Question

В классе 21 шестиклассник, среди них два друга - Митя и Петя. Класс случайным образом делят на три группы, по 7 человек в каждой. Найдите вероятность того, что Митя и Петя окажутся в одной и той же группе.

Answer 1

Для наглядности : пусть ребята тянут жребий. 
К примеру Митя попал в какую-то группу. Группа номер 1. 
Осталось 20 человек. Нужна вероятность, что Петя тоже попадет в группу номер 1. 
Т.к. это жребий - всего 20 бумажек. Из них на шести написанна "группа номер 1".
следовательно вероятность того, что Петя вытянет эту бумажку - 6/20=3/10.
Это и ответ.  

Answer 2

Problem Analysis

We have a class of 21 sixth-grade students, including two friends named Mitia and Petya. The class is randomly divided into three groups, with 7 students in each group. We need to find the probability that Mitia and Petya end up in the same group.

Solution

To solve this problem, we can consider the total number of ways the class can be divided into three groups and the number of ways Mitia and Petya can be in the same group.

1. Total number of ways to divide the class into three groups: - We have 21 students in total, and we want to divide them into three groups of 7 students each. - The order of the groups does not matter, so we can use combinations to calculate the number of ways to divide the class. - The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen. - In this case, we have 21 students and we want to choose 7 students for the first group, 7 students for the second group, and the remaining 7 students for the third group. - Therefore, the total number of ways to divide the class into three groups is: C(21, 7) * C(14, 7) = 116,280.

2. Number of ways Mitia and Petya can be in the same group: - Since Mitia and Petya are friends, we can consider them as a single entity. - We need to choose 6 more students to form a group with Mitia and Petya. - The remaining 14 students can be divided into two groups of 7 students each. - Therefore, the number of ways Mitia and Petya can be in the same group is: C(6, 6) * C(14, 7) = 3003.

3. Probability that Mitia and Petya end up in the same group: - The probability is given by the number of favorable outcomes (Mitia and Petya in the same group) divided by the total number of possible outcomes (total number of ways to divide the class into three groups). - Therefore, the probability is: 3003 / 116,280 ≈ 0.0258.

Answer

The probability that Mitia and Petya will end up in the same group is approximately 0.0258.