At random, a symmetrical coin is tossed twice. Mathematics and us. Combination enumeration method
Task Formulation: In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads (tails) will not fall out even once (it will fall out exactly / at least 1, 2 times).
The task is included in the USE in mathematics of the basic level for grade 11 at number 10 (Classical definition of probability).
Let's see how such problems are solved with examples.
Task 1 example:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads never come up.
OO OR RO RR
There are 4 such combinations in total. We are only interested in those of them in which there is not a single eagle. There is only one such combination (PP).
P = 1 / 4 = 0.25
Answer: 0.25
Task 2 example:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that it comes up heads exactly twice.
Consider all the possible combinations that can fall out if the coin is tossed twice. For convenience, we will denote the eagle with the letter O, and tails with the letter P:
OO OR RO RR
There are 4 such combinations in total. We are only interested in those combinations in which the heads appear exactly 2 times. There is only one such combination (OO).
P = 1 / 4 = 0.25
Answer: 0.25
Task 3 example:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that it comes up heads exactly once.
Consider all the possible combinations that can fall out if the coin is tossed twice. For convenience, we will denote the eagle with the letter O, and tails with the letter P:
OO OR RO RR
In total, there are 4 such combinations. We are only interested in those of them in which heads fell out exactly 1 time. There are only two such combinations (OP and RO).
Answer: 0.5
Task 4 example:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will come up at least once.
Consider all the possible combinations that can fall out if the coin is tossed twice. For convenience, we will denote the eagle with the letter O, and tails with the letter P:
OO OR RO RR
There are 4 such combinations in total. We are only interested in those combinations in which the heads fall out at least once. There are only three such combinations (OO, OR and RO).
P = 3 / 4 = 0.75
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 the coin has three sides - obverse, reverse and edge.
And among those, and among others, few people know what a symmetrical coin is. But they know about it (well, or should know :), those who are preparing to take the exam.
In general, this article will focus on an 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 has no edge, that is, it really has only two sides. The main property of a symmetrical coin is that under such conditions the probability of falling heads or tails is exactly the same. And they came up with a symmetrical coin for thought experiments.
The most popular problem with a symmetrical coin sounds like this - "In a random experiment, a symmetrical coin is tossed twice (three times, four times, etc.). It is required to determine the probability that one of the sides will fall out a certain number of times.
Solving the problem with a symmetrical coin
It is clear that as a result of the toss, the coin will fall 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. Either heads or tails will come up. Those. we have two possible outcomes, one of which satisfies us - 1/2=50%
Twothrow
For two throws can fall:
two eagles
two tails
heads, then tails
tails, then heads
Those. only four options are possible. Problems with more than one throw are easiest to solve by making 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 fall once, you just need to count the number of suitable options in the table - i.e. those lines where the eagle occurs once. There are two such lines. So the probability of getting one heads in two tosses of a symmetrical coin is 2/4=50%
The probability of getting heads twice in two tosses is 1/4=25%
Three roses
We make a table of options:
000
001
010
011
100
101
110
111
Those who are familiar with binary calculus understand what we have come to. :) Yes, they are binary numbers from "0" to "7". This way it's easier not to get confused with the options.
Let's solve the problem from the previous paragraph - we calculate the probability that the eagle will fall out once. There are three lines where "0" occurs once. So the probability of getting one heads in three tosses of a symmetrical coin is 3/8=37.5%
The probability that heads in three throws will fall out twice is 3/8=37.5%, i.e. absolutely the same.
The probability that the head in three throws will fall out three times is 1/8 = 12.5%.
Four throws
We make a table of options:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
The probability that heads comes up once. There are only three rows where "0" occurs once, just as in the case of three throws. But, there are already sixteen options. So the probability of getting one heads in four tosses of a symmetrical coin is 3/16=18.75%
The probability that the eagle will fall out twice in three throws is 6/8=75%,.
The probability that heads will come up three times in three tosses 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 an appropriate progression, the number of options increases.
Description of the presentation on 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: Review different types problems in probability theory and methods for their solution. Lesson objectives: to teach to recognize various types of problems in probability theory and improve logical thinking schoolchildren.
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Task 1. In a random experiment, a symmetrical coin is tossed 2 times. Find the probability of getting the same number of heads and tails.
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Task 2. A coin is tossed four times. Find the probability that it never comes up tails.
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Problem 3. In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads come up 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 choose favorable outcomes from them (favorable outcomes are outcomes that meet the requirements of the problem). In our case, those outcomes will be favorable in which, with two tosses of a symmetrical coin, heads will fall out 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 a symmetrical coin is tossed twice, heads will fall out only once, is equal to: P \u003d 2/4 \u003d 0.5 \u003d 50% Answer: the probability that as a result of the above experiment the heads will fall out only once is 50 %. Number of experiment 1st roll 2nd roll 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. A dice was thrown once. What is the probability that the number of points rolled is greater than 4. Solution: Random experiment - rolling a die. An elementary event is a number on a dropped edge. Answer: 1/3 Total faces: 1, 2, 3, 4, 5, 6 Elementary events: N=6 N(A)=2
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Task 5. The biathlete shoots at the targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hit the targets the first three times and missed the last two. Round the result to the nearest hundredth. Solution: Probability of hitting = 0.8 Probability of missing = 1 - 0.8 = 0.2 А=(hit, hit, hit, missed, missed) 0.8 ∙ 0.2 ∙ 0.2 P (A) \u003d 0.512 ∙ 0.04 \u003d 0.02048 ≈ 0.02 Answer: 0.02
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Task 6. In a random experiment, two dice are thrown. Find the probability that the sum of the points rolled is 6. Round your answer to the nearest hundredth. Solution: The elementary outcome in this experiment is an ordered pair of numbers. The first number will fall on the first die, the second on the second. The set of elementary outcomes is conveniently represented by a table. The rows correspond to the number of points on the first die, the columns correspond to the second die. There are n = 36 elementary events in total. Let's write in each cell the sum of the dropped points and color the cells where the sum is 6. There are 5 such cells. Hence, the event A = (the sum of the dropped points 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 the coin be tossed n times. Then the probability that heads will fall out exactly k times can be found by the formula: Where Cnk is the number of combinations of n elements by k, which is calculated by the formula:
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Problem 7. A coin is tossed four times. Find the probability that heads will come up exactly three times. Solution According to the condition of the problem, there were n =4 throws in total. Required number of eagles: k =3. Substitute n and k into the formula: With the same success, you can count the number of tails: 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 it never comes up tails. 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 tails, 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 come up more times than tails. Solution: In order for there to be more heads than tails, they must fall out either 3 times (then there will be 1 tails) or 4 (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 fall 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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Task 10. Before the start of the volleyball match, the team captains draw a fair lot 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. It is required to find the probability of product of three events: "Stator" starts the first game, does not start the second game, starts the third game. The probability of producing independent events is equal to the product of the probabilities of these events. The probability of each of them is equal to 0.5, whence we find: 0.5 0.5 0.5 = 0.125. Answer: 0.125.
In probability theory, there is a group of problems, for the solution of which it is enough to know the classical definition of probability and visualize the proposed situation. These problems are most coin toss problems and dice toss problems. 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 that favor 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 enumeration of all possible options (combinations) and direct calculation.
From the table we see that the number of possible elementary outcomes is n=4. Favorable outcomes of the event A = (eagle falls out 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 Р(А)=m/n=2/4=0.5
Task 2 . In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will never come up.
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 = (eagle will not fall out even once) correspond to variant No. 4 of the experiment (see the table in task 1). There is only one such option, so m=1.
Find the probability of the event Р(А)=m/n=1/4=0.25
Task 3 . In a random experiment, a symmetrical coin is tossed three times. Find the probability that it comes up heads exactly 2 times.
Solution . Possible options for three coin tosses (all possible combinations of heads and tails) are presented 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 2 times) correspond to options No. 5, 6 and 7 of the experiment. There are three such options, so m=3.
Find the probability of the event Р(А)=m/n=3/8=0.375
Task 4 . In a random experiment, a symmetrical coin is tossed four times. Find the probability that it comes up heads exactly 3 times.
Solution . Possible variants of four coin tosses (all possible combinations of heads and tails) are presented in the form of a table:
| option number | 1st throw | 2nd roll | 3rd roll | 4th roll | option number | 1st throw | 2nd roll | 3rd roll | 4th roll |
| 1 | Eagle | Eagle | Eagle | Eagle | 9 | Tails | Eagle | Tails | Eagle |
| 2 | Eagle | Tails | Tails | Tails | 10 | Eagle | Tails | Eagle | Tails |
| 3 | Tails | Eagle | Tails | Tails | 11 | Eagle | Tails | Tails | Eagle |
| 4 | Tails | Tails | Eagle | Tails | 12 | Eagle | Eagle | Eagle | Tails |
| 5 | Tails | Tails | Tails | Eagle | 13 | Tails | Eagle | Eagle | Eagle |
| 6 | Eagle | Eagle | Tails | Tails | 14 | Eagle | Tails | Eagle | Eagle |
| 7 | Tails | Eagle | Eagle | Tails | 15 | Eagle | Eagle | Tails | Eagle |
| 8 | Tails | Tails | Eagle | Eagle | 16 | Tails | Tails | Tails | Tails |
From the table we see that the number of possible elementary outcomes is n=16. Favorable outcomes of the event A = (eagle falls out 3 times) correspond to options No. 12, 13, 14 and 15 of the experiment, which means m=4.
Find the probability of the event Р(А)=m/n=4/16=0.25
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Determining Probability in Dice Problems
Task 5 . Determine the probability that more than 3 points will fall out when a dice (correct die) is thrown.
Solution
. When throwing a dice (a regular die), any of its six faces can fall out, i.e. to occur any of the elementary events - loss from 1 to 6 points (points). So the number of possible elementary outcomes is n=6.
Event A = (more than 3 points fell out) means that 4, 5 or 6 points (points) fell out. So the number of favorable outcomes m=3.
Probability of the event Р(А)=m/n=3/6=0.5
Task 6 . Determine the probability that when a dice is thrown, a number of points does not exceed 4. Round the result to the nearest thousandth.
Solution
. When throwing a dice, any of its six faces can fall out, i.e. to occur any of the elementary events - loss from 1 to 6 points (points). So the number of possible elementary outcomes is n=6.
Event A = (no more than 4 points fell out) means that 4, 3, 2 or 1 points (point) fell out. So the number of favorable outcomes m=4.
Probability of the event Р(А)=m/n=4/6=0.6666…≈0.667
Task 7 . A die is thrown twice. Find the probability that both numbers are less than 4.
Solution . Since a dice (dice) is thrown twice, we will argue as follows: if one point fell on the first die, then 1, 2, 3, 4, 5, 6 can fall out on the second. We get 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:
| 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 |
Favorable outcomes of the event A = (both times a number less than 4 fell out) (they are highlighted in bold) will be calculated and we will get m=9.
Find the probability of the event Р(А)=m/n=9/36=0.25
Task 8 . A die is thrown twice. Find the probability that the largest of the two numbers drawn is 5. Round your answer to the nearest thousandth.
Solution . All possible outcomes of two throws dice present in the table:
| 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 |
From the table we see that the number of possible elementary outcomes is n=6*6=36.
Favorable outcomes of the event A = (the largest of the two numbers drawn is 5) (they are highlighted in bold) are calculated and we get m=8.
Find the probability of the event Р(А)=m/n=8/36=0.2222…≈0.222
Task 9 . A die is thrown twice. Find the probability that a number less than 4 is rolled at least once.
Solution . All possible outcomes of two throws of a dice are presented in the table:
| 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 |
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 fell out” means “a number less than 4 fell out once or twice”, then the number of favorable outcomes of the event A = (at least once a number less than 4 fell out) (they are in bold) m=27.
Find the probability of the event Р(А)=m/n=27/36=0.75
