Question

Из 30 студентов 10 имеют спортивные разряды. Какова вероятность того,что три наугад выбраные студенты будут разрядниками?

Answer 1

Вероятность - это отношение благоприятного числа исходов к общему числу исходов.
Общее число исходов - выбор трёх студентов из 30:
$C_{30}^3= \frac{30!}{3!(30-3)!}= \frac{30!}{3!27!}= \frac{28*29*30}{1*2*3}=5*28*29=4060$

Благоприятное число исходов - выбор трёх студентов из числа десяти разрядников
$C_{10}^3= \frac{10!}{3!(10-3)!}= \frac{10!}{3!7!}= \frac{8*9*10}{1*2*3}=120$

Вероятность события  $P(A)= \frac{120}{4060}= \frac{6}{203} \approx 0,03$

Answer 2

Calculating the Probability of Selecting Three Students with Sports Ranks

To calculate the probability of selecting three students with sports ranks out of 30 students, we can use the formula for calculating combinations. The formula for combinations is:

nCr = n! / (r! * (n - r)!)

Where: - n is the total number of items - r is the number of items to choose

In this case, n = 30 (total students) and r = 3 (students to be chosen). The number of students with sports ranks is 10.

Applying the Formula

Using the formula for combinations, the probability can be calculated as follows:

P = (Number of ways to choose 3 students with sports ranks) / (Total number of ways to choose 3 students)

Number of ways to choose 3 students with sports ranks (m): - nCr = 10C3 = 10! / (3! * (10 - 3)!) = 120

Total number of ways to choose 3 students (n): - nCr = 30C3 = 30! / (3! * (30 - 3)!) = 4060

P = m / n = 120 / 4060 ≈ 0.0296

Conclusion

The probability of randomly selecting three students with sports ranks from a group of 30 students is approximately 0.0296.