This problem is about probability involving two types of boxes. Here's a breakdown of the given situation:

There are 10 boxes containing products:

• 7 boxes are of type 1, with 7 high-quality items (chính phẩm) and 3 defective items (phế phẩm).
• 3 boxes are of type 2, with 8 high-quality items and 2 defective items.

You are asked two questions:

Part (a)

You randomly choose 1 box and draw 2 items from that box. What is the probability that both items are high-quality?

Part (b)

Given that the two items drawn are both high-quality, what is the probability that the selected box was of type 2?

Let's start solving this step-by-step.

Part (a)

Step 1: Probability of selecting a type 1 or type 2 box

• Probability of choosing a type 1 box: $P(\text{type 1}) = \frac{7}{10}$
• Probability of choosing a type 2 box: $P(\text{type 2}) = \frac{3}{10}$

Step 2: Probability of drawing 2 high-quality items from each type of box

1. Type 1 box: The probability of drawing 2 high-quality items from a type 1 box (7 high-quality and 3 defective items): $P(\text{2 high-quality | type 1}) = \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{\frac{7 \times 6}{2}}{\frac{10 \times 9}{2}} = \frac{42}{90} = \frac{7}{15}$

2. Type 2 box: The probability of drawing 2 high-quality items from a type 2 box (8 high-quality and 2 defective items): $P(\text{2 high-quality | type 2}) = \frac{\binom{8}{2}}{\binom{10}{2}} = \frac{\frac{8 \times 7}{2}}{\frac{10 \times 9}{2}} = \frac{56}{90} = \frac{28}{45}$

Step 3: Total probability of drawing 2 high-quality items

Using the law of total probability, the overall probability of drawing 2 high-quality items is: $P(\text{2 high-quality}) = P(\text{type 1}) \times P(\text{2 high-quality | type 1}) + P(\text{type 2}) \times P(\text{2 high-quality | type 2})$ Substitute the values: $P(\text{2 high-quality}) = \frac{7}{10} \times \frac{7}{15} + \frac{3}{10} \times \frac{28}{45}$ Simplifying: $P(\text{2 high-quality}) = \frac{49}{150} + \frac{84}{450} = \frac{49}{150} + \frac{28}{150} = \frac{77}{150}$

So, the probability that both items are high-quality is $\frac{77}{150}$.

Part (b)

Given that both drawn items are high-quality, find the probability that the selected box was of type 2. We use Bayes' Theorem for this:

$P(\text{type 2 | 2 high-quality}) = \frac{P(\text{2 high-quality | type 2}) \times P(\text{type 2})}{P(\text{2 high-quality})}$ Substitute the known values: $P(\text{type 2 | 2 high-quality}) = \frac{\frac{28}{45} \times \frac{3}{10}}{\frac{77}{150}} = \frac{\frac{84}{450}}{\frac{77}{150}} = \frac{84}{450} \times \frac{150}{77} = \frac{28}{77}$

Thus, the probability that the box is of type 2, given that both drawn items are high-quality, is $\frac{28}{77}$.

Would you like more details or explanations on any part of this solution? Here are some additional related questions:

1. What is the probability of drawing at least one defective item?
2. How would the result change if there were 9 high-quality items and 1 defective item in type 2 boxes?
3. What is the conditional probability of drawing two defective items from a type 1 box?
4. Can the probability that the items are high-quality be generalized for different numbers of high-quality and defective items?
5. What would happen if the number of type 1 and type 2 boxes changed?

Tip: When dealing with conditional probability, Bayes' Theorem is your go-to tool!