Probability theory Unified State Exam profile level. Probability theory on the exam in mathematics
Attention to applicants! Several USE tasks are discussed here. The rest, more interesting, are in our free video. Watch and do!
We will start with simple problems and basic concepts of probability theory.
Random An event that cannot be accurately predicted in advance is called. It can either happen or not.
You won the lottery - a random event. You invited friends to celebrate your win, and on the way to you they got stuck in the elevator - also a random event. True, the master turned out to be nearby and freed the entire company within ten minutes - and this can also be considered a happy accident...
Our life is full of random events. About each of them we can say that it will happen with some probability. Most likely, you are intuitively familiar with this concept. We will now give the mathematical definition of probability.
Let's start from the very beginning simple example. You flip a coin. Heads or tails?
Such an action, which can lead to one of several results, is called in probability theory test.
Heads and tails - two possible outcome tests.
Heads will fall out in one case out of two possible. They say that probability that the coin will land on heads is .
Let's throw a dice. The die has six sides, so there are also six possible outcomes.
For example, you wished that three points would appear. This is one outcome out of six possible. In probability theory it will be called favorable outcome.
The probability of getting a three is equal (one favorable outcome out of six possible).
The probability of four is also
But the probability of a seven appearing is zero. After all, there is no edge with seven points on the cube.
The probability of an event is equal to the ratio of the number of favorable outcomes to total number outcomes.
Obviously, the probability cannot be greater than one.
Here's another example. There are apples in a bag, some of them are red, the rest are green. The apples do not differ in shape or size. You put your hand into the bag and take out an apple at random. The probability of drawing a red apple is equal to , and the probability of drawing a green apple is equal to .
Probability of getting red or green apple equal to .
Let us analyze problems in probability theory included in the collections for preparing for the Unified State Exam.
. At a taxi company in at the moment free cars: red, yellow and green. One of the cars that happened to be closest to the customer responded to the call. Find the probability that a yellow taxi will come to her.
There are a total of cars, that is, one out of fifteen will come to the customer. There are nine yellow ones, which means that the probability of a yellow car arriving is equal to , that is.
. (Demo version) In the collection of tickets on biology of all tickets, in two of them there is a question about mushrooms. During the exam, the student receives one randomly selected ticket. Find the probability that this ticket will not contain a question about mushrooms.
Obviously, the probability of drawing a ticket without asking about mushrooms is equal to , that is.
. The Parents Committee purchased puzzles for graduation gifts for children. academic year, of which with paintings famous artists and with images of animals. Gifts are distributed randomly. Find the probability that Vovochka will get a puzzle with an animal.
The problem is solved in a similar way.
Answer: .
. Athletes from Russia, from the USA, and the rest from China are participating in the gymnastics championship. The order in which the gymnasts perform is determined by lot. Find the probability that the last athlete to compete is from China.
Let's imagine that all the athletes simultaneously approached the hat and pulled out pieces of paper with numbers from it. Some of them will get number twenty. The probability that a Chinese athlete will pull it out is equal (since the athletes are from China). Answer: .
. The student was asked to name the number from to. What is the probability that he will name a number that is a multiple of five?
Every fifth a number from this set is divisible by . This means the probability is equal to .
A die is thrown. Find the probability of getting an odd number of points.
Odd numbers; - even. The probability of an odd number of points is .
Answer: .
. The coin is tossed three times. What is the probability of two heads and one tail?
Note that the problem can be formulated differently: three coins were thrown at the same time. This will not affect the decision.
How many possible outcomes do you think there are?
We toss a coin. This action has two possible outcomes: heads and tails.
Two coins - already four outcomes:
Three coins? That's right, outcomes, since .
Two heads and one tails appear three out of eight times.
Answer: .
. In a random experiment, two dice are rolled. Find the probability that the total will be points. Round the result to hundredths.
We throw the first die - six outcomes. And for each of them six more are possible - when we throw the second die.
We find that this action - throwing two dice - has a total of possible outcomes, since .
And now - favorable outcomes:
The probability of getting eight points is .
>. The shooter hits the target with probability. Find the probability that he hits the target four times in a row.
If the probability of a hit is equal, then the probability of a miss is . We reason in the same way as in the previous problem. The probability of two hits in a row is equal. And the probability of four hits in a row is equal.
Probability: brute force logic.
Here is the problem from diagnostic work, which many found difficult.
Petya had coins worth rubles and coins worth rubles in his pocket. Petya, without looking, transferred some coins to another pocket. Find the probability that the five-ruble coins are now in different pockets.
We know that the probability of an event is equal to the ratio of the number of favorable outcomes to the total number of outcomes. But how to calculate all these outcomes?
You can, of course, designate five-ruble coins with numbers, and ten-ruble coins with numbers - and then count how many ways you can select three elements from the set.
However, there is a simpler solution:
We code the coins with numbers: , (these are five-ruble coins), (these are ten-ruble coins). The problem condition can now be formulated as follows:
There are six chips with numbers from to . In how many ways can they be distributed equally into two pockets, so that the chips with numbers do not end up together?
Let's write down what we have in our first pocket.
To do this, we will compose all possible combinations from the set. A set of three chips will be a three-digit number. Obviously, in our conditions and are the same set of chips. In order not to miss anything or repeat ourselves, we have the appropriate three digit numbers ascending:
All! We went through all possible combinations starting with . Let's continue:
Total possible outcomes.
We have a condition - chips with numbers should not be together. This means, for example, that the combination does not suit us - it means that both chips ended up not in the first, but in the second pocket. Outcomes that are favorable for us are those where there is either only , or only . Here they are:
134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256 – total favorable outcomes.
Then the required probability is equal to .
What tasks await you on the Unified State Examination in mathematics?
Let's look at one of them complex tasks according to probability theory.
To enter the institute for the specialty "Linguistics", applicant Z. must score at least 70 points on the Unified State Examination in each of three subjects - mathematics, Russian language and a foreign language. To enroll in the specialty "Commerce", you need to score at least 70 points in each of three subjects - mathematics, Russian language and social studies.
The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in foreign language- 0.7 and in social studies - 0.5.
Find the probability that Z. will be able to enroll in at least one of the two mentioned specialties.
Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at once and receive two diplomas. Here we need to find the probability that Z. will be able to enroll in at least one of these two specialties - that is, he will score the required number of points.
In order to enter at least one of the two specialties, Z. must score at least 70 points in mathematics. And in Russian. And also - social studies or foreign.
The probability for him to score 70 points in mathematics is 0.6.
The probability of scoring points in mathematics and Russian is 0.6 0.8.
Let's deal with foreign and social studies. The options that suit us are when the applicant has scored points in social studies, foreign studies, or both. The option is not suitable when he did not score any points in either language or “society”. This means that the probability of passing social studies or foreign language with at least 70 points is equal to
1 – 0,5 0,3.
As a result, the probability of passing mathematics, Russian and social studies or foreign is equal
0.6 0.8 (1 - 0.5 0.3) = 0.408. This is the answer.
The probability of an event $A$ is the ratio of the number of outcomes favorable for $A$ to the number of all equally possible outcomes
$P(A)=(m)/(n)$, where $n$ – total quantity possible outcomes, and $m$ is the number of outcomes favorable to event $A$.
The probability of an event is a number from the segment $$
The taxi company has $50$ cars in stock. $35$ of them are black, the rest are yellow. Find the probability that a yellow car will respond to a random call.
Let's find the number of yellow cars:
There are $50$ cars in total, that is, one in fifty will respond to a call. Yellow cars are $15$, therefore, the probability of a yellow car arriving is $(15)/(50)=(3)/(10)=0.3$
Answer: $0.3$
Opposite events
Two events are called opposite if this test they are incompatible and one of them is bound to happen. The probabilities of opposite events add up to 1. An event opposite to event $A$ is written $((A))↖(-)$.
$P(A)+P((A))↖(-)=1$
Independent events
Two events $A$ and $B$ are called independent if the probability of occurrence of each of them does not depend on whether the other event occurred or not. Otherwise, the events are called dependent.
The probability of the product of two independent events $A$ and $B$ is equal to the product of these probabilities:
$P(A·B)=P(A)·P(B)$
Ivan Ivanovich bought two different lottery tickets. Probability that the first one will win lottery ticket, is equal to $0.15$. The probability that the second lottery ticket will win is $0.12$. Ivan Ivanovich participates in both draws. Assuming that the draws are held independently of each other, find the probability that Ivan Ivanovich will win in both draws.
Probability $P(A)$ - the first ticket will win.
Probability $P(B)$ - the second ticket will win.
Events $A$ and $B$ are independent events. That is, to find the probability that both events will occur, you need to find the product of the probabilities
$P(A·B)=P(A)·P(B)$
$Р=0.15·0.12=0.018$
Answer: $0.018$
Incompatible events
Two events $A$ and $B$ are called incompatible if there are no outcomes that favor both event $A$ and event $B$. (Events that cannot happen at the same time)
The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:
$P(A+B)=P(A)+P(B)$
On an algebra exam, a student gets one question out of all the exam questions. Chances are this is a question about " Quadratic equations", is equal to $0.3$. Chances are this is a question about " Irrational equations", is equal to $0.18$. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in the exam.
These events are called incompatible, since the student will get a question EITHER on the topic “Quadratic Equations” OR on the topic “Irrational Equations”. Topics cannot be found at the same time. The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:
$P(A+B)=P(A)+P(B)$
$P = 0.3+0.18=0.48$
Answer: $0.48$
Joint events
Two events are called joint if the occurrence of one of them does not exclude the occurrence of the other in the same trial. Otherwise, the events are called incompatible.
The probability of the sum of two joint events $A$ and $B$ is equal to the sum of the probabilities of these events minus the probability of their product:
$P(A+B)=P(A)+P(B)-P(A B)$
In the cinema hall, two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is $0.6$. The probability that both machines will run out of coffee is $0.32$. Find the probability that by the end of the day at least one of the machines will run out of coffee.
Let us denote the events:
$A$ = coffee will run out in the first machine,
$B$ = coffee will run out in the second machine.
$A·B =$ the coffee will run out in both machines,
$A + B =$ coffee will run out in at least one machine.
According to the condition, $P(A) = P(B) = 0.6; P(A·B) = $0.32.
Events $A$ and $B$ are joint, the probability of the sum of two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their product:
$P(A + B) = P(A) + P(B) − P(A B) = 0.6 + 0.6 − 0.32 = 0.88$
Brought to date in open jar Unified State Examination problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is the classical definition of probability.
The easiest way to understand the formula is with examples.
Example 1. There are 9 red balls and 3 blue balls in the basket. The balls differ only in color. We take out one of them at random (without looking). What is the probability that the ball chosen in this way will be blue?
Comment. In problems in probability theory, something happens (in in this case our action of pulling out the ball), which can have different result- outcome. It should be noted that the result can be looked at in different ways. “We pulled out some kind of ball” is also a result. “We pulled out the blue ball” - the result. “We pulled out exactly this ball from all possible balls” - this least generalized view of the result is called an elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability.
Solution. Now let's calculate the probability of choosing the blue ball.
Event A: “the selected ball turned out to be blue”
Total number of all possible outcomes: 9+3=12 (the number of all balls that we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25
For the same problem, let's calculate the probability of choosing a red ball.
The total number of possible outcomes will remain the same, 12. Number of favorable outcomes: 9. Probability sought: 9/12=3/4=0.75
The probability of any event always lies between 0 and 1.
Sometimes in everyday speech (but not in probability theory!) the probability of events is estimated as a percentage. The transition between math and conversational scores is accomplished by multiplying (or dividing) by 100%.
So,
Moreover, the probability is zero for events that cannot happen - incredible. For example, in our example this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0, if calculated using the formula)
Probability 1 has events that are absolutely certain to happen, without options. For example, the probability that “the selected ball will be either red or blue” is for our task. (Number of favorable outcomes: 12, P(A)=12/12=1)
We looked at a classic example illustrating the definition of probability. All similar Unified State Examination tasks According to probability theory, they are solved by using this formula.
In place of the red and blue balls there may be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on a certain topic (prototypes,), defective and high-quality bags or garden pumps (prototypes,) - the principle remains the same.
They differ slightly in the formulation of the problem of the probability theory of the Unified State Examination, where you need to calculate the probability of some event occurring on a certain day. ( , ) As in previous problems, you need to determine what is the elementary outcome, and then apply the same formula.
Example 2. The conference lasts three days. On the first and second days there are 15 speakers, on the third day - 20. What is the probability that Professor M.’s report will fall on the third day if the order of reports is determined by drawing lots?
What is the elementary outcome here? – Assigning the professor’s report one of all possible serial numbers for the speech. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report may receive one of 50 issues. This means there are only 50 elementary outcomes.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4
The drawing of lots here represents the establishment of a random correspondence between people and ordered places. In example 2, matching was considered from the point of view of which of the seats a particular person could occupy. You can approach the same situation from the other side: which of the people with what probability could get to a specific place (prototypes , , , ):
Example 3. The draw includes 5 Germans, 8 French and 3 Estonians. What is the probability that the first (/second/seventh/last – it doesn’t matter) will be a Frenchman.
The number of elementary outcomes is the number of all possible people who could get into a given place by drawing lots. 5+8+3=16 people.
Favorable outcomes - French. 8 people.
Required probability: 8/16=1/2=0.5
Answer: 0.5
The prototype is slightly different. There are still problems about coins () and dice (), which are somewhat more creative. The solution to these problems can be found on the prototype pages.
Here are a few examples of tossing a coin or dice.
Example 4. When we toss a coin, what is the probability of landing on heads?
There are 2 outcomes – heads or tails. (it is believed that the coin never lands on its edge) A favorable outcome is tails, 1.
Probability 1/2=0.5
Answer: 0.5.
Example 5. What if we toss a coin twice? What is the probability of getting heads both times?
The main thing is to determine what elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP – both times it came up heads
2) PO – first time heads, second time heads
3) OP – heads the first time, tails the second time
4) OO – heads came up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one, 1, is favorable.
Probability: 1/4=0.25
Answer: 0.25
What is the probability that two coin tosses will result in tails?
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5
In such problems, another formula may be useful.
If during one toss of a coin possible options we have 2 results, then for two throws the results will be 2 2 = 2 2 = 4 (as in example 5), for three throws 2 2 2 = 2 3 = 8, for four: 2 2 2 2 =2 4 =16, ... for N throws possible results will be 2·2·...·2=2 N .
So, you can find the probability of getting 5 heads out of 5 coin tosses.
Total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR – heads all 5 times)
Probability: 1/32=0.03125
The same is true for dice. With one throw, there are 6 possible results. So, for two throws: 6 6 = 36, for three 6 6 6 = 216, etc.
Example 6. We throw the dice. What is the probability that an even number will be rolled?
Total outcomes: 6, according to the number of sides.
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5
Example 7. We throw two dice. What is the probability that the total will be 10? (round to the nearest hundredth)
For one die there are 6 possible outcomes. This means that for two, according to the above rule, 6·6=36.
What outcomes will be favorable for the total to roll 10?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. This means that the following options are possible for the cubes:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
Total, 3 options. Required probability: 3/36=1/12=0.08
Answer: 0.08
Other types of B6 problems will be discussed in a future How to Solve article.
Probability. Problems of the profile Unified State Examination in mathematics.
Prepared by a mathematics teacher at MBOU “Lyceum No. 4”, Ruzaevka
Ovchinnikova T.V.
Definition of probability
Probability events A are called number ratio m outcomes favorable to this event to the total number n all equally possible incompatible events that can occur as a result of one test or observation:
m
n
Let k – the number of coin tosses, then the number of possible outcomes: n=2 k .
Let k – the number of dice rolls, then the number of possible outcomes: n=6 k .
In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads will appear exactly once.
Solution.
There are only 4 options: O; o o; p p; p p; O .
Favorable 2: O; r And p; O .
The probability is 2/4 = 1/2 = 0,5 .
Answer: 0.5.
In a random experiment, two dice are rolled. Find the probability that the total will be 8 points. Round the result to hundredths.
Solution.
Dice are cubes with 6 sides. The first die can roll 1, 2, 3, 4, 5 or 6 points. Each scoring option corresponds to 6 scoring options on the second die.
Those. total various options 6x6 = 36.
The options (experiment outcomes) will be as follows:
1; 1 1; 2 1; 3 1; 4 1; 5 1; 6
2; 1 2; 2 2; 3 2; 4 2; 5 2; 6
etc. ...............................
6; 1 6; 2 6; 3 6; 4 6; 5 6; 6
Let's count the number of outcomes (options) in which the sum of the points of two dice is 8.
2; 6 3; 5; 4; 4 5; 3 6; 2.
There are 5 options in total.
Let's find the probability: 5/36 = 0.138 ≈ 0.14.
Answer: 0.14.
There are only 55 tickets in the collection of tickets for biology, 11 of them contain a question on botany. Find the probability that a student will get a question on botany on a randomly selected exam ticket.
Solution:
The probability that a student will get a question on botany on a randomly selected exam ticket is 11/55 = 1/5 = 0.2.
Answer: 0.2.
20 athletes are participating in the gymnastics championship: 8 from Russia, 7 from the USA, the rest from China. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete competing first is from China.
Solution.
A total of 20 athletes are participating,
of which 20 – 8 – 7 = 5 athletes from China.
The probability that the athlete competing first will be from China is 5/20 = 1/4 = 0.25.
Answer: 0.25.
Scientific conference carried out in 5 days. A total of 75 reports are planned - the first three days contain 17 reports, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by drawing lots. What is the probability that Professor M.'s report will be scheduled for the last day of the conference?
Solution:
On the last day of the conference it is planned
(75 – 17 × 3) : 2 = 12 reports.
The probability that Professor M.'s report will be scheduled for the last day of the conference is 12/75 = 4/25 = 0.16.
Answer: 0.16.
Before the start of the first round of the badminton championship, participants are randomly divided into playing pairs using lots. In total, 26 badminton players are participating in the championship, including 10 participants from Russia, including Ruslan Orlov. Find the probability that in the first round Ruslan Orlov will play with any badminton player from Russia?
Solution:
It must be taken into account that Ruslan Orlov must play with some badminton player from Russia. And Ruslan Orlov himself is also from Russia.
The probability that Ruslan Orlov will play with any badminton player from Russia in the first round is 9/25 = 36/100 = 0.36.
Answer: 0.36.
Dasha throws the dice twice. She got a total of 8 points. Find the probability that on the first roll you get 2 points.
Solution.
A total of 8 points should appear on the two dice. This is possible if there are the following combinations:
There are 5 options in total. Let's count the number of outcomes (options) in which 2 points were obtained on the first throw.
This is option 1.
Let's find the probability: 1/5 = 0.2.
Answer: 0.2.
There are 20 teams participating in the World Championship. Using lots, they need to be divided into five groups of four teams each. There are cards with group numbers mixed in the box:
1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5.
Team captains draw one card each. What is the probability that the Russian team will be in the third group.
Solution:
There are 20 teams in total, 5 groups.
Each group has 4 teams.
So, there are 20 total outcomes, the ones we need are 4, which means the probability of getting the desired outcome is 4/20 = 0.2.
Answer: 0.2.
Two factories produce the same glass for car headlights. The first factory produces 45% of these glasses, the second – 55%. The first factory produces 3% of defective glass, and the second – 1%. Find the probability that glass accidentally purchased in a store will be defective.
Solution:
The probability that the glass was purchased at the first factory and is defective:
r 1 = 0.45 · 0.03 = 0.0135.
The probability that the glass was purchased from a second factory and is defective:
r 2 = 0.55 · 0.01 = 0.0055.
Therefore, according to the total probability formula, the probability that glass accidentally purchased in a store will be defective is equal to
p = p 1 + p 2 = 0,0135 + 0,0055 = 0,019.
Answer: 0.019.
If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.52. If A. plays black, then A. wins against B. with probability 0.3.
Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.
Solution:
The possibility of winning the first and second games are independent of each other. The probability of a product of independent events is equal to the product of their probabilities:
p = 0.52 · 0.3 = 0.156.
Answer: 0.156.
A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hits the targets the first three times and misses the last two times. Round the result to hundredths.
Solution:
The result of each next shot does not depend on the previous ones. Therefore, the events “hit on the first shot,” “hit on the second shot,” etc. independent.
The probability of each hit is 0.8. This means that the probability of a miss is 1 – 0.8 = 0.2.
1 shot: 0.8
2 shot: 0.8
3 shot: 0.8
4 shot: 0.2
5 shot: 0.2
Using the formula for multiplying the probabilities of independent events, we find that the desired probability is equal to:
0,8 ∙ 0,8 ∙ 0,8 ∙ 0,2 ∙ 0,2 = 0,02048 ≈ 0,02.
Answer: 0.02.
There are two payment machines in the store. Each of them can be faulty with a probability of 0.05, regardless of the other machine. Find the probability that at least one machine is working.
Solution:
Let's find the probability that both machines are faulty.
These events are independent, the probability of their occurrence is equal to the product of the probabilities of these events:
0.05 · 0.05 = 0.0025.
An event consisting in the fact that at least one machine is working, the opposite.
Therefore, its probability is equal
1 − 0,0025 = 0,9975.
Answer: 0.9975.
Cowboy John has a 0.9 chance of hitting a fly on the wall if he fires a zeroed revolver. If John fires an unfired revolver, he hits the fly with probability 0.2. There are 10 revolvers on the table, only 4 of which have been shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots the fly. Find the probability that John misses.
Solution:
The probability that John will miss if he grabs a zeroed revolver is:
0.4 (1 − 0.9) = 0.04
The probability that John will miss if he grabs an unfired revolver is:
0.6 · (1 − 0.2) = 0.48
These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events:
0,04 + 0,48 = 0,52.
Answer: 0.52.
During artillery fire, the automatic system fires a shot at the target. If the target is not destroyed, the system fires a second shot. Shots are repeated until the target is destroyed. The probability of destroying a certain target with the first shot is 0.4, and with each subsequent shot it is 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98?
Solution:
You can solve the problem “by action”, calculating the probability of surviving after a series of consecutive mistakes:
P(1) = 0.6;
P(2) = P(1) 0.4 = 0.24;
P(3) = P(2) 0.4 = 0.096;
P(4) = P(3) 0.4 = 0.0384;
P(5) = P(4) 0.4 = 0.01536.
The latter probability is less than 0.02, so five shots at the target is enough.
Answer: 5.
There are 26 people in the class, among them two twins - Andrey and Sergey. The class is randomly divided into two groups of 13 people each. Find the probability that Andrey and Sergey will be in the same group.
Solution:
Let one of the twins be in some group.
Together with him, 12 people from the 25 remaining classmates will be in the group.
The probability that the second twin will be among these 12 people is
P = 12: 25 = 0.48.
Answer: 0.48.
The picture shows a labyrinth. The spider crawls into the maze at the Entrance point. The spider cannot turn around and crawl back, so at each branch the spider chooses one of the paths along which it has not yet crawled. Assuming that the choice of the further path is purely random, determine with what probability the spider will come to exit D.
Solution:
At each of the four marked forks, the spider can choose either the path leading to exit D or another path with probability 0.5. These are independent events, the probability of their occurrence (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of arriving at exit D is (0.5) 4 = 0,0625.