To get through next round competition, the football team needs to score
at least 9 points in two games. If the team wins, they get 5 points,
in case of a draw - 4 points if loses - 0 points. Find the probability
that the team will be able to reach the next round of the competition. Consider
that in every game the probabilities of winning and losing are equal 0,4 .

Obviously, the team can not lose. Both draws won't suit her either. What's left?
1) Win both times. 2) Win only once and draw the second game.

The probability of winning is 0,4 . The probability of winning both times is 0.4 0.4 = 0.16.

The probability of a draw is 1 - 0,4 - 0,4 = 0,2 . What is the probability of once
draw and win once? 0.4 0.2? No, it is equal 0.4 0.2 + 0.2 0.4.
The fact is that you can win the first game, or you can win the second game, this is important.
We now consider the probability of reaching the next round: 0,16 + 0,08 + 0,08 = 0,32 .

Answer: 0,32

We illustrate the solution graphically using the table 10 x 10 from 100 cells:

Red indicates victory, marsh indicates loss, and blue indicates draw.

Gray cell: first game - loss, second game - loss.
Red cage: the first game is a loss, the second game is a victory.
Green cell: the first game is a win, the second game is a draw.
Blue cell: first game - draw, second game - draw.

In this diagram, we will color both victories in yellow,
in blue - one win and one draw.

And one more illustrative scheme. At the first moment, the team has
three scenarios: win, draw and lose.

In each case, there are three options for the outcome of the second game.

We will leave only those branches that the team is satisfied with.

Calculate the probability of each branch and add them.

Quest Source: Task 4. To go to the next round of the competition, the football team needs to score

Task 4. To advance to the next round of the competition, a football team needs to score at least 4 points in two games. If a team wins it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Decision.

Since the probabilities of winning and losing are 0.4 each, the probability of a draw is 1-0.4-0.4=0.2. Thus, a football team can advance to the next round with the following non-joint outcomes:

Won the first game and won the second game;

Draw the first game and win the second game;

Won the first game and drew the second game.

The probability of the first outcome is . Probability of the second outcome . Probability of the third outcome . The desired probability of reaching the next round of the competition is equal to the sum of the probabilities of these three independent outcomes.

"Problems about a circle and a circle" - 3. The perimeter of a regular triangle inscribed in a circle is 6 | / 3 dm. Find the area of ​​the shaded figure. Problem solving. What is the area of ​​the circular sector corresponding to the given arc? Circumference and area of ​​a circle.

"Circle and circle geometry" - Did you know: A figure bounded by a circle is called a circle. Circle. A circle. L=2?R. Area of ​​a circle. History reference. Circle and circle. Circumference.

"Problems in Euler circles" - 8 people speak English and German at the same time, German. 70 children rested in the children's camp. English. This means that 10 - 3 = 7 (persons) speak English and French. 11. So, English and German are spoken by 8 - 3 = 5 (persons). In England and Italy - five, in England and France - 6, in all three countries - 5 employees.

"Circumference and Circle" - Circle. MATHEMATICS-5 Thematic planning Course of the lesson Author Resources. Favorite activity is reading. Training exercises. The point is called the center of the circle. Category - the highest. Part of a circle is called an arc. Arc.

"Circle and circle lesson" - Circle and circle methodical development. Additional tasks. Updating of basic knowledge. Find the radius of the circle passing through the centers of these circles. Conclusion. Equipment: board, chalk, drawing tools, cards with additional tasks. Tasks. Learning new material Consolidation of the studied material Summing up the lesson.

Bets on the passage of the team in the line of bookmakers are very common. Perhaps, now all bookmakers offer bets on the passage to the following types sports:

• Football. Basically, these are major world-class competitions: World Championship, European Championship, Confederations Cup, Club World Championship, Champions League, Europa League, Cup competitions of different football countries, etc.
• Basketball. A bet on the passage of a basketball team means the victory of one of the basketball teams over its opponent, taking into account overtime. It could also mean winning by the points difference the club needs to advance to the next round of the cup competition.
• Hockey. Similarly to basketball betting, the team wins in overtime in case of a draw in regular time. If we are talking about the playoffs, then the passage of the team to the next round is the object of the so-called bet on the passage (team to qualify).

Let's consider in more detail the bets on the pass in football. Bookmakers offer this type of bet only on matches that are played according to the Olympic system, i.e. right through. Such bets are not accepted for matches of regular championships, and there are no such bets in the betting lines. Cup competitions can consist of one match - for example, the FA Cup, the Italian Cup or two games - the Spanish Cup, etc. Accordingly, the bet on the team's passage to the next round will be made taking into account one or two matches, including a penalty shootout.

At major international tournaments, a group tournament is short-lived and a player can place a bet in the office not only on the knockout stage (1/8, 1/4), but also on the exit of the selected team from the group. By and large, this category of bets can also be attributed to bets on the passage.

Another feature of bets on the passage of the team to the next stage in football are the odds that bookmakers set in their own. The odds for winning two matches in football can be an order of magnitude higher than in hockey or basketball. For example, if one of the teams won the first match, then the odds for the second club to advance to the next stage of the competition will be overstated, which allows the player to earn more on a successful bet.

Pass betting in basketball or hockey is different from football due to the rules of the game. In basketball and hockey matches, a draw can only be in regular time, and the winner is determined in overtime (or in a shootout in hockey).

In basketball and hockey, you can bet on winning a series of games that start in the playoffs. According to the regulations of the league, cup or championship, a series can go up to 3 or 4 victories of one of the teams, respectively, and the bet will cover all these games.

In hockey or basketball, bets on the run are a kind of insurance for a player who is not sure that the team will win in regular time. The odds of bookmakers will be lower than for the main outcome, but the chances that the bet will play will increase.

TB(4)

What does a sports bet on total over 4 mean? What is TB(4) in bookmaker bets? How to understand what is total...

Challenge B10 Prototype (#320188) To advance to the next round of the competition, a football team needs to score at least 4 points in two games. If a team wins it gets 3 points, in case of a draw - 1 point, if it loses - 0 points. Find the probability that the team will be able to advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Task B10 (No. 321491) There are 33 students in the class, two of them are friends - Mikhail and Vadim. The class is randomly divided into 3 equal groups. Find the probability that Mikhail and Vadim will be in the same group.

Decision. According to the question of the problem, we are interested in the distribution of two guys into three groups (for convenience, we number these groups: group 1, group 2 and group 3). Therefore, the possible outcomes of the experiment under consideration are:

U 1 \u003d (Mikhail in the first group, Vadim in the second group) \u003d (M1, B2),

U 2 \u003d (Mikhail in the first group, Vadim in the third group) \u003d (M1, B3),

U 3 \u003d (Mikhail in the first group, Vadim in the first group) \u003d (M1, B1),

U 4 \u003d (Mikhail in the second group, Vadim in the first group) \u003d (M2, B1),

U 5 \u003d (Mikhail in the second group, Vadim in the second group) \u003d (M2, B2),

U 6 \u003d (Mikhail in the second group, Vadim in the third group) \u003d (M2, B3),

U 7 \u003d (Mikhail in the third group, Vadim in the first group) \u003d (M3, B1),

U 8 \u003d (Mikhail in the third group, Vadim in the second group) \u003d (M3, B2),

U 9 ​​\u003d (Mikhail in the third group, Vadim in the third group) \u003d (M3, B3),

Thus, the set U of all outcomes of the experiment under consideration consists of nine elements U= (U 1 , U 2 , U 3 ,… U 7 , U 9 ), and the event A - "Mikhail and Vadim were in the same group" - is favored by only three outcomes - U 3 , U 5 and U 9 . Let's find the probability of each of these outcomes. Since, according to the condition of the problem, a class of 33 people is randomly divided into three equal groups, then in each such group there will be 11 students of this class. Solely for the sake of convenience in solving the problem, imagine 33 chairs arranged in a row, on the seats of which numbers are written: the number 1 is written on the first 11 chairs, the number 2 is written on the next 11 chairs, and the number 3 is written on the last eleven chairs. The probability that Mikhail will get a chair with the number 1, equal to (11 chairs with the number 1 out of the total number of chairs). After Mikhail sat down on the chair with the number 1, there are only 32 chairs left, among which there are only 10 chairs with the number 1, therefore, the probability that Vadim will get the chair with the same number 1 is . Therefore, the probability of the outcome U 3 =(Mikhail in the first group, Vadim in the first group)=(M1, B1) is equal to the product and is equal to . Arguing in a similar way, we find the probabilities of the outcomes U 5 and U 9 . We have, P(U 5)=P(U 9)=P(U 3)=.

Thus, P(A)=P(U 3)+P(U 5)+P(U 9)=.

Answer. 0.3125.

Comment. Many students, having compiled a set U of possible outcomes of the experiment under consideration, find the desired probability as a quotient of dividing the number of outcomes U 3 , U 5 and U 9 that favor the event A to the number of possible outcomes U 1 , U 2 , U 3 ,… U 7 , U 9 , i.e. P(A)=. The fallacy of such a decision lies in the fact that the outcomes of the experiment under consideration are not equally probable. Indeed, P(U 1)=, and P(U 3)=.

Decision. According to the condition of the problem, the team plays two games, and the result of each such game can be either a win, or a loss, or a draw. So, the possible outcomes of this experience are: U 1 \u003d (B; B), hereinafter B - the team won the game, P - the team lost the game, H - the team played a draw, U 2 \u003d (B; H), U 3 = (V; P), U 4 = (P; V), U 5 = (P; N), U 6 = (P; P), U 7 = (N; N), U 8 = (N; P), U 8 \u003d (N; V). Thus, the set of possible outcomes of the experiment under consideration consists of 9 elements, and the event C - “the football team went to the next round of competitions” is favored by the outcomes U 1 = (B; B), U 2 = (B; H) and U 8 = ( N; C), since the occurrence of each of these outcomes guarantees the required number of points to enter the next round of the competition. Let's find the probabilities of outcomes U 1 = (B; B), U 2 = (B; H) and U 8 = (H; B). According to the condition of the problem, the probabilities of winning and losing are equal to 0.4, since the result of one game can be either a win, or a loss, or a draw, then the probability of a draw is equal to the difference 1-(U 2 +U 8) and is equal to 0.2. So, according to the theorem on the probability of the product of independent events, P(U 1)=0.40.4=0.16 and P(U 2)=P(U 8)=0.40.2=0.08. So, the desired probability is: P (C) \u003d P (U 1) + P (U 2) + P (U 8) \u003d 0.16 + 0.08 + 0.08 \u003d 0.32.