In a random experiment, a symmetrical coin is tossed 4. In a random experiment, a symmetrical coin is tossed twice. Solving the problem with a symmetrical coin
Problem formulation: In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads (tails) will not appear even once (will appear exactly/at least 1, 2 times).
The problem is part of the Unified State Examination in basic level mathematics for grade 11 under number 10 (Classical definition of probability).
Let's look at how such problems are solved using examples.
Example task 1:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will not come up even once.
OO OR RO RR
There are 4 such combinations in total. We are only interested in those that do not contain 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 tossed twice. Find the probability of getting heads exactly twice.
Let's consider all the possible combinations that can occur if a coin is tossed 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 in which heads appear 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 tossed twice. Find the probability that heads come up exactly once.
Let's consider all the possible combinations that can occur if a coin is tossed 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 in which heads came up exactly 1 time. There are only two such combinations (OR and RO).
Answer: 0.5
Example task 4:
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear at least once.
Let's consider all the possible combinations that can occur if a coin is tossed 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 in which heads appear at least once. There are only three such combinations (OO, OP and RO).
P = 3 / 4 = 0.75
Coin toss problems are considered quite difficult. And before solving them, a little explanation is required. Think about it, any problem in probability theory ultimately comes 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 can be solved using this formula literally in one line - just read the condition. But in the case of tossing coins, this formula is useless, since 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 difficulty lies.
However, there are at least two fundamentally different solution methods:
- 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;
- 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!
Combination search method
This method is also called the “solution ahead”. Consists of three steps:
- We write down 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 conditions 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 probability of error is approaching 100%. Take a look at the examples and you will understand everything yourself:
Task. In a random experiment, a symmetrical coin is tossed twice. Find the probability that you get 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 let’s write down the options that suit the conditions of the problem:
There were k = 2 such options. Find the probability:
Task. The coin is tossed four times. Find the probability that you will never get heads.
Again we write down all possible combinations of heads and tails:
OOOO OOOP OOPO OOPP OPOO OPOP OPPO OPPP
POOO POOP POPO POPP PPOO PPOP PPPO PPPP
In total there were n = 16 options. It seems like I haven’t forgotten anything. Of these options, we are only satisfied with the “OOOO” combination, which does not contain tails at all. Therefore, k = 1. It remains to find the probability:
As you can see, in the last problem I had to write out 16 options. Are you sure you can write them out without making a single mistake? Personally, I'm not sure. So let's look at the second solution.
Special probability formula
So, coin problems 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 tossed n times. Then the probability that heads will appear exactly k times can be found using 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 coin problem, you need two numbers: 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 you count: tails or heads. The answer will be the same.
At first glance, the theorem seems too cumbersome. But once you practice a little, you will no longer want to return to the standard algorithm described above.
Task. The coin is tossed four times. Find the probability of getting heads exactly three times.
According to the conditions 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 you will never get heads.
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 C 3 0 = 1.
Task. In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will appear more times than tails.
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 p 1 be the probability that heads will appear 3 times. Then n = 4, k = 3. We have:
Now let's find p 2 - the probability that heads will appear all 4 times. In this case n = 4, k = 4. We have:
To get the answer, all that remains is 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
Condition
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear at least once.
Solution
- We will solve this problem using 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 possible outcomes.
- Let's apply this theory to our problem:
A – an event when heads appear at least once;
P(A) – the probability that heads will appear at least once.
- Let's define m and n:
m is the number of outcomes favorable to this event, that is, the number of outcomes when heads appear 1 time. In the experiment, a coin is tossed twice, which has 2 sides: tails (P) and heads (O). We need heads to come up at least once, and this is possible when the following combinations come up: OP, PO and OO, that is, it turns out that
m = 3, since there are 3 possible ways of getting at least 1 head;
n is the total number of possible outcomes, that is, to determine n we need to find the number of all possible combinations that can occur when tossing a coin twice. When throwing a coin for the first time, it can come up either tails or heads, that is, two options are possible. When throwing a coin a second time, exactly the same options are possible. It turns out that
Condition
In a random experiment, a symmetrical coin is tossed twice. Find the probability that the same thing will come out the second time as the first time.
Solution
- We will solve this problem using 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 possible outcomes.
- Let's apply this theory to our problem:
A – an event when the same thing comes up for the second time as the first time;
P(A) – the probability that the same thing will come up the second time as the first time.
- Let's define m and n:
m is the number of outcomes favorable to this event, that is, the number of outcomes when the same thing happens the second time as the first. In the experiment, a coin is tossed twice, which has 2 sides: tails (P) and heads (O). We need the same thing to come up the second time as the first time, and this is possible when the following combinations come up: OO or PP, that is, it turns out that
m = 2, since there are 2 possible options, when the same thing comes up the second time as the first time;
n is the total number of possible outcomes, that is, to determine n we need to find the number of all possible combinations that can occur when tossing a coin twice. When throwing a coin for the first time, it can come up either tails or heads, that is, two options are possible. When throwing a coin a second time, exactly the same options are possible. It turns out that
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, if 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.
We will consider the outcome to be three numbers: 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
