https://r2.erweima.ai/mathbot/image/172693737152121xzemqe.jpg

Let's go through the math problems from the image step by step:

### Q17:
This question involves a statistical series \( X = (x_1, x_2, \dots, x_n) \) with finite values in the intervals \( ]-\infty, -1] \cup [1, +\infty[ \). The question asks about the standard deviation of \( X \), and you marked the answer as **C**: "prend deux valeurs opposées" (takes two opposite values). This option seems to refer to the characteristics of the dataset, and the answer could be based on the specific context of the data distribution.

### Q18:
This is a probability question about a limited-access competition with 75 candidates. The probability of randomly selecting a candidate who passes the exam is calculated as follows:

- 20 candidates are called, and 4 times this number (i.e., 80 candidates) are selected in total. 
- Out of the 80, only 75 candidates succeed.
- The probability is given as \( P = \frac{20}{80} = 0.25 \).

Thus, the answer you selected, **A: 0.2500**, is correct.

### Q19:
This is a Bayesian probability question. It involves three machines, \( M_1 \), \( M_2 \), and \( M_3 \), producing bolts with respective defect rates of 0.3%, 0.8%, and 1%. The total number of bolts from machines \( M_1 \) and \( M_2 \) in the urn is 500 and 350, respectively. The probability of picking a defective bolt is 0.0058. The question asks for the probability that a defective bolt comes from machine \( M_1 \).

Using Bayes' Theorem:

\[
P(M_1 | D) = \frac{P(D | M_1) P(M_1)}{P(D)}
\]

Where:
- \( P(D | M_1) \) is the probability of a defect from machine 1 (0.003),
- \( P(M_1) = \frac{500}{1000} = 0.5 \),
- \( P(D) = 0.0058 \) is the total probability of a defective bolt.

The result should align with the answer choices, and you selected **A: 0.2586**, which looks reasonable.

### Q20:
This problem deals with a simple probability involving red and black balls. The total number of balls is unknown, but we know that there are 16 red balls, and the probability of selecting a red ball is \( \frac{1}{4} \).

Let the total number of balls be \( x \). Since the probability of selecting a red ball is \( \frac{1}{4} \), we have:

\[
\frac{16}{x} = \frac{1}{4}
\]

Solving for \( x \):

\[
x = 16 \times 4 = 64
\]

Thus, the total number of balls is 64. The number of black balls is:

\[
64 - 16 = 48
\]

Therefore, the correct answer is **C: 48**.

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Would you like further clarification on any of these questions?

Here are 5 related questions to explore:

1. How do you compute the variance of a statistical series?
2. What is the standard deviation's relationship with the variance?
3. How does Bayes' theorem work in probability problems involving defective parts?
4. Can you explain the difference between conditional probability and total probability?
5. What are the key principles behind combinatorial probability calculations?

**Tip**: In probability problems, carefully identifying given and conditional probabilities is crucial to applying the right theorem or formula.

The questions cover topics in statistics, probability, and Bayesian theorem, specifically focusing on standard deviation, probability of passing a contest, defective products from machines, and probability involving red and black balls.

Step-by-step solutions to problems involving standard deviation, passing probability in a contest, Bayesian analysis of defective products, and probability involving colored balls. Understand key mathematical concepts like Bayes' Theorem and combinatorial probability with these problems suitable for Grades 10-12.

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