In probability theory, there is a group of problems for which it is enough to know the classical definition of probability and visually represent the proposed situation. Such problems include most coin toss problems and dice rolling problems. Let us recall the classical definition of probability.

Probability of event A (the objective possibility of an event occurring in numerical terms) is equal to the ratio of the number of outcomes favorable to this event to the total number of all equally possible incompatible elementary outcomes: P(A)=m/n, Where:

• m is the number of elementary test outcomes favorable to the occurrence of event A;
• n is the total number of all possible elementary test outcomes.

It is convenient to determine the number of possible elementary test outcomes and the number of favorable outcomes in the problems under consideration by enumerating all possible options (combinations) and direct counting.

From the table we see that the number of possible elementary outcomes is n=4. Favorable outcomes of the event A = (heads appear 1 time) correspond to option No. 2 and No. 3 of the experiment, there are two such options m = 2.
Find the probability of the event P(A)=m/n=2/4=0.5

Problem 2 . In a random experiment, a symmetrical coin is tossed twice. Find the probability that you will get no heads at all.

Solution . Since the coin is tossed twice, then, as in problem 1, the number of possible elementary outcomes is n=4. Favorable outcomes of the event A = (heads will not appear even once) correspond to option No. 4 of the experiment (see the table in problem 1). There is only one such option, which means m=1.
Find the probability of the event P(A)=m/n=1/4=0.25

Problem 3 . In a random experiment, a symmetrical coin is tossed three times. Find the probability that heads will appear exactly 2 times.

Solution . We present the possible options for three coin tosses (all possible combinations of heads and tails) in the form of a table:

From the table we see that the number of possible elementary outcomes is n=8. Favorable outcomes of the event A = (heads appear 2 times) correspond to options No. 5, 6 and 7 of the experiment. There are three such options, which means m=3.
Find the probability of the event P(A)=m/n=3/8=0.375

Problem 4 . In a random experiment, a symmetrical coin is tossed four times. Find the probability of getting heads exactly 3 times.

Solution . We present the possible options for four coin tosses (all possible combinations of heads and tails) in the form of a table:

From the table we see that the number of possible elementary outcomes is n=16. Favorable outcomes of the event A = (heads will appear 3 times) correspond to options No. 12, 13, 14 and 15 of the experiment, which means m = 4.
Find the probability of the event P(A)=m/n=4/16=0.25

Determining Probability in Dice Problems

Problem 5 . Determine the probability that when throwing a dice (a fair dice) you will get more than 3 points.

Solution . When throwing a die (a regular dice), any of its six faces can fall out, i.e. any of the elementary events occur - the loss of 1 to 6 dots (points). This means the number of possible elementary outcomes is n=6.
Event A = (more than 3 points rolled) means that 4, 5 or 6 points (points) rolled. This means the number of favorable outcomes is m=3.
Probability of event P(A)=m/n=3/6=0.5

Problem 6 . Determine the probability that when throwing a dice you get a number of points no greater than 4. Round the result to the nearest thousandth.

Solution . When throwing a die, any of its six faces can fall out, i.e. any of the elementary events occur - the loss of 1 to 6 dots (points). This means the number of possible elementary outcomes is n=6.
Event A = (no more than 4 points rolled) means that 4, 3, 2 or 1 point (point) was rolled. This means the number of favorable outcomes is m=4.
Probability of event Р(А)=m/n=4/6=0.6666…≈0.667

Problem 7 . The dice are thrown twice. Find the probability that the number rolled is less than 4 both times.

Solution . Since the dice (dice) are thrown twice, we will reason as follows: if one point is rolled up on the first die, then 1, 2, 3, 4, 5, 6 can come up on the second die. We get the pairs (1;1), (1;2), (1;3), (1;4), (1;5), (1;6) and so on with each face. We present all cases in the form of a table of 6 rows and 6 columns:

We calculate the favorable outcomes of the event A = (both times the number was less than 4) (they are highlighted in bold) and we get m=9.
Find the probability of the event P(A)=m/n=9/36=0.25

Problem 8 . The dice are thrown twice. Find the probability that the larger of the two numbers drawn is 5. Round your answer to the nearest thousand.

Solution . We present all possible outcomes of two dice throws in the table:

From the table we see that the number of possible elementary outcomes is n=6*6=36.
We calculate the favorable outcomes of the event A = (the largest of the two numbers drawn is 5) (they are highlighted in bold) and get m=8.
Find the probability of the event P(A)=m/n=8/36=0.2222…≈0.222

Problem 9 . The dice are thrown twice. Find the probability that a number less than 4 is rolled at least once.

Solution . We present all possible outcomes of two dice throws in the table:

From the table we see that the number of possible elementary outcomes is n=6*6=36.
The phrase “at least once a number less than 4 came up” means “a number less than 4 came up once or twice”, then the number of favorable outcomes of the event A = (at least once a number less than 4 came up) (they are highlighted in bold) m=27.
Find the probability of the event P(A)=m/n=27/36=0.75

Description of the presentation by individual slides:

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Solving problems in probability theory. Mathematics teacher MBOU Nivnyanskaya secondary school, Nechaeva Tamara Ivanovna

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Lesson objectives: consider different types problems in probability theory and methods for solving them. Objectives of the lesson: to teach students to recognize different types of problems in probability theory and improve the logical thinking of schoolchildren.

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Problem 1. In a random experiment, a symmetrical coin is tossed 2 times. Find the probability that you get the same number of heads and tails.

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Problem 2. A coin is tossed four times. Find the probability that you will never get heads.

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Problem 3. In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear exactly once. Solution: In order to find the probability of a specified event, it is necessary to consider all possible outcomes of the experiment, and then select favorable outcomes from them (favorable outcomes are outcomes that satisfy the requirements of the problem). In our case, favorable outcomes will be those in which, with two tosses of a symmetrical coin, heads come up only once. The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of outcomes. Therefore, the probability that when throwing a symmetrical coin twice, heads will appear only once is equal to: P = 2/4 = 0.5 = 50% Answer: the probability that, as a result of the above experiment, heads will appear only once is 50 %. Experiment number 1st throw 2nd throw Number of times heads 1 Heads Heads 2 2 Tails Tails 0 3 Heads Tails 1 4 Tails Heads 1

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Task 4. Dice thrown once. What is the probability that the number of points rolled is greater than 4. Solution: Random experiment - throwing a die. The elementary event is the number on the dropped side. Answer: 1/3 Total faces: 1, 2, 3, 4, 5, 6 Elementary events: N=6 N(A)=2

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Problem 5. A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hits the targets the first three times and misses the last two times. Round the result to hundredths. Solution: Probability of hit = 0.8 Probability of miss = 1 - 0.8 = 0.2 A = (hit, hit, hit, missed, missed) According to the formula for multiplying probabilities P(A) = 0.8 ∙ 0.8 ∙ 0.8 ∙ 0.2 ∙ 0.2 P(A) = 0.512 ∙ 0.04 = 0.02048 ≈ 0.02 Answer: 0.02

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Problem 6. In a random experiment, two dice are thrown. Find the probability that the sum of the points drawn is 6. Round the answer to the nearest hundredth Solution: The elementary 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 represent the set of elementary outcomes in a table. The rows correspond to the number of points on the first die, the columns on the second die. There are n = 36 elementary events in total. Let us write in each cell the sum of the points drawn 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 elementary outcomes. Therefore, m = 5. Therefore, P(A) = 5/36 = 0.14. Answer: 0.14. 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12

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Probability formula Theorem Let a coin be tossed n times. Then the probability that heads will appear exactly k times can be found by the formula: Where Cnk is the number of combinations of n elements in k, which is calculated by the formula:

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Problem 7. A coin is tossed four times. Find the probability of getting heads exactly three times. Solution According to the conditions of the problem, there were n = 4 throws in total. Required number of eagles: k =3. We substitute n and k into the formula: With the same success, we can count the number of heads: k = 4 − 3 = 1. The answer will be the same. Answer: 0.25

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Problem 8. A coin is tossed three times. Find the probability that you will never get heads. Solution We write down the numbers n and k again. Since the coin is tossed 3 times, n = 3. And since there should be no heads, k = 0. It remains to substitute the numbers n and k into the formula: Let me remind you that 0! = 1 by definition. Therefore C30 = 1. Answer: 0.125

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Problem 9. In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will appear more times than tails. Solution: For there to be more heads than tails, they must appear either 3 times (then there will be 1 tails) or 4 times (then there will be no tails at all). Let's find the probability of each of these events. Let p1 be the probability of getting heads 3 times. Then n = 4, k = 3. We have: Now let’s find p2 - the probability that heads will land all 4 times. In this case, n = 4, k = 4. We have: To get the answer, it remains to add the probabilities p1 and p2. Remember: you can only add probabilities for mutually exclusive events. We have: p = p1 + p2 = 0.25 + 0.0625 = 0.3125 Answer: 0.3125

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Problem 10. Before the start of a volleyball match, team captains draw fair lots to determine which team will start the game with the ball. The “Stator” team takes turns playing with the “Rotor”, “Motor” and “Starter” teams. Find the probability that Stator will start only the first and last games. Solution. You need to find the probability of three events happening: “Stator” starts the first game, does not start the second game, and starts the third game. The probability of a product of independent events is equal to the product of the probabilities of these events. The probability of each of them is 0.5, from which we find: 0.5·0.5·0.5 = 0.125. Answer: 0.125.

In a random experiment, a symmetrical coin is tossed...

As a preface.
Everyone knows that a coin has two sides - heads and tails.
Numismatists believe that a coin has three sides - obverse, reverse and edge.
Both among those and among others, few people know what a symmetrical coin is. But those who are preparing to take the Unified State Exam know about this (well, or should know:).

In general, this article will talk about unusual coin, which has nothing to do with numismatics, but at the same time is the most popular coin among schoolchildren.

So.
Symmetrical coin- this is an imaginary mathematically ideal coin without size, weight, diameter, etc. As a result, such a coin also does not have an edge, that is, it really only has two sides. The main property of a symmetrical coin is that under such conditions the probability of heads or tails appearing is absolutely the same. And they came up with a symmetrical coin to conduct thought experiments.
The most popular symmetrical coin problem is: “In a random experiment, a symmetrical coin is tossed twice (three times, four times, etc.). The problem is to determine the probability that one side will land a certain number of times.

Solving the problem with a symmetrical coin

It is clear that as a result of a toss, the coin will land on either heads or tails. How many times depends on how many throws to make. The probability of getting heads or tails is calculated by dividing the number of outcomes that satisfy the condition by the total number of possible outcomes.

One throw

Everything is simple here. It will either be heads or tails. Those. we have two possible outcomes, one of which satisfies us - 1/2=50%

Two-throw

In two throws you can get:
two eagles
two heads
heads then tails
tails, then heads
Those. There are only four possible options. Problems with more than one roll are most easily solved by drawing up a table of possible options. For simplicity, let's denote heads as "0" and tails as "1". Then the table of possible outcomes will look like this:
00
01
10
11
If, for example, you need to find the probability that heads will appear once, you simply need to count the number of suitable options in the table - i.e. those lines where the eagle appears once. There are two such lines. This means that the probability of getting one head in two tosses of a symmetrical coin is 2/4 = 50%
The probability that heads will appear twice in two throws is 1/4=25%

Three roska

Let's create a table of options:
000
001
010
011
100
101
110
111
Those familiar with binary calculus understand what we've come to. :) Yes, these are binary digits from "0" to "7". This makes it easier not to get confused with the options.
Let's solve the problem from the previous paragraph - calculate the probability that heads will appear once. There are three lines where "0" appears once. This means that the probability of getting one head in three tosses of a symmetrical coin is 3/8 = 37.5%
The probability that heads will appear twice in three throws is 3/8 = 37.5%, i.e. absolutely the same.
The probability that heads will appear three times in three throws is 1/8 = 12.5%.

Four throws

Let's create a table of options:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
The probability that heads will appear once. There are only three lines where "0" appears once, just like in the case of three throws. But there are already sixteen options. This means that the probability of getting one head in four tosses of a symmetrical coin is 3/16 = 18.75%
The probability that heads will appear twice in three throws is 6/8 = 75%.
The probability that heads will appear three times in three throws is 4/8 = 50%.

So, with an increase in the number of throws, the principle of solving the problem does not change at all - only, in a corresponding progression, the number of options increases.

In the problems on probability theory, which are presented in the Unified State Exam number 4, in addition to, there are problems on tossing a coin and throwing a dice. We'll look at them today.

Coin toss problems

Task 1. A symmetrical coin is tossed twice. Find the probability that heads will appear exactly once.

In such problems, it is convenient to write down all possible outcomes, writing them using the letters P (tails) and O (heads). So, the outcome of the OP means that on the first throw it came up heads, and on the second throw it came up tails. In the problem under consideration, there are 4 possible outcomes: RR, RO, OR, OO. The event “tails will appear exactly once” is favored by 2 outcomes: RO and OP. The required probability is equal to .

Answer: 0.5.

Task 2. A symmetrical coin is tossed three times. Find the probability that it lands on heads exactly twice.

There are 8 possible outcomes in total: RRR, RRO, ROR, ROO, ORR, ORO, OOR, OOO. The event “heads will appear exactly twice” is favored by 3 outcomes: ROO, ORO, OOR. The required probability is equal to .

Answer: 0.375.

Task 3. Before the start of a football match, the referee flips a coin to determine which team will start with the ball. The Emerald team plays three matches with different teams. Find the probability that in these games “Emerald” will win the lot exactly once.

This task is similar to the previous one. Let each time landing heads mean winning the lot with the “Emerald” (this assumption does not affect the calculation of probabilities). Then 8 outcomes are possible: RRR, RRO, ROR, ROO, ORR, ORO, OOR, OOO. The event “tails will appear exactly once” is favored by 3 outcomes: ROO, ORO, OOR. The required probability is equal to .

Answer: 0.375.

Problem 4. A symmetrical coin is tossed three times. Find the probability that the ROO outcome will occur (the first time it lands heads, the second and third times it lands heads).

As in previous tasks, there are 8 outcomes: RRR, RRO, ROR, ROO, ORR, ORO, OOR, OOO. The probability of the ROO outcome occurring is equal to .

Answer: 0.125.

Dice rolling problems

Task 5. The dice are thrown twice. How many elementary outcomes of the experiment favor the event “the sum of points is 8”?

Problem 6. Two dice are thrown at the same time. Find the probability that the total will be 4 points. Round the result to hundredths.

In general, when dice are thrown, there are equally possible outcomes. The same number of outcomes is obtained if the same die is rolled several times in a row.

The event “the total number is 4” is favored by the following outcomes: 1 – 3, 2 – 2, 3 – 1. Their number is 3. The required probability is .

To calculate the approximate value of a fraction, it is convenient to use angle division. Thus, approximately equal to 0.083..., rounded to the nearest hundredth we have 0.08.

Answer: 0.08

Problem 7. Three dice are thrown at the same time. Find the probability that the total will be 5 points. Round the result to hundredths.

The outcome will be considered to be three numbers: the points rolled on the first, second and third dice. There are all equally possible outcomes. The following outcomes are favorable for the “total 5” event: 1–1–3, 1–3–1, 3–1–1, 1–2–2, 2–1–2, 2–2–1. Their number is 6. The required probability is . To calculate the approximate value of a fraction, it is convenient to use angle division. Approximately we get 0.027..., rounding to hundredths, we have 0.03. Source “Preparation for the Unified State Exam. Mathematics. Probability theory". Edited by F.F. Lysenko, S.Yu. Kulabukhova