Problem 19 ( OGE - 2015, Yashchenko I.V.)
Olya, Denis, Vitya, Arthur and Rita cast lots as to who should start the game. Find the probability that Rita should start the game.
Solution
A total of 5 people can start the game.
Answer: 0.2.
Problem 19 ( OGE - 2015, Yashchenko I.V.)
Misha had four candies in his pocket - "Grillage", "Mask", "Squirrel" and "Little Red Riding Hood", as well as the keys to the apartment. While taking out the keys, Misha accidentally dropped one piece of candy. Find the probability that the Mask candy is lost.
Solution
There are 4 options in total.
The probability that Misha dropped the Mask candy is equal to
Answer: 0.25.
Problem 19 ( OGE - 2015, Yashchenko I.V.)
The dice (dice) are thrown once. What is the probability that the number rolled is no less than 3?
Solution
There are a total of 6 different options for scoring points on a die.
The number of points, not less than 3, can be: 3,4,5,6 - that is, 4 options.
This means the probability is P = 4/6 = 2/3.
Answer: 2/3.
Problem 19 ( OGE - 2015, Yashchenko I.V.)
The grandmother decided to give her grandson Ilyusha some randomly selected fruit for the trip. She had 3 green apples, 3 green pears and 2 yellow bananas. Find the probability that Ilya will receive a green fruit from his grandmother.
Solution
3+3+2 = 8 - total fruits. Of these, 6 are green (3 apples and 3 pears).
Then the probability that Ilya will receive a green fruit from his grandmother is equal to
P = 6/8 = 3/4 = 0.75.
Answer: 0.75.
Problem 19 ( OGE - 2015, Yashchenko I.V.)
The dice are thrown twice. Find the probability that a number greater than 3 is rolled both times.
Solution
6*6 = 36 - total number of possible numbers when throwing two dice.
The options that suit us are:
There are 9 such options in total.
This means that the probability that a number greater than 3 is rolled both times is equal to
P = 9/36 = 1/4 = 0.25.
Answer: 0.25.
Problem 19 ( OGE - 2015, Yashchenko I.V.)
The dice (dice) are thrown 2 times. Find the probability that one time a number greater than 3 is rolled, and another time a number less than 3 appears.
Solution
Total options: 6*6 = 36.
The following outcomes suit us:
Explain the principle of solving the problem. The dice were thrown once. What is the probability of rolling less than 4 points? and got the best answer
Answer from Divergent[guru]
50 percent
The principle is extremely simple. Total outcomes 6: 1,2,3,4,5,6
Of these, three satisfy the condition: 1,2,3, and three do not: 4,5,6. Therefore the probability is 3/6=1/2=0.5=50%
Answer from I am superman[guru]
There may be six options in total (1,2,3,4,5,6)
And of these options 1, 2, and 3 are less than four
So 3 answers out of 6
To calculate the probability, we divide the favorable distribution to everything, i.e. 3 by 6 = 0.5 or 50%
Answer from Oriy Dovbysh[active]
50%
divide 100% by the number of numbers on the dice,
and then multiply the percentage received by the amount that you need to find out, that is, by 3)
Answer from Ivan Panin[guru]
I don’t know for sure, I’m preparing for the GIA, but the teacher told me something today, only about the probability of cars, since I understood that the ratio is shown as a fraction, at the top the number is favorable, and at the bottom, in my opinion, it’s generally general, well, we had something like that about cars : The taxi company currently has 3 black, 3 yellow and 14 green cars available. One of the cars drove out to the customer. Find the probability that a yellow taxi will come to him. So, there are 3 yellow taxis and out of the total number of cars there are 3 of them, it turns out that we write 3 on top of the fraction, since this is a favorable number of cars, and on the bottom we write 20, since there are 20 cars in total in the taxi fleet, so we get the probability 3 to 20 or 3/20 as a fraction, well, that’s how I understood it.... I don’t know exactly how to deal with bones, but maybe it helped in some way...
Answer from 3 answers[guru]
Hello! Here is a selection of topics with answers to your question: Explain the principle of solving the problem. The dice were thrown once. What is the probability of rolling less than 4 points?
Tasks for probability of dice 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 equal to 10, or the number of points will be 4, or the product of the number of points, or the product of the number of points divided by 2 etc.
The application 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 dice. 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 the experiment with throwing a bone or cube.
Problem 1. The dice are thrown once. What is the probability of getting an even number of points?
Since the die is a cube (or it is also called a regular die, the die will land on all sides with equal probability, since it is balanced), the die has 6 sides (the number of points from 1 to 6, which are usually indicated by dots), this means that the problem has a total number of outcomes: n=6. The event is favored only by outcomes in which the side with even points 2,4 and 6 appears; the die has the following sides: m=3. Now we can determine the desired probability of the dice: P=3/6=1/2=0.5.
Task 2. The dice are thrown once. What is the probability that you will get at least 5 points?
This problem is solved by analogy with the example given above. When throwing a dice, the total number of equally possible outcomes is: n=6, and only 2 outcomes satisfy the condition of the problem (at least 5 points rolled up, that is, 5 or 6 points rolled out), which means m=2. Next, we find the required probability: P=2/6=1/3=0.333.
Two dice, probability.
When solving problems involving throwing 2 dice, it is very convenient to use a special scoring table. On it, the number of points that fell on the first dice is displayed horizontally, and the number of points that fell on the second dice is displayed vertically. 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 problem is about the sum of points, then the sum is written there, and if it’s about the difference, then the difference is written down, 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 die - 6 outcomes of the experiment. But when there are already two dice, the possible outcomes can be represented as ordered pairs of numbers of the form (x, y), where x shows how many points were rolled on the first dice (from 1 to 6), and y - how many points were rolled on the second dice (from 1 until 6). There will be a total of such number pairs: n=6*6=36 (in the table of outcomes they correspond exactly to 36 cells).
Now you can fill out the table; to do this, the number of points that fell on the first and second dice is entered in each cell. The completed table looks like this:
Using the table, we will determine the number of outcomes that favor the event “a total of less than 5 points will appear.” Let's count the number of cells in which the sum value will be less than the number 5 (these are 2, 3 and 4). For convenience, we paint over such cells; there will be m=6 of them:
Considering the table data, probability of dice 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 highlight the 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 problem) and the number of favorable outcomes (the number of cells that are shaded in the table) m=20. The probability of the event is: P=20/36=5/9.
Problem 5. The dice are thrown twice. What is the probability that the difference in the number of points on the first and second dice will be from 2 to 5?
To determine probability of dice Let's write down a table of point differences and select in it those cells whose difference value will be between 2 and 5:
The number of favorable outcomes (the number of cells shaded in the table) is 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.
Lesson objectives:
Students should know:
- determining the probability of a random event;
- be able to solve problems to find the probability of a random event;
- be able to apply theoretical knowledge in practice.
Lesson objectives:
Educational: create conditions for students to master a system of knowledge, skills and abilities with the concepts of probability of an event.
Educational: to form a scientific worldview in students
Developmental: to develop students’ cognitive interest, creativity, will, memory, speech, attention, imagination, perception.
Methods of organizing educational and cognitive activities:
- visual,
- practical,
- by mental activity: inductive,
- according to the assimilation of material: partially search, reproductive,
- by degree of independence: independent work,
- stimulating: encouragement,
- types of control: checking independently solved problems.
Lesson Plan
- Oral exercises
- Learning new material
- Solving tasks.
- Independent work.
- Summing up the lesson.
- Commenting on homework.
Equipment: multimedia projector (presentation), cards ( independent work)
During the classes
I. Organizational moment.
Organization of the class throughout the lesson, students' readiness for the lesson, order and discipline.
Setting learning goals for students, both for the entire lesson and for its individual stages.
Determine the significance of the material being studied, both in this topic and in the entire course.
II. Repetition
1. What is probability?
Probability is the possibility of something happening or being feasible.
2. What definition is given by the founder of modern probability theory A.N. Kolmogorov?
Mathematical probability is a numerical characteristic of the degree of possibility of the occurrence of a certain event in certain certain conditions that can be repeated an unlimited number of times.
3. What is the classic definition of probability given by the authors of school textbooks?
The probability P(A) of event A in a trial with equally possible elementary outcomes is the ratio of the number of outcomes m favorable to event A to the number n of all outcomes of the trial.
Conclusion: in mathematics, probability is measured by number.
Today we will continue to consider the mathematical model of “dice”.
The subject of research in probability theory is events that appear under certain conditions and that can be reproduced an unlimited number of times. Each occurrence of these conditions is called a test.
The test is throwing a die.
Event – rolling a six or rolling an even number of points.
When rolling a die multiple times, each side has the same probability (the die is fair).
III. Oral problem solving.
1. The dice (dice) were thrown once. What is the probability that a 4 is rolled?
Solution. A random experiment is throwing a die. Event – a number on the dropped side. There are only six faces. Let's list all the events: 1, 2, 3, 4, 5, 6. So P= 6. Event A = (4 points rolled) is favored by one event: 4. Therefore T= 1. Events are equally possible, since it is assumed that the die is fair. Therefore P(A) = t/n= 1/6 = 0,17.
2. The dice (dice) were thrown once. What is the probability that no more than 4 points are rolled?
P= 6. Event A = (no more than 4 points rolled) is favored by 4 events: 1, 2, 3, 4. Therefore T= 4. Therefore P(A) = t/n= 4/6 = 0,67.
3. The dice (dice) were thrown once. What is the probability of rolling less than 4 points?
Solution. A random experiment is throwing a die. Event – a number on the dropped side. Means P= 6. Event A = (less than 4 points rolled) is favored by 3 events: 1, 2, 3. Therefore T= 3. P(A) = t/n= 3/6 = 0,5.
4. The dice (dice) were thrown once. What is the probability that an odd number of points is rolled?
Solution. A random experiment is throwing a die. Event – a number on the dropped side. Means P= 6. Event A = (an odd number of points is rolled) is favored by 3 events: 1,3,5. That's why T= 3. P(A) = t/n= 3/6 = 0,5.
IV. Learning new things
Today we will consider problems when in a random experiment two dice are used or two or three throws are performed.
1. In a random experiment, two dice are rolled. Find the probability that the sum of the points drawn is 6. Round the answer to the nearest hundredth .
Solution. The outcome in this experiment is an ordered pair of numbers. The first number will appear on the first die, the second on the second. It is convenient to present a set of outcomes in a table.
The rows correspond to the number of points on the first die, the columns - on the second die. Total elementary events P= 36.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Let's write the sum of the rolled points in each cell and color in the cells where the sum is 6.
There are 5 such cells. This means that the event A = (the sum of the points drawn is 6) is favored by 5 outcomes. Hence, T= 5. Therefore, P(A) = 5/36 = 0.14.
2. In a random experiment, two dice are rolled. Find the probability that the total will be 3 points. Round the result to hundredths .
P= 36.
Event A = (sum equals 3) is favored by 2 outcomes. Hence, T= 2.
Therefore, P(A) = 2/36 = 0.06.
3. In a random experiment, two dice are rolled. Find the probability that the total will be more than 10 points. Round the result to hundredths .
Solution. The outcome in this experiment is an ordered pair of numbers. Total events P= 36.
Event A = (a total of more than 10 points will be rolled) is favored by 3 outcomes.
Hence, T
4. Lyuba throws the dice twice. In total, she scored 9 points. Find the probability that one of the throws results in 5 points .
Solution The outcome in this experiment is an ordered pair of numbers. The first number will appear on the first throw, the second on the second. It is convenient to present a set of outcomes in a table.
The rows correspond to the result of the first throw, the columns - the result of the second throw.
Total events for which the total score is 9 P= 4. Event A = (one of the throws resulted in 5 points) is favored by 2 outcomes. Hence, T= 2.
Therefore, P(A) = 2/4 = 0.5.
5. Sveta throws the dice twice. In total, she scored 6 points. Find the probability that one of the throws results in 1 point.
First throw |
Second toss |
Sum of points |
||
There are 5 equally possible outcomes.
The probability of the event is p = 2/5 = 0.4.
6. Olya throws the dice twice. She got a total of 5 points. Find the probability that on the first roll you get 3 points.
First throw |
Second toss |
Sum of points |
||
| + | = | |||
| + | = | |||
| + | = | |||
| + | = |
There are 4 equally possible outcomes.
Favorable outcomes – 1.
Probability of event R= 1/4 = 0,25.
7. Natasha and Vitya are playing dice. They roll the dice once.
The one who throws more points wins. If the points are equal, then there is a draw. There are 8 points in total. Find the probability that Natasha won.
Sum of points |
||||
| + | = | |||
| + | = | |||
| + | = | |||
| + | = | |||
| + | = |
There are 5 equally possible outcomes.
Favorable outcomes – 2.
Probability of event R= 2/5 = 0,4.
8. Tanya and Natasha are playing dice. They roll the dice once. The one who throws more points wins. If the points are equal, then there is a draw. A total of 6 points were rolled. Find the probability that Tanya lost.
| Tanya | Natasha | Sum of points | ||
| + | = | |||
| + | = | |||
| + | = | |||
| + | = | |||
| + | = |
There are 5 equally possible outcomes.
Favorable outcomes – 2.
Probability of event R= 2/5 = 0,4.
9. Kolya and Lena are playing dice. They roll the dice once. The one who throws more points wins. If the points are equal, then there is a draw. Kolya was the first to throw, and he got 3 points. Find the probability that Lena does not win.
Kolya got 3 points.
Lena has 6 equally possible outcomes.
There are 3 favorable outcomes for losing (at 1 and at 2 and at 3).
Probability of event R= 3/6 = 0,5.
10. Masha throws the dice three times. What is the probability of getting even numbers all three times?
Masha has 6 6 6 = 216 equally possible outcomes.
There are 3 · 3 · 3 = 27 favorable outcomes for losing.
Probability of event R= 27/216 = 1/8 = 0,125.
11. In a random experiment, three dice are rolled. Find the probability that the total will be 16 points. Round the result to hundredths.
Solution.
| Second | Third | Sum of points | ||||
| + | + | = | ||||
| + | + | = | ||||
| + | + | = | ||||
| + | + | = | ||||
| + | + | = | ||||
| + | + | = |
Equally possible outcomes – 6 6 6 = 216.
Favorable outcomes – 6.
Probability of event R= 6/216 = 1/36 = 0.277... = 0.28. Hence, T= 3. Therefore, P (A) = 3/36 = 0.08.
V. Independent work.
Option 1.
- The dice (dice) are thrown once. What is the probability that you rolled at least 4 points? (Answer:0.5)
- In a random experiment, two dice are rolled. Find the probability that the total will be 5 points. Round the result to hundredths. (Answer: 0.11)
- Anya rolls the dice twice. She got a total of 3 points. Find the probability that on the first roll you get 1 point. (Answer:0.5)
- Katya and Ira are playing dice. They roll the dice once. The one who throws more points wins. If the points are equal, then there is a draw. The total is 9 points. Find the probability that Ira lost. (Answer:0.5)
- In a random experiment, three dice are rolled. Find the probability that the total will be 15 points. Round the result to hundredths. (Answer:0.05)
Option 2.
- The dice (dice) are thrown once. What is the probability that no more than 3 points are rolled? (Answer:0.5)
- In a random experiment, two dice are rolled. Find the probability that the total will be 10 points. Round the result to hundredths. (Answer:0.08)
- Zhenya throws the dice twice. She got a total of 5 points. Find the probability that on the first roll you get 2 points. (Answer:0.25)
- Masha and Dasha are playing dice. They roll the dice once. The one who throws more points wins. If the points are equal, then there is a draw. There were 11 points in total. Find the probability that Masha won. (Answer:0.5)
- In a random experiment, three dice are rolled. Find the probability that the total will be 17 points. Round the result
VI. Homework
- In a random experiment, three dice are rolled. There are 12 points in total. Find the probability that on the first roll you get 5 points. Round the result to the nearest hundredth.
- Katya throws the dice three times. What is the probability that the same numbers will come up all three times?
VII. Lesson summary
What do you need to know to find the probability of a random event?
To calculate classical probability, you need to know all possible outcomes of an event and favorable outcomes.
The classical definition of probability is applicable only to events with equally likely outcomes, which limits its scope.
Why do we study probability theory at school?
Many phenomena in the world around us can only be described using probability theory.
Literature
- Algebra and the beginnings of mathematical analysis. Grades 10-11: textbook. for general educational institutions: basic level / [Sh.A. Alimov, Yu.M. Kolyagin, M.V. Tkacheva, etc.]. – 16th ed., revised. – M.: Education, 2010. – 464 p.
- Semenov A.L. Unified State Exam: 3000 problems with answers in mathematics. All tasks of group B / – 3rd ed., revised. and additional – M.: Publishing house “Exam”, 2012. – 543 p.
- Vysotsky I.R., Yashchenko I.V. Unified State Exam 2012. Mathematics. Problem B10. Probability theory. Workbook/Ed. A.L. Semenov and I.V. Yashchenko. – M.: MCSHMO, 2012. – 48 p.