Exodus. In a random experiment, a symmetrical coin is thrown twice.
Answer: 0.25. 34. Solution. There are 4 options in total: o; oh oh; p p; p p; O. Favorable 1: o; R. The probability is 1/4 = 0.25. V random experiment a symmetrical coin is thrown twice. Find the probability of an OP outcome (heads up the first time, tails the second time).
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Condition
In a random experiment, a symmetrical coin is thrown twice. Find the probability that the second time will be the same as the first.
Solution
- This problem will be solved by the formula:
Where P (A) is the probability of event A, m is the number of favorable outcomes for this event, n is the total number of all possible outcomes.
- Let's apply this theory to our problem:
A - an event when the second time falls out the same as the first;
P (A) - the probability that the second time will be the same as the first.
- Let's define m and n:
m - the number of outcomes favorable to this event, that is, the number of outcomes when the second time the same occurs as the first. In the experiment, a coin is thrown twice, which has 2 sides: tails (P) and heads (O). We need the second time to drop the same as the first, and this is possible when the following combinations fall out: OO or PP, that is, it turns out that
m = 2, since 2 options are possible, when the second time the same will fall out as the first;
n is the total number of all possible outcomes, that is, to determine n, we need to find the number of all possible combinations that can occur when the coin is thrown twice. When throwing a coin for the first time, either tails or heads can fall out, that is, two options are possible. When tossing a coin a second time, exactly the same options are possible. It turns out that
In the theory of probability, there is a group of problems, for the solution of which it is enough to know the classical definition of probability and to visualize the proposed situation. Such problems are most coin toss and dice throw problems. Let us recall the classical definition of probability.
The probability of event A (the objective possibility of the occurrence of an event 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 enumerating 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 = (heads fall 1 time) correspond to options 2 and 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
Task 2 ... In a random experiment, a symmetrical coin is thrown twice. Find the probability that it will never land heads.
Solution
... Since the coin is thrown twice, then, as in Problem 1, the number of possible elementary outcomes is n = 4. Favorable outcomes of the event A = (heads will not fall out even once) correspond to option 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 P (A) = m / n = 1/4 = 0.25
Problem 3 ... In a random experiment, a symmetrical coin is thrown three times. Find the probability that heads will come up exactly 2 times.
Solution ... Possible variants of 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 fall 2 times) correspond to options 5, 6 and 7 of the experiment. There are three such options, which means m = 3.
We find the probability of the event P (A) = m / n = 3/8 = 0.375
Problem 4 ... In a random experiment, a symmetrical coin is thrown four times. Find the probability that heads will land 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 No. | 1st throw | 2nd throw | 3rd throw | 4th throw | Option No. | 1st throw | 2nd throw | 3rd throw | 4th throw |
| 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 = (heads fall 3 times) correspond to options 12, 13, 14 and 15 of the experiment, so m = 4.
We find the probability of the event P (A) = m / n = 4/16 = 0.25
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Determination of Probability in Dice Problems
Problem 5 ... Determine the probability that more than 3 points will be rolled when rolling the dice (correct die).
Solution
... When throwing a dice (correct dice), any of its six faces can fall out, i.e. any of the elementary events occur - a drop from 1 to 6 points (points). Hence, the number of possible elementary outcomes is n = 6.
Event А = (more than 3 points dropped out) means that 4, 5 or 6 points (points) fell out. Hence, the number of favorable outcomes is m = 3.
The probability of an event P (A) = m / n = 3/6 = 0.5
Problem 6 ... Determine the probability that the number of points dropped out when throwing the dice is not more than 4. Round the result to thousandths.
Solution
... When throwing a dice, any of its six faces can fall out, i.e. any of the elementary events occur - a drop from 1 to 6 points (points). Hence, the number of possible elementary outcomes is n = 6.
Event А = (no more than 4 points dropped out) means that 4, 3, 2 or 1 point (point) fell out. Hence, the number of favorable outcomes is m = 4.
The probability of an event P (A) = m / n = 4/6 = 0.6666 ... ≈0.667
Problem 7 ... The dice are rolled twice. Find the probability that both times the number is less than 4.
Solution ... Since the dice ( dice( 1; 3), (1; 4), (1; 5), (1; 6) and so on with each face. All cases are presented 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 dropped out) (they are highlighted in bold), we calculate and get m = 9.
We find the probability of the event P (A) = m / n = 9/36 = 0.25
Problem 8 ... The dice are rolled twice. Find the probability that the largest of the two drawn numbers is 5. Round your answer to the nearest thousand.
Solution ... Everything possible outcomes 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 drawn numbers is 5) (they are highlighted in bold) will be calculated and we will get m = 8.
We find the probability of the event P (A) = m / n = 8/36 = 0.2222 ... ≈0.222
Problem 9 ... The dice are rolled twice. Find the probability that at least once a number dropped out less than 4.
Solution ... All possible outcomes of two throws of the 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 highlighted in bold) m = 27.
Find the probability of the event P (A) = m / n = 27/36 = 0.75
The coin flip tasks are considered to be quite difficult. And before solving them, a little clarification is required. Think about it, any problem in probability theory ultimately comes down to a 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 the whole difficulty.
However, there are at least two fundamentally different solution methods:
- The brute force method is a standard algorithm. All combinations of heads and tails are written out, after which the necessary ones are selected;
- A special probability formula is 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 is taught in schools. Let's not repeat school mistakes. So let's go!
The brute force method
This method is also called “head-to-head solution”. Consists of three steps:
- Write out all possible combinations of heads and tails. For example: OP, RO, OO, RR. The number of such combinations is n;
- Among the combinations obtained, we mark 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 shots. Because with each new throw, the number of combinations doubles. For example, for 2 coins, you have to write out only 4 combinations. For 3 coins, there are already 8, and for 4 - 16, and the error probability approaches 100%. Take a look at the examples - and you yourself will understand everything:
Task. In a random experiment, a symmetrical coin is thrown 2 times. Find the probability that the same number of heads and tails will come up.
So, the coin is thrown twice. Let's write out all possible combinations (O - heads, P - tails):
Total n = 4 options. Now let's write out those options that fit the problem statement:
There were k = 2 such options. Find the probability:
Task. The coin is thrown four times. Find the probability that it will never come up tails.
Again, write out all the possible combinations of heads and tails:
OOOO OOOP OOPO OOPP OPOO OPOP OPPO OPPP
POOO POOP POPO POPP PPOO PPOP PPPO PPPP
There were n = 16 options in total. Didn't seem to have forgotten anything. Of these options, we are satisfied with only 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 problem we had to write out 16 options. Are you sure 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 formulate it in the form of a theorem. Take a look:
Theorem. Let the coin be thrown n times. Then the probability that heads will land exactly k times can be found by the formula:
Where C n k - 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 eagles. Most often, these numbers are given directly in the text of the problem. Moreover, it doesn't matter what you count: tails or heads. The answer will be the same.
At first glance, the theorem seems too cumbersome. But with a little practice, you don't want to go back to the standard algorithm described above.
Task. The coin is thrown four times. Find the probability that it will be heads exactly three times.
By the condition of the problem, the total number of throws was n = 4. The required number of heads: k = 3. Substitute n and k into the formula:
Task. The coin is thrown three times. Find the probability that it will never come up tails.
We write out the numbers n and k again. Since the coin is thrown 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 thrown 4 times. Find the probability that heads will come up more times than tails.
To have more heads than tails, they must come up 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 that heads will land 3 times. Then n = 4, k = 3. We have:
Now we will find p 2 - the probability that heads will come up 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 = p 1 + p 2 = 0.25 + 0.0625 = 0.3125
Problem statement: In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads (tails) will not come up even once (it will come up exactly / at least 1, 2 times).
The task is included in the exam in mathematics basic level for grade 11 at number 10 (Classical definition of probability).
Let's see how similar tasks are solved using examples.
Example task 1:
In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads will never come up.
OO OR RO RR
There were 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
Example task 2:
In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads will come up exactly two times.
Consider all the possible combinations that can occur if a coin is thrown twice. For convenience, we will denote heads by the letter O, and tails by the letter P:
OO OR RO RR
There are 4 such combinations in total. We are only interested in those of them in which the heads fall exactly 2 times. There is only one such combination (OO).
P = 1/4 = 0.25
Answer: 0.25
Example task 3:
In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads will come up exactly once.
Consider all the possible combinations that can occur if a coin is thrown twice. For convenience, we will denote heads by the letter O, and tails by the letter P:
OO OR RO RR
There were 4 such combinations in total. We are only interested in those of them in which the heads fell exactly 1 time. There are only two such combinations (OP and RO).
Answer: 0.5
Example task 4:
In a random experiment, a symmetrical coin is thrown twice. Find the probability that heads will come up at least once.
Consider all the possible combinations that can occur if a coin is thrown twice. For convenience, we will denote heads by the letter O, and tails by the letter P:
OO OR RO RR
There were 4 such combinations in total. We are only interested in those of them in which heads fall at least 1 time. There are only three such combinations (OO, OP and RO).
P = 3/4 = 0.75