The coin is thrown twice. In a random experiment, a symmetric coin is thrown twice. Special formula probability
Task wording: In random experiment symmetric coin Throw twice. Find the chance that Orel (Rushka) will not fall out (Rivne / at least 1, 2 times).
The task is part of the exam in mathematics basic level For grade 11, number 10 (classical probability definition).
Consider how similar tasks are solved on the examples.
Example problem 1:
In a random experiment, a symmetric coin is thrown twice. Find the chance that the eagle will never fall.
Oo op ro po
In total, it turned out to be 4. We are only interested in those of them, in which there is not a single eagle. Such a combination is only one (PP).
P \u003d 1/4 \u003d 0.25
Answer: 0.25
Example Task 2:
In a random experiment, a symmetric coin is thrown twice. Find the chance that the eagle falls exactly twice.
Consider all possible combinations that may fall if the coin is thrown twice. For convenience, we will denote the eagle of the letter O, and the decision - the letter R:
Oo op ro po
In total, there were only such combinations. We are only interested in those of which the eagle falls exactly 2 times. Such a combination is only one (oo).
P \u003d 1/4 \u003d 0.25
Answer: 0.25
Example Task 3:
In a random experiment, a symmetric coin is thrown twice. Find the chance that the eagle will fall out exactly once.
Consider all possible combinations that may fall if the coin is thrown twice. For convenience, we will denote the eagle of the letter O, and the decision - the letter R:
Oo op ro po
All such combinations it turned out. 4. We are only interested in those of them in which the eagle fell exactly 1 time. There are only two combinations (OR and PO).
Answer: 0.5.
Example Task 4:
In a random experiment, a symmetric coin is thrown twice. Find the chance that the eagle will fall out at least once.
Consider all possible combinations that may fall if the coin is thrown twice. For convenience, we will denote the eagle of the letter O, and the decision - the letter R:
Oo op ro po
All such combinations it turned out. 4. We are only interested in those of which the eagle will fall out at least 1 time. There are only three such combinations (oo, op and ro).
P \u003d 3/4 \u003d 0.75
Answer: 0.25. 34. Decision. Total 4 options: o; O O; p p; p p; about. Favorable 1: o; R. The probability is 1/4 \u003d 0.25. In a random experiment, a symmetric coin is thrown twice. Find the likelihood that the outcome of the OR (the first time the eagle falls, in the second - the rush).
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Condition
In a random experiment, a symmetric coin is thrown twice. Find the likelihood that the second time will fall out as in the first.
Decision
- This task will be solved by the formula:
Where P (a) is the probability of event A, M - the number of conducive outcomes of this event, n is the total number of all sorts of outcomes.
- Apply this theory to our task:
A - event when the same thing as in the first time will fall as in the second time;
P (a) is the likelihood that the same thing as in the first time will fall out.
- Determine M and N:
m is the number of outcomes conducive to this event, that is, the number of outcomes when the same thing as in the first time falls. In the experiment, they throw a coin twice, which has 2 sides: Rushka (P) and Eagle (o). We needed that the second time will fall out the same as in the first, and this is possible when the following combinations fall: oo or pp, that is, it turns out that
m \u003d 2, since it is possible 2 options when the same thing as in the first one will fall asleep.
n is the total number of all sorts of outcomes, that is, to determine N, we need to find the number of all possible combinations that can fall when throwing a coin twice. You throwing the first time the coin can fall or the rush, or the eagle, that is, there are two options. When throwing the second time, the coin is possible exactly the same options. Turns out that