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The image shows a probability question written in Hebrew. The problem states that there are two white balls, two black balls, and two red balls in a box. The question asks for the probability of drawing a black ball from one of the random draws.

To solve this:

- The total number of balls is 6 (2 white + 2 black + 2 red).
- The probability of drawing a black ball is the number of black balls (2) divided by the total number of balls (6).

Thus, the probability is:

\[
P(\text{black}) = \frac{2}{6} = \frac{1}{3}.
\]

However, the closest option in the answer choices is \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{4}{9}\), and \(\frac{5}{9}\), which means the setup might involve more complexity than the simple calculation above, possibly based on conditional probabilities or repeated draws.

Would you like me to provide a more detailed breakdown or explore additional conditions? Here are some related questions:

1. How does conditional probability affect these types of problems?
2. What is the significance of replacement in probability?
3. How does this problem relate to combinatorics?
4. How would the probability change if more balls were added?
5. What happens if the number of draws increases?

**Tip:** Always clarify whether a problem involves replacement or multiple trials to avoid confusion in probability calculations.

What is the probability of drawing a black ball from a set of two white balls, two black balls, and two red balls?

Learn how to calculate the probability of drawing a black ball from a set containing two white, two black, and two red balls. This problem involves basic probability and is suitable for students in Grades 6-8.

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