gambling where probability theory was applied to determine the winning prospects. This formula is used for experiments with a finite number of equally possible outcomes.
Probability of event = Number of favorable outcomes / number of all equally possible outcomes
So the probability is a number between 0 and 1.
The probability is 0 if the event is impossible.
The probability is 1 if the event is reliable.
Let's solve the problem orally: There are 20 books on the bookshelf, of which 3 are reference books. What is the likelihood that a book taken off the shelf will not turn out to be a reference?
Solution:
The total number of equally possible outcomes - 20
The number of favorable outcomes - 20 - 3 = 17
Answer: 0.85.
2. Obtaining new knowledge.
And now let's return to the topic of our lesson: “Probabilities of events”, we will sign it in our notebooks.
The purpose of the lesson: to learn how to solve problems of finding the probability when throwing a dice or 2 dice.
Our topic today is related to the dice, or it is also called the dice. The dice has been known since antiquity. The dice game is one of the oldest, the first prototypes of dice were found in Egypt, and they date back to the 20th century BC. NS. There are many varieties, from simple ones (the one who threw large quantity points) to complex ones, in which you can use various tactics of the game.
The oldest bones date back to the 20th century BC. e., found in Thebes. Initially, the bones served as a tool for fortune telling. According to archaeological excavations, dice were played everywhere in all corners of the globe. The name comes from the original material - animal bones.
The ancient Greeks believed that the Lydians invented bones, fleeing hunger, in order to at least occupy their minds with something.
The dice game was reflected in ancient Egyptian, Greco-Roman, Vedic mythology. Mentioned in the Bible, Iliad, Odyssey, Mahabharata, collection of Vedic hymns “Rigveda”. In the pantheons of gods, at least one god was the owner of dice as an integral attribute http://ru.wikipedia.org/wiki/%CA%EE%F1%F2%E8_%28%E8%E3%F0%E0%29 - cite_note-2 .
After the fall of the Roman Empire, the game spread throughout Europe, especially during the Middle Ages. Since dice were used not only for playing, but also for fortune telling, the church repeatedly tried to ban the game, for this purpose the most sophisticated punishments were invented, but all attempts ended in failure.
According to archeological data, dice were also played in pagan Rus. After the baptism, the Orthodox Church tried to eradicate the game, but among the common people it remained popular, in contrast to Europe, where the highest nobility and even the clergy sinned by playing dice.
War declared by the authorities different countries the game of dice has spawned many different cheating tricks.
In the Age of Enlightenment, the hobby for playing dice gradually declined, people developed new hobbies, they became more interested in literature, music and painting. Nowadays dice are not so widespread.
Correct bones provide equal chances of edge falling out. To do this, all faces must be the same: smooth, flat, have the same area, fillets (if any), holes must be drilled to the same depth. The sum of points on opposite sides is 7.
The mathematical die, which is used in probability theory, is the mathematical image of a regular die. Mathematical bone has no size, color, weight, etc.
When throwing playing bones(cubes) any of its six faces can drop out, i.e. happen any of events- dropping out from 1 to 6 points (points). But no two and more faces cannot appear at the same time. Such developments called inconsistent.
Consider the case where 1 die is rolled. Let's execute No. 2 in the form of a table.
Now consider the case where 2 dice are rolled.
If the first die has one point, then the second can drop 1, 2, 3, 4, 5, 6. We get pairs (1; 1), (1; 2), (1; 3), (1; 4) , (1; 5), (1; 6) and so on with each face. All cases can be presented as a table of 6 rows and 6 columns:
Elementary Events Table
You have an envelope on your desk.
Take the assignment sheet from the envelope.
Now you will complete the practice using the elementary events table.
Show by shading the events favorable to the events:
Task 1. "The same number of points dropped out";
| 1; 1 | 2; 1 | 3; 1 | 4; 1 | 5; 1 | 6; 1 |
| 1; 2 | 2; 2 | 3; 2 | 4; 2 | 5; 2 | 6; 2 |
| 1; 3 | 2; 3 | 3; 3 | 4; 3 | 5; 3 | 6; 3 |
| 1; 4 | 2; 4 | 3; 4 | 4; 4 | 5; 4 | 6; 4 |
| 1; 5 | 2; 5 | 3; 5 | 4; 5 | 5; 5 | 6; 5 |
| 1; 6 | 2; 6 | 3; 6 | 4; 6 | 5; 6 | 6; 6 |
Task 2. “The sum of points is equal to 7”;
| 1; 1 | 2; 1 | 3; 1 | 4; 1 | 5; 1 | 6; 1 |
| 1; 2 | 2; 2 | 3; 2 | 4; 2 | 5; 2 | 6; 2 |
| 1; 3 | 2; 3 | 3; 3 | 4; 3 | 5; 3 | 6; 3 |
| 1; 4 | 2; 4 | 3; 4 | 4; 4 | 5; 4 | 6; 4 |
| 1; 5 | 2; 5 | 3; 5 | 4; 5 | 5; 5 | 6; 5 |
| 1; 6 | 2; 6 | 3; 6 | 4; 6 | 5; 6 | 6; 6 |
Task 3. "The sum of points is not less than 7".
What does “no less” mean? (The answer is "greater than, or equal")
| 1; 1 | 2; 1 | 3; 1 | 4; 1 | 5; 1 | 6; 1 |
| 1; 2 | 2; 2 | 3; 2 | 4; 2 | 5; 2 | 6; 2 |
| 1; 3 | 2; 3 | 3; 3 | 4; 3 | 5; 3 | 6; 3 |
| 1; 4 | 2; 4 | 3; 4 | 4; 4 | 5; 4 | 6; 4 |
| 1; 5 | 2; 5 | 3; 5 | 4; 5 | 5; 5 | 6; 5 |
| 1; 6 | 2; 6 | 3; 6 | 4; 6 | 5; 6 | 6; 6 |
And now we will find the probabilities of events for which in practical work shaded favorable events.
Let's write down in notebooks №3
Exercise 1.
The total number of outcomes - 36
Answer: 1/6.
Task 2.
The total number of outcomes - 36
Favorable outcomes - 6
Answer: 1/6.
Task 3.
The total number of outcomes is 36
Favorable outcomes - 21
P = 21/36 = 7/12.
Answer: 7/12.
№4. Sasha and Vlad are playing dice. Each one rolls the dice twice. The winner is the one who has the highest amount of points. If the points are equal, the game ends in a draw. Sasha was the first to throw the dice, and he got 5 points and 3 points. Now Vlad is throwing the dice.
a) In the table of elementary events, indicate (by shading) the elementary events favorable to the event “Vlad wins”.
b) Find the probability of the event “Vlad wins”.
3. Physical education.
If the event is reliable, we all clap together,
If the event is impossible, we all stomp together,
If the event is random - shake our head / left-right
“There are 3 apples in the basket (2 red, 1 green).
3 red ones were pulled out of the basket - (impossible)
A red apple was pulled out of the basket - (random)
A green apple was pulled out of the basket - (random)
They pulled 2 red and 1 green from the basket - (reliable)
Let's solve the next number.
The correct die is rolled twice. Which event is more likely:
A: “Both times got 5 points”;
Q: “The first time I got 2 points, the second 5 points”;
S: “Once dropped 2 points, once 5 points”?
Let's analyze event A: the total number of outcomes is 36, the number of favorable outcomes is 1 (5; 5)
Let's analyze event B: the total number of outcomes is 36, the number of favorable outcomes is 1 (2; 5)
Let's analyze event C: the total number of outcomes is 36, the number of favorable outcomes is 2 (2; 5 and 5; 2)
Answer: event C.
4. Statement of homework.
1. Cut out the reamer, glue the cubes. Bring to next lesson.
2. Perform 25 throws. Record the results in a table: (in the next lesson, you can enter the concept of frequency)
3. Solve the problem: Throw two dice. Calculate the probability:
a) “The sum of points is equal to 6”;
b) “The sum of points is not less than 5”;
c) "There are more points on the first die than on the second."
What is the probability that an even number of points will fall on one die roll?
54. Katya and Anya are writing a dictation. The probability that Katya will make a mistake is 60%, and Anya's probability of a mistake is 40%. Find the probability that both girls will write the dictation without mistakes.
55. The plant produces 15% of products of the highest grade, 25% - of the first grade, 40% - of the second grade, and the rest is defective. Find the probability that the selected product will not be defective.
What is the probability that the baby will be born on the 7th?
57. Each of the three shooters shoots at the target once, with the first shooter hitting 90%, the second 80%, and the third 70%. Find the probability that all three arrows hit the target?
There are 7 white and 9 black balls in the box. Take the ball out at random and return it. Then take out the ball again. What is the probability that both balls are white
What is the likelihood that at least one coat of arms will appear when flipping two coins?
IN toolbox there are 15 standard and 5 defective parts. One piece is taken out of the box at random. Find the probability that this part is standard
The control panel has three independently installed alarm devices. The probability that in the event of an accident the first is triggered is 0.9, the second is 0.7, the third is 0.8. Find the probability that no alarm will go off in case of an accident
62. Nikolai and Leonid perform test... The probability of an error in calculations for Nikolai is 70%, and for Leonid - 30%. Find the probability that Leonidas will make a mistake and Nikolai will not.
63. The School of Music is recruiting students. The likelihood of not being credited during a musical ear test is 40%, and a sense of rhythm is 10%. What is the likelihood of testing positive?
64. Each of the three shooters shoots at the target once, and the probability of hitting 1 shooter is 80%, the second - 70%, the third - 60%. Find the probability that only the second shooter will hit the target.
65. The basket contains fruits, including 30% bananas and 60% apples. What is the likelihood that a fruit chosen at random will be a banana or an apple?
The box contains 4 blue, 3 red, 9 green, 6 yellow balls. What is the probability that the selected ball will not be green?
There are 1000 tickets in the lottery, among which 20 are winning. One ticket is purchased. What is the probability that this ticket is not a winning ticket?
68. There are 6 textbooks, of which 3 are bound. They take 2 textbooks at random. The probability that both textbooks taken will end up in a binding is….
69. The workshop employs 7 men and 3 women. 3 people are chosen at random by personnel numbers. The probability that all those selected will be men is….
70. There are 10 balls in the box, 6 of which are colored. 4 balls are removed at random without returning them. The probability that all balls taken out will be colored is….
71. There are 4 red and 2 blue balls in the box. Three balls are taken from it at random. The probability that all these three balls are red is….
72. The student knows 20 questions out of 25 discipline questions. He is offered 3 questions. The probability that the student knows them is….
73. There are 4 white and 3 black balls in the urn. Take out two balls at the same time. The probability that both balls are white is….
74. Roll 3 dice at once. The probability that 3 sixes will be dropped is….
The local doctor received 35 patients within a week, of which five patients were diagnosed with a stomach ulcer. Determine the relative frequency of an appointment with a patient with stomach disease.