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The problem in the image is asking to find the probability that a randomly chosen ball is white, given two boxes containing balls of different colors.

**Problem details:**
- The first box contains **1 white ball** and **2 black balls**.
- The second box contains **3 white balls** and **4 black balls**.
- A box is chosen randomly, and a ball is picked from that box. We need to calculate the probability that the chosen ball is white.

**Solution approach:**
We will use the **total probability theorem** to solve this.

1. **Probability of choosing a box:**
   - Since a box is chosen at random, the probability of choosing either the first or the second box is \( \frac{1}{2} \).

2. **Probability of drawing a white ball from each box:**
   - From the first box, the probability of drawing a white ball is \( P(\text{white | first box}) = \frac{1}{3} \) because there are 1 white and 2 black balls (3 total balls).
   - From the second box, the probability of drawing a white ball is \( P(\text{white | second box}) = \frac{3}{7} \) because there are 3 white and 4 black balls (7 total balls).

3. **Total probability of drawing a white ball (using total probability formula):**
   \[
   P(\text{white}) = P(\text{first box}) \cdot P(\text{white | first box}) + P(\text{second box}) \cdot P(\text{white | second box})
   \]
   Substituting the known values:
   \[
   P(\text{white}) = \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{3}{7}
   \]
   \[
   P(\text{white}) = \frac{1}{6} + \frac{3}{14}
   \]
   To add these fractions, we find a common denominator (42):
   \[
   P(\text{white}) = \frac{7}{42} + \frac{9}{42} = \frac{16}{42} = \frac{8}{21}
   \]

Thus, the probability of drawing a white ball is **\( \frac{8}{21} \)**.

The correct answer from the options is \( \frac{8}{21} \).

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Would you like more details or have any further questions?

Here are 5 related questions for practice:
1. What is the probability of drawing a black ball in this scenario?
2. How would the probability change if both boxes contained an equal number of white and black balls?
3. How is the total probability theorem applied in this problem?
4. What is the expected number of white balls if you draw 10 times?
5. How does changing the number of black balls in the second box affect the outcome?

**Tip:** When solving probability problems involving multiple events, breaking them into smaller conditional probabilities simplifies the process!

В первом ящике 1 белый и 2 черных шара, во втором - 3 белых и 4 черных. Наугад выбран ящик и из него взят шар. Найти вероятность, что он белый.

Calculate the probability of drawing a white ball from two boxes, one containing 1 white and 2 black balls, and the other containing 3 white and 4 black balls. This problem covers key probability concepts, including conditional probability and the total probability theorem. Suitable for students in Grades 9-12.

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