In a random Expert Advisor, a symmetrical coin is tossed twice. Mathematics and us
In the tasks on the theory of probability, which are presented in the Unified State Examination by number No. 4, in addition to, there are tasks for tossing a coin and about throwing a dice. Today we will analyze them.
Coin Toss Problems
Task 1. A symmetrical coin is tossed twice. Find the probability that it comes up tails exactly once.
In such problems, it is convenient to write down all possible outcomes, writing them using the letters P (tails) and O (heads). Thus, the outcome of the OR means that the first throw came up heads, and the second came up tails. In the problem under consideration, 4 outcomes are possible: PP, RO, OR, OO. Favor the event "tails come up exactly once" 2 outcomes: RO and OR. The required probability is .
Answer: 0.5.
Task 2. A symmetrical coin is tossed three times, Find the probability that heads will come up exactly twice.
In total, 8 outcomes are possible: PRR, RRO, ROR, ROO, ORR, ORO, OOR, LLC. Favor the event "heads exactly twice" 3 outcomes: ROO, ORO, OOR. The required probability is .
Answer: 0.375.
Task 3. Before the start of a football match, the referee tosses a coin to determine which team will start 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 the loss of tails means the winning of the lot by "Emerald" (such an assumption does not affect the calculation of probabilities). Then 8 outcomes are possible: PRR, RRO, ROR, ROO, ORR, ORO, OOR, LLC. There are 3 outcomes favoring the event “tails come up exactly once”: POO, ORO, OOP. The required probability is .
Answer: 0.375.
Task 4. A symmetrical coin is tossed three times. Find the probability that the outcome of the ROO will come (the first time it comes up tails, the second and third - heads).
As in the previous tasks, there are 8 outcomes here: PPP, PPO, POP, POO, OPP, ORO, OOP, OOO. The probability of the outcome of ROO is equal to .
Answer: 0.125.
Dice Roll Problems
Task 5. The dice is thrown twice. How many elementary outcomes of experience favor the event "the sum of points is 8"?
Task 6. Two dice are thrown at the same time. Find the probability that the total will be 4. Round the result to the nearest hundredth.
In general, if dice (dice) are thrown, then there are equally possible outcomes. The same number of outcomes is obtained if the same die is thrown once in a row.
The following outcomes favor the event “4 in total”: 1 - 3, 2 - 2, 3 - 1. Their number is 3. The desired probability is .
To calculate the approximate value of a fraction, it is convenient to use division by a corner. Thus, it is approximately equal to 0.083 ..., rounded to hundredths, we have 0.08.
Answer: 0.08
Task 7. Three dice are thrown at the same time. Find the probability of getting 5 points in total. Round the result to the nearest hundredth.
We will consider the outcome as a triple of numbers: the points that fell on the first, second and third dice. In total there are equally possible outcomes. The following outcomes favor the “5 in total” event: 1-1-3, 1-3-1, 3-1-1, 1-2-2, 2-1-2, 2-2-1. Their number is 6. The desired probability is . To calculate the approximate value of a fraction, it is convenient to use division by a corner. Approximately we get 0.027 ..., rounded to hundredths, we have 0.03. Source “Preparation for the exam. Mathematics. Probability theory”. Edited by F.F. Lysenko, S.Yu. Kulabukhov
Task Formulation: IN random experiment symmetrical coin thrown 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 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.
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 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 . Because dice(dice) is thrown twice, then 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 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.
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
Coin toss puzzles are considered quite difficult. And before solving them, a little explanation is required. Think about it, any task in probability theory ultimately boils down to the standard formula:
where p is the desired probability, k is the number of events that suit us, n is the total number of possible events.
Most B6 problems are solved using this formula literally in one line - just read the condition. But in the case of tossing coins, this formula is useless, because from the text of such problems it is not at all clear what the numbers k and n are equal to. This is where the complexity lies.
However, there are at least two fundamental different method solutions:
- The method of enumerating combinations is a standard algorithm. All combinations of heads and tails are written out, after which the necessary ones are selected;
- Special probability formula - a standard definition of probability, specially rewritten so that it is convenient to work with coins.
To solve problem B6, you need to know both methods. Unfortunately, only the first one is taught in schools. Let's not repeat school mistakes. So let's go!
Combination enumeration method
This method is also known as "breakthrough decision". Consists of three steps:
- We write out all possible combinations of heads and tails. For example: OR, RO, OO, RR. The number of such combinations is n;
- Among the combinations obtained, we note those that are required by the condition of the problem. We count the marked combinations - we get the number k;
- It remains to find the probability: p = k : n .
Unfortunately, this method only works for a small number of throws. Because with each new throw, the number of combinations doubles. For example, for 2 coins, you will have to write out only 4 combinations. For 3 coins there are already 8 of them, and for 4 - 16, and the error probability approaches 100%. Take a look at the examples and you will understand everything yourself:
Task. In a random experiment, a symmetrical coin is tossed 2 times. Find the probability of getting the same number of heads and tails.
So the coin is tossed twice. Let's write down all possible combinations (O - heads, P - tails):
Total n = 4 options. Now we write out those options that are suitable for the condition of the problem:
There were k = 2 such options. We find the probability:
Task. The coin is tossed four times. Find the probability that it never comes up tails.
Again, we write out all possible combinations of heads and tails:
OOOO OOOP OOPO OOPP OPOO OPOP OPPO OPPP
POOO POOP POPO POPP PPOO PPOP PPPO PPPP
There are n = 16 options in total. It seems like he didn't forget anything. Of these options, we are satisfied only with the combination “OOOO”, in which there are no tails at all. Therefore, k = 1. It remains to find the probability:
As you can see, in the last task I had to write out 16 options. Are you sure that you can write them out without a single mistake? Personally, I'm not sure. So let's look at the second solution.
Special Probability Formula
So, problems with coins have their own probability formula. It is so simple and important that I decided to put it in the form of a theorem. Take a look:
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 C n k is the number of combinations of n elements by k , which is calculated by the formula:
Thus, to solve the problem with coins, two numbers are needed: the number of tosses and the number of heads. Most often, these numbers are given directly in the text of the problem. Moreover, it does not matter what exactly to count: tails or eagles. The answer will be the same.
At first glance, the theorem seems too cumbersome. But it's worth a little practice - and you no longer want to return to the standard algorithm described above.
Task. The coin is tossed four times. Find the probability that heads will come up exactly three times.
According to the condition of the problem, there were n = 4 total throws. The required number of heads: k = 3. Substitute n and k into the formula:
Task. The coin is tossed three times. Find the probability that it never comes up tails.
Again we write out the numbers n and k. 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 C 3 0 = 1.
Task. In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will come up more times than tails.
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 p 1 be the probability of getting heads 3 times. Then n = 4, k = 3. We have:
Now we find p 2 - 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 p 1 and p 2 . Remember: you can only add probabilities for mutually exclusive events. We have:
p \u003d p 1 + p 2 \u003d 0.25 + 0.0625 \u003d 0.3125
