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.