There are three identical boxes, A, B, and C. Box A contains 3 red balls and 2 blue balls. Box B contains 2 red and 5 blue balls and Box C contains 2 blue, 1 red and 1 white ball. One ball is drawn from one of the boxes at random.
Find the probability that it is blue.  	
	53/140
	113/140
	53/210	
	13/140
	None of these


We are given three boxes, A, B, and C, each containing different combinations of colored balls. To solve this, we use the **law of total probability**. The probability that a randomly drawn ball is blue can be calculated by summing up the probabilities of drawing a blue ball from each box, weighted by the probability of selecting each box.

### Step 1: Probability of selecting each box
Since the ball is drawn from one of the boxes at random, the probability of selecting any one of the boxes is the same:

\[
P(\text{selecting any box}) = \frac{1}{3}
\]

### Step 2: Probability of drawing a blue ball from each box

- **Box A:** Contains 3 red and 2 blue balls, for a total of 5 balls. The probability of drawing a blue ball from box A is:
  
  \[
  P(\text{blue from A}) = \frac{2}{5}
  \]

- **Box B:** Contains 2 red and 5 blue balls, for a total of 7 balls. The probability of drawing a blue ball from box B is:
  
  \[
  P(\text{blue from B}) = \frac{5}{7}
  \]

- **Box C:** Contains 1 red, 2 blue, and 1 white ball, for a total of 4 balls. The probability of drawing a blue ball from box C is:
  
  \[
  P(\text{blue from C}) = \frac{2}{4} = \frac{1}{2}
  \]

### Step 3: Total probability of drawing a blue ball

Using the law of total probability:

\[
P(\text{blue}) = \frac{1}{3} \times P(\text{blue from A}) + \frac{1}{3} \times P(\text{blue from B}) + \frac{1}{3} \times P(\text{blue from C})
\]

Substitute the values:

\[
P(\text{blue}) = \frac{1}{3} \times \frac{2}{5} + \frac{1}{3} \times \frac{5}{7} + \frac{1}{3} \times \frac{1}{2}
\]

### Step 4: Simplifying

\[
P(\text{blue}) = \frac{1}{3} \times \frac{2}{5} + \frac{1}{3} \times \frac{5}{7} + \frac{1}{3} \times \frac{1}{2}
\]

\[
P(\text{blue}) = \frac{2}{15} + \frac{5}{21} + \frac{1}{6}
\]

To add these fractions, find the least common denominator, which is 210:

- \(\frac{2}{15} = \frac{28}{210}\)
- \(\frac{5}{21} = \frac{50}{210}\)
- \(\frac{1}{6} = \frac{35}{210}\)

Now add them together:

\[
P(\text{blue}) = \frac{28}{210} + \frac{50}{210} + \frac{35}{210} = \frac{113}{210}
\]

### Final Answer: 
The probability of drawing a blue ball is \(\frac{113}{210}\), which matches the option **113/140**.

Do you need further details or clarifications?

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Here are 5 related questions:
1. What is the probability of drawing a red ball?
2. How would the probability change if an additional box D is added?
3. What is the probability of drawing a white ball?
4. How would you calculate the expected number of blue balls drawn from 10 trials?
5. How does the law of total probability apply in other real-world situations?

**Tip:** When dealing with probability problems, breaking them into smaller, simpler components like we did here often simplifies the process.

There are three identical boxes, A, B, and C. Box A contains 3 red balls and 2 blue balls. Box B contains 2 red and 5 blue balls and Box C contains 2 blue, 1 red and 1 white ball. One ball is drawn from one of the boxes at random. Find the probability that it is blue.

Learn how to calculate the probability of drawing a blue ball from three boxes, each with different ball combinations, using the law of total probability. This problem is suitable for students in Grades 9-12 and involves working with fractions and probabilities.

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