Question

Из подготовки к ЕГЭ! Помогите, пожалуйста :) Перед началом первого тура чемпионата по настольному теннису участников разбивают на игровые пары случайным образом с помощью жребия. Всего в чемпионате участвует 26 спортсменов, среди которых 7 спортсменов из России, в том числе Георгий Бочкин. Найдите вероятность того, что в первом туре Георгий Бочкин будет играть с каким–либо спортсменом из России.

Answer 1

Тут сложнее)
26-1=25-спортсменов,не считая Георгия Бочкина
7-1=6 спортсменов России,не считая Георгия Бочкина
6:25=0,24
Ответ:0,24 вероятность

Answer 2

Probability of Georgiy Bochkin playing against a Russian athlete in the first round

To find the probability of Georgiy Bochkin playing against a Russian athlete in the first round of the table tennis championship, we need to determine the number of possible pairings that include Georgiy Bochkin and a Russian athlete, and then divide it by the total number of possible pairings.

Given that there are 26 participants in the championship, including 7 athletes from Russia, we can calculate the probability as follows:

1. Determine the number of possible pairings that include Georgiy Bochkin and a Russian athlete: - Georgiy Bochkin can be paired with any of the 25 remaining participants (excluding himself). - There are 7 Russian athletes to choose from. - Therefore, the number of possible pairings with Georgiy Bochkin and a Russian athlete is 25 * 7 = 175.

2. Determine the total number of possible pairings: - Each participant can be paired with any of the remaining 25 participants (excluding themselves). - Therefore, the total number of possible pairings is 25 * 24 = 600.

3. Calculate the probability: - Divide the number of possible pairings that include Georgiy Bochkin and a Russian athlete (175) by the total number of possible pairings (600). - The probability is 175/600, which can be simplified to 7/24.

Therefore, the probability of Georgiy Bochkin playing against a Russian athlete in the first round is 7/24.

Please note that the information provided is based on the assumption that the pairings are randomly determined using a draw, as mentioned in the question.