Solving dice throwing problems.
In all tasks B6 on probability theory, which are presented in Open task bank for, you need to find probability any event.
You only need to know one formula, which is used to calculate probability:
In this formula p - probability of event,
k- the number of events that “satisfy” us, in language probability theory they're called favorable outcomes.
n- the number of all possible events, or number of all possible outcomes.
Obviously, the number of all possible events is greater than the number of favorable outcomes, so probability is a value that is less than or equal to 1.
If probability event value is 1, which means that this event will definitely happen. Such an event is called reliable. For example, the fact that after Sunday there will be Monday is, unfortunately, a reliable event and its probability is equal to 1.
The greatest difficulties in solving problems arise precisely with finding the numbers k and n.
Of course, as when solving any problems, when solving problems on probability theory You need to carefully read the condition in order to correctly understand what is given and what you need to find.
Let's look at several examples of solving problems from from Open Bank tasks for .
Example 1. In a random experiment, two dice are rolled. Find the probability that the total will be 8 points. Round the result to hundredths.
Let the first die roll one point, then the second die can roll a 6 various options. So since the first die has 6 different sides, total number different options equals 6x6=36.
But we are not satisfied with everything. According to the conditions of the problem, the sum of the points drawn should be equal to 8. Let's create a table of favorable outcomes:
We see that the number of outcomes that suit us is 5.
Thus, the probability that a total of 8 points will appear is 5/36=0.13(8).
Once again we read the question of the problem: we need to round the result to hundredths.
Let's remember rounding rule.
We need to round to the nearest hundredth. If in the next place after the hundredths (that is, in the thousandths place) there is a number that is greater than or equal to 5, then we add 1 to the number in the hundredths place; if this number is less than 5, then we leave the number in the hundredths place unchanged.
In our case, the number in the thousandths place is 8, so we increase the number 3, which is in the hundredths place, by 1.
So, p=5/36 ≈0.14
Answer: 0.14
Example 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.
In this problem, the number of possible outcomes is 20 - this is the number of all athletes.
Let's find the number of favorable outcomes. It is equal to the number of female athletes from China.
Thus,
Answer: 0.25
Example 3: On average, out of 1000 garden pumps sold, 5 leak. Find the probability that one pump randomly selected for control does not leak.
In this problem n=1000.
We are interested in pumps that do not leak. Their number is 1000-5=995. Those.
Left a reply Guest
With one dice the situation is indecently simple. Let me remind you that the probability is found by the formula P=m/n
P
=
m
n
, where n
n
is the number of all equally possible elementary outcomes of an experiment involving tossing a cube or dice, and m
m
- the number of outcomes that favor the event.
Example 1: The die is thrown once. What is the probability that an even number of points is rolled?
Since a die is a cube (they also say a regular die, that is, a balanced die so that it lands on all sides with equal probability), the cube has 6 faces (with a number of points from 1 to 6, usually indicated by points), then and the total number of outcomes in the problem n=6
n
=
6
. The only outcomes that favor the event are those where a side with 2, 4 or 6 points (even numbers only) appears; there are m=3 such sides
m
=
3
. Then the required probability is P=3/6=1/2=0.5
P
=
3
6
=
1
2
=
0.5
.
Example 2. A die is thrown. Find the probability of rolling at least 5 points.
We reason the same way as in the previous example. The total number of equally possible outcomes when throwing a die n=6
n
=
6
, and the condition “at least 5 points rolled”, that is, “either 5 or 6 points rolled” is satisfied by 2 outcomes, m=2
m
=
2
. The required probability is P=2/6=1/3=0.333
P
=
2
6
=
1
3
=
0.333
.
I don’t even see the point in giving more examples, let’s move on to two dice, where everything gets more interesting and complicated.
Two dice
When we're talking about For problems involving rolling 2 dice, it is very convenient to use a scoring table. Let us plot horizontally the number of points that fell on the first dice, and vertically the number of points that fell on the second dice. Let's get something like this (I usually do it in Excel, you can download the file below):
table of points for rolling 2 dice
What's in the table cells, you ask? And this depends on what problem we will solve. There will be a task about the sum of points - we will write the sum there, about the difference - we will write the difference and so on. Let's get started?
Example 3. 2 dice are thrown at the same time. Find the probability that the total will be less than 5 points.
First, let's look at the total number of outcomes of the experiment. when we threw one die, everything was obvious, 6 sides - 6 outcomes. There are already two dice here, so the outcomes can be represented as ordered pairs of numbers of the form (x,y)
x
,
y
, where x
x
- how many points were rolled on the first die (from 1 to 6), y
y
- how many points were rolled on the second dice (from 1 to 6). Obviously, there will be n=6⋅6=36 such pairs of numbers
n
=
6
⋅
6
=
36
(and exactly 36 cells in the table of outcomes correspond to them).
Now it's time to fill out the table. In each cell we enter the sum of the number of points rolled on the first and second dice and we get the following picture:
table of the sum of points when throwing 2 dice
Now this table will help us find the number of outcomes favorable to the event “a total of less than 5 points will appear.” To do this, we count the number of cells in which the sum value is less than 5 (that is, 2, 3 or 4). For clarity, let's color these cells, there will be m=6
m
=
6
:
table of total points less than 5 when throwing 2 dice
Then the probability is: P=6/36=1/6
P
=
6
36
=
1
6
.
Example 4. Two dice are thrown. Find the probability that the product of the number of points is divisible by 3.
We create a table of the products of the points rolled on the first and second dice. We immediately highlight in it those numbers that are multiples of 3:
Table of the product of points when throwing 2 dice
All that remains is to write down that the total number of outcomes is n=36
n
=
36
(see the previous example, the reasoning is the same), and the number of favorable outcomes (the number of shaded cells in the table above) m=20
m
=
20
. Then the probability of the event will be equal to P=20/36=5/9
P
=
20
36
=
5
9
.
As you can see, this type of problem, with proper preparation (let’s look at a couple more problems), can be solved quickly and simply. For variety, let’s do one more task with a different table (all tables can be downloaded at the bottom of the page).
Example 5: A die is thrown twice. Find the probability that the difference in the number of points on the first and second dice will be from 2 to 5.
Let's write down a table of point differences, highlight the cells in it in which the difference value will be between 2 and 5:
Table of difference of points when throwing 2 dice
So, the total number of equally possible elementary outcomes is n=36
n
=
36
, and the number of favorable outcomes (the number of shaded cells in the table above) m=10
m
=
10
. Then the probability of the event will be equal to P=10/36=5/18
P
=
10
36
=
5
18
.
So, in the case when we are talking about throwing 2 dice and a simple event, you need to build a table, select the necessary cells in it and divide their number by 36, this will be the probability. In addition to problems on the sum, product and difference of the number of points, there are also problems on the modulus of the difference, the smallest and largest number of points drawn (you will find suitable tables in the Excel file).
Problems 1.4 - 1.6
Problem condition 1.4
Indicate the error in the “solution” of the problem: two dice are thrown; find the probability that the sum of the points drawn is 3 (event A). "Solution". There are two possible outcomes of the test: the sum of the points drawn is 3, the sum of the points drawn is not equal to 3. Event A is favored by one outcome, the total number of outcomes is two. Therefore, the desired probability is equal to P(A) = 1/2.
Solution to Problem 1.4
The error in this “solution” is that the outcomes in question are not equally possible. Correct solution: The total number of equally possible outcomes is equal (each number of points rolled on one die can be combined with all numbers of points rolled on another die). Among these outcomes, only two outcomes favor the event: (1; 2) and (2; 1). This means that the required probability 
Answer:
Problem condition 1.5
Two dice are thrown. Find the probabilities of the following events: a) the sum of the points drawn is seven; b) the sum of the points drawn is eight, and the difference is four; c) the sum of the points drawn is eight, if it is known that their difference is four; d) the sum of the rolled points is five, and the product is four.
Solution to problem 1.5
a) Six options on the first die, six on the second. Total options: (according to the product rule). Options for a sum equal to 7: (1.6), (6.1), (2.5), (5.2), (3.4), (4.3) - six options in total. Means,
b) Only two suitable options: (6.2) and (2.6). Means,
c) There are only two suitable options: (2,6), (6,2). But in total possible options 4: (2.6), (6.2), (1.5), (5.1). Means, .
d) For a sum equal to 5, the following options are suitable: (1.4), (4.1), (2.3), (3.2). The product is 4 for only two options. Then
Answer: a) 1/6; b) 1/18; c) 1/2; d) 1/18
Problem condition 1.6
A cube, all edges of which are colored, is sawn into a thousand cubes of the same size, which are then thoroughly mixed. Find the probability that the cube drawn by luck has colored faces: a) one; b) two; at three o'clok.
Solution to problem 1.6
A total of 1000 cubes were formed. Cubes with three colored faces: 8 (these are corner cubes). With two colored faces: 96 (since there are 12 edges of a cube with 8 cubes on each edge). Dice with colored edges: 384 (since there are 6 faces and there are 64 cubes on each face). All that remains is to divide each quantity found by 1000.
Answer: a) 0.384; b) 0.096 c) 0.008
Another popular problem in probability theory (along with the coin toss problem) is dice tossing problem.
Usually the task sounds like this: one or more dice are thrown (usually 2, less often 3). You need to find the probability that the number of points is 4, or the sum of the points is 10, or the product of the number of points is divisible by 2, or the numbers of points differ by 3, and so on.
The main method for solving such problems is to use the classical probability formula, which we will analyze using examples below.
After familiarizing yourself with the solution methods, you can download a super-useful solution for throwing 2 dice (with tables and examples).
One dice
With one dice the situation is indecently simple. Let me remind you that the probability is found by the formula $P=m/n$, where $n$ is the number of all equally possible elementary outcomes of an experiment with tossing a cube or dice, and $m$ is the number of those outcomes that favor the event.
Example 1. The die is thrown once. What is the probability that an even number of points is rolled?
Since the die is a cube (they also say fair dice, that is, the cube is balanced, so it lands on all sides with the same probability), the cube has 6 sides (with a number of points from 1 to 6, usually designated points), then the total number of outcomes in the problem is $n=6$. The only outcomes that favor the event are those where a side with 2, 4 or 6 points (even ones only) appears; there are $m=3$ of such sides. Then the desired probability is equal to $P=3/6=1/2=0.5$.
Example 2. The dice are thrown. Find the probability of rolling at least 5 points.
We reason the same way as in the previous example. The total number of equally possible outcomes when throwing a die is $n=6$, and the condition “at least 5 points rolled up”, that is, “either 5 or 6 points rolled up” are satisfied by 2 outcomes, $m=2$. The required probability is $P=2/6=1/3=0.333$.
I don’t even see the point in giving more examples, let’s move on to two dice, where everything gets more interesting and complicated.
Two dice
When it comes to problems involving rolling 2 dice, it is very convenient to use points table. Let us plot horizontally the number of points that fell on the first dice, and vertically the number of points that fell on the second dice. Let's get something like this (I usually do it in Excel, you can download the file):
What's in the table cells, you ask? And this depends on what problem we will solve. There will be a task about the sum of points - we will write the sum there, about the difference - we will write the difference and so on. Let's get started?
Example 3. 2 dice are thrown at the same time. Find the probability that the total will be less than 5 points.
First, let's look at the total number of outcomes of the experiment. when we threw one die, everything was obvious, 6 sides - 6 outcomes. There are already two dice here, so the outcomes can be represented as ordered pairs of numbers of the form $(x,y)$, where $x$ is how many points fell on the first dice (from 1 to 6), $y$ is how many points fell on the second dice (from 1 to 6). Obviously, the total number of such pairs of numbers will be $n=6\cdot 6=36$ (and they correspond to exactly 36 cells in the table of outcomes).
Now it's time to fill out the table. In each cell we enter the sum of the number of points rolled on the first and second dice and we get the following picture:
Now this table will help us find the number of outcomes favorable to the event “a total of less than 5 points will appear.” To do this, we count the number of cells in which the sum value is less than 5 (that is, 2, 3 or 4). For clarity, let’s color these cells, there will be $m=6$:
Then the probability is equal to: $P=6/36=1/6$.
Example 4. Two dice are thrown. Find the probability that the product of the number of points is divisible by 3.
We create a table of the products of the points rolled on the first and second dice. We immediately highlight in it those numbers that are multiples of 3:
All that remains is to write down that the total number of outcomes is $n=36$ (see the previous example, the reasoning is the same), and the number of favorable outcomes (the number of shaded cells in the table above) is $m=20$. Then the probability of the event will be equal to $P=20/36=5/9$.
As you can see, this type of problem, with proper preparation (let’s look at a couple more problems), can be solved quickly and simply. For variety, let’s do one more task with a different table (all tables can be downloaded at the bottom of the page).
Example 5. The dice are thrown twice. Find the probability that the difference in the number of points on the first and second dice will be from 2 to 5.
Let's write down a table of point differences, highlight the cells in it in which the difference value will be between 2 and 5:
So, the total number of equally possible elementary outcomes is $n=36$, and the number of favorable outcomes (the number of shaded cells in the table above) is $m=10$. Then the probability of the event will be equal to $P=10/36=5/18$.
So, in the case when we are talking about throwing 2 dice and a simple event, you need to build a table, select the necessary cells in it and divide their number by 36, this will be the probability. In addition to problems on the sum, product and difference of the number of points, there are also problems on the modulus of the difference, the smallest and largest number of points drawn (you will find suitable tables in).
Other problems about dice and cubes
Of course, the matter is not limited to the two classes of problems about throwing dice discussed above (they are simply the most frequently encountered in problem books and training manuals), there are others. For variety and understanding of the approximate solution method, we will analyze three more typical examples: for throwing 3 dice, for conditional probability and for Bernoulli’s formula.
Example 6. 3 dice are thrown. Find the probability that the total is 15 points.
In the case of 3 dice, tables are drawn up less often, since you will need as many as 6 pieces (and not one, as above), they get by by simply searching through the required combinations.
Let's find the total number of outcomes of the experiment. Outcomes can be represented as ordered triplets of numbers of the form $(x,y,z)$, where $x$ is how many points fell on the first die (from 1 to 6), $y$ is how many points fell on the second die (from 1 to 6), $z$ - how many points rolled on the third die (from 1 to 6). Obviously, the total number of such triples of numbers will be $n=6\cdot 6\cdot 6=216$ .
Now let’s select outcomes that give a total of 15 points.
$$ (3,6,6), (6,3,6), (6,6,3),\\ (4,5,6), (4,6,5), (5,4,6), (6,5,4), (5,6,4), (6,4,5),\\ (5,5,5). $$
We got $m=3+6+1=10$ outcomes. The required probability is $P=10/216=0.046$.
Example 7. 2 dice are thrown. Find the probability that the first die rolls no more than 4 points, provided that the total number of points is even.
The easiest way to solve this problem is to use the table again (everything will be clear), as before. We write out a table of the sums of points and select only cells with even values:
We get that, according to the conditions of the experiment, there are not 36, but $n=18$ outcomes (when the sum of points is even).
Now from these cells Let’s select only those that correspond to the event “no more than 4 points rolled on the first die” - that is, in fact, the cells in the first 4 rows of the table (highlighted in orange), there will be $m=12$.
The required probability $P=12/18=2/3.$
The same task can be done decide differently using the conditional probability formula. Let's enter the events:
A = The sum of the number of points is even
B = No more than 4 points rolled on the first die
AB = The sum of the number of points is even and no more than 4 points were rolled on the first die
Then the formula for the desired probability has the form: $$ P(B|A)=\frac(P(AB))(P(A)). $$ Finding probabilities. The total number of outcomes is $n=36$, for event A the number of favorable outcomes (see tables above) is $m(A)=18$, and for event AB - $m(AB)=12$. We get: $$ P(A)=\frac(m(A))(n)=\frac(18)(36)=\frac(1)(2); \quad P(AB)=\frac(m(AB))(n)=\frac(12)(36)=\frac(1)(3);\\ P(B|A)=\frac(P (AB))(P(A))=\frac(1/3)(1/2)=\frac(2)(3). $$ The answers were the same.
Example 8. The dice is thrown 4 times. Find the probability that an even number of points will appear exactly 3 times.
In the case when the dice throws several times, and the event is not about the sum, product, etc. integral characteristics, but only about number of drops of a certain type, you can use it to calculate the probability
Tasks for probability of dice no less popular than coin toss problems. The condition of such a problem usually sounds like this: when throwing one or more dice (2 or 3), what is the probability that the sum of the points will be equal to 10, or the number of points will be 4, or the product of the number of points, or the product of the number of points divided by 2 etc.
The application of the classical probability formula is the main method for solving problems of this type.
One die, probability.
The situation is quite simple with one dice.
is determined by the formula: P=m/n, where m is the number of outcomes favorable to the event, and n is the number of all elementary equally possible outcomes of the experiment with throwing a bone or cube.
Problem 1. The dice are thrown once. What is the probability of getting an even number of points?
Since the die is a cube (or it is also called a regular die, the die will land on all sides with equal probability, since it is balanced), the die has 6 sides (the number of points from 1 to 6, which are usually indicated by dots), this means that the problem has a total number of outcomes: n=6. The event is favored only by outcomes in which the side with even points 2,4 and 6 appears; the die has the following sides: m=3. Now we can determine the desired probability of the dice: P=3/6=1/2=0.5.
Task 2. The dice are thrown once. What is the probability that you will get at least 5 points?
This problem is solved by analogy with the example given above. When throwing a dice, the total number of equally possible outcomes is: n=6, and only 2 outcomes satisfy the condition of the problem (at least 5 points rolled up, that is, 5 or 6 points rolled out), which means m=2. Next, we find the required probability: P=2/6=1/3=0.333.
Two dice, probability.
When solving problems involving throwing 2 dice, it is very convenient to use a special scoring table. On it, the number of points that fell on the first dice is displayed horizontally, and the number of points that fell on the second dice is displayed vertically. The workpiece looks like this:
But the question arises, what will be in the empty cells of the table? It depends on the problem that needs to be solved. If the problem is about the sum of points, then the sum is written there, and if it’s about the difference, then the difference is written down, and so on.
First, you need to figure out what the total number of outcomes of the experiment will be. Everything was obvious when throwing one die, 6 sides of the die - 6 outcomes of the experiment. But when there are already two dice, the possible outcomes can be represented as ordered pairs of numbers of the form (x, y), where x shows how many points were rolled on the first dice (from 1 to 6), and y - how many points were rolled on the second dice (from 1 until 6). There will be a total of such number pairs: n=6*6=36 (in the table of outcomes they correspond exactly to 36 cells).
Now you can fill out the table; to do this, the number of points that fell on the first and second dice is entered in each cell. The completed table looks like this:
Using the table, we will determine the number of outcomes that favor the event “a total of less than 5 points will appear.” Let's count the number of cells in which the sum value will be less than the number 5 (these are 2, 3 and 4). For convenience, we paint over such cells; there will be m=6 of them:
Considering the table data, probability of dice equals: P=6/36=1/6.
Problem 4. Two dice were thrown. Determine the probability that the product of the number of points will be divisible by 3.
To solve the problem, let's make a table of the products of the points that fell on the first and second dice. In it, we immediately highlight the numbers that are multiples of 3:
We write down the total number of outcomes of the experiment n=36 (the reasoning is the same as in the previous problem) and the number of favorable outcomes (the number of cells that are shaded in the table) m=20. The probability of the event is: P=20/36=5/9.
Problem 5. The dice are thrown twice. What is the probability that the difference in the number of points on the first and second dice will be from 2 to 5?
To determine probability of dice Let's write down a table of point differences and select in it those cells whose difference value will be between 2 and 5:
The number of favorable outcomes (the number of cells shaded in the table) is m=10, the total number of equally possible elementary outcomes will be n=36. Determines the probability of the event: P=10/36=5/18.
In the case of a simple event and when throwing 2 dice, you need to build a table, then select the necessary cells in it and divide their number by 36, this will be considered a probability.
