An example of solving problems in probability theory from the USE

Mathematics is quite a versatile subject. Now we propose to consider an example of solving problems in probability theory, which is one of the directions of mathematics. Let me immediately say that the ability to solve such tasks will be a big plus when passing a unified state examination. Problems in the theory of probability USE contains in part B, which, accordingly, is evaluated higher than the test tasks of group A.

Random events and their probability

It is this group that is being studied by this science. What is a random event? When conducting any experiment, we get the result. There are tests that have a certain result with a probability of one hundred percent or zero percent. Such events are called reliable and impossible, respectively. We are interested in those that can happen or not, that is, random. To find the probability of an event, use the formula P = m / n, where m are the options that satisfy us, and n - all possible outcomes. Now consider an example of solving problems in probability theory.

Combinatorics. Tasks

Probability theory includes the next section, assignments of this type are often found on the exam. Condition: the student group consists of twenty-three people (ten men and thirteen girls). It is necessary to choose two people. How many ways can I choose two guys or girls? By the condition, we need to find two girls or two men. We see that the wording is prompted by the right decision:

1. Find the number of ways to choose men.
2. Then the girls.
3. We add the obtained results.

Perform the first action: = 45. Next girls: and we get 78 ways. Last action: 45 + 78 = 123. It turns out that there are 123 ways to choose a same-sex couples such as a village elder and a deputy, not important to girls or men.

Classical tasks

We have considered an example from combinatorics, we proceed to the next stage. Let us consider an example of solving problems in probability theory to find the classical probability of the origin of an event.

Condition: Before you is a box, inside there are balls of different colors, namely fifteen white, five red and ten black. You are offered to pull one at random. What is the probability that you will take the ball: 1) white; 2) red; 3) black.

Our advantage is the calculation of all possible options, in this example we have thirty of them. Now we have found n. Denote by the letter A the extracted white ball, we get m equals fifteen - these are the successful outcomes. Using the basic rule of finding the probability, we find: P = 15/30, that is, 1/2. With such a probability we will get a white ball.

In a similar way, we find B - red balls and C - black. P (B) will be 1/6, and the probability of the event C = 1/3. To check if the problem is solved correctly, you can use the rule of the sum of probabilities. Our complex consists of events A, B and C, in the sum they must be one. As a result of the test, we got the very desired value, and therefore the task was solved correctly. Answer: 1) 0.5; 2) 0.17; 3) 0.33.

Unified State Examination

Let us consider an example of solving problems in probability theory from USE tickets. Often there are examples with a coin toss. We offer to disassemble one of them. The coin is thrown three times, what is the probability that the eagle will fall twice and once the tails. Let's reformulate the task: throw three coins at the same time. For simplicity, we compile tables. For one coin, everything is clear:

Two coins:

With two coins, we already have four outcomes, but with three, the task is slightly more complicated, and there are eight outcomes.

Now we count the options that suit us: 2; 3; 4. We get that three of the eight options satisfy us, that is, the answer is 3/8.