Question

Помогите пожалуйста.На экзамен вынесено 20 вопросов, из которых студент выучил 10. Для успешной сдачи экзамена достаточно ответить на 2 из 3 предложенных вопроса. Какова вероятность успешной сдачи экзамена этим студентом?

Answer 1

$P=\frac{2}{10}*\frac{1}{20-10}+\frac{3}{10}=\frac{2}{100}+\frac{3}{10}=\\\\=0,02+0,3=0,32$

***Пояснение:
Для успешной сдачи экзамена достаточно ответить на 2 вопроса из 3-х. Значит при успешной сдаче студенту достанется  2 вопроса из 10-ти выученных и один вопрос из 20-10=10 невыученных или все три вопроса из десяти выученных.

Answer 2

Calculation of the Probability of Passing the Exam

To calculate the probability of passing the exam, we need to consider the number of questions the student has studied and the number of questions they need to answer correctly.

Given: - Total number of questions: 20 - Number of questions studied: 10 - Number of questions needed to pass: 2 out of 3

To calculate the probability, we can use the concept of combinations. The formula for calculating combinations is:

C(n, r) = n! / (r!(n-r)!)

Where: - n is the total number of items - r is the number of items chosen

In this case, we need to calculate the probability of answering 2 out of 3 questions correctly. Let's break down the calculation step by step:

1. Calculate the number of ways to choose 2 questions out of the 3 offered: - C(3, 2) = 3! / (2!(3-2)!) = 3

2. Calculate the number of ways to choose the remaining questions from the questions studied: - C(10, 0) = 10! / (0!(10-0)!) = 1 (since we don't need to choose any questions from the studied ones)

3. Calculate the number of ways to choose the remaining questions from the questions not studied: - C(10, 1) = 10! / (1!(10-1)!) = 10

4. Calculate the total number of possible combinations: - Total combinations = C(3, 2) * C(10, 0) * C(10, 1) = 3 * 1 * 10 = 30

5. Calculate the probability of passing the exam: - Probability = Number of successful combinations / Total combinations - Probability = 3 / 30 = 0.1 or 10%

Therefore, the probability of this student passing the exam is 10%.