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Let's solve the problem step by step based on the table and the questions given.

### Problem Interpretation:
We are given the joint probability distribution for a product's evaluation in two rounds (Round 1: \(X\), Round 2: \(Y\)), where the product is evaluated with ratings of 0, 1, or 2. The product is **qualified** if it has a rating of 1 or 2, and **not qualified** if the rating is 0.

#### Joint Probability Distribution:
The table provides the joint probabilities for \(X\) and \(Y\), where \(X\) is the rating in Round 1 and \(Y\) is the rating in Round 2:

| Round 1 (X) \ Round 2 (Y) | 0     | 1     | 2     |
|----------------------------|-------|-------|-------|
| 0                          | 0.139 | 0.16  | 0.15  |
| 1                          | 0.299 | 0.108 | 0.041 |
| 2                          | 0.054 | 0.014 | 0.035 |

### 1. **Find the probability that the product is evaluated as qualified in both rounds.**

The product is qualified in both rounds if \(X \in \{1, 2\}\) and \(Y \in \{1, 2\}\). We sum the joint probabilities where both \(X\) and \(Y\) are either 1 or 2:

\[
P(\text{Qualified in both rounds}) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=2, Y=1) + P(X=2, Y=2)
\]
From the table:

\[
P(\text{Qualified in both rounds}) = 0.108 + 0.041 + 0.014 + 0.035 = 0.198
\]

### 2. **Find the probability that the product is qualified in the second round given that it is not qualified in the first round.**

The product is not qualified in the first round if \(X = 0\), and it is qualified in the second round if \(Y \in \{1, 2\}\).

We are looking for:
\[
P(Y \in \{1, 2\} \mid X = 0) = \frac{P(X=0, Y \in \{1, 2\})}{P(X=0)}
\]

Where:

- \(P(X=0, Y \in \{1, 2\}) = P(X=0, Y=1) + P(X=0, Y=2) = 0.16 + 0.15 = 0.31\)
- \(P(X=0) = P(X=0, Y=0) + P(X=0, Y=1) + P(X=0, Y=2) = 0.139 + 0.16 + 0.15 = 0.449\)

Thus:

\[
P(Y \in \{1, 2\} \mid X=0) = \frac{0.31}{0.449} \approx 0.69
\]

### 3. **Compute \(E(XY)\).**

The expected value \(E(XY)\) is calculated as the sum of \(X \times Y \times P(X, Y)\) for all possible values of \(X\) and \(Y\):

\[
E(XY) = \sum_{X, Y} X \times Y \times P(X, Y)
\]

We calculate each term individually using the probabilities from the table:

- \(E(XY) = (0 \times 0 \times 0.139) + (0 \times 1 \times 0.16) + (0 \times 2 \times 0.15)\)
- \(+ (1 \times 0 \times 0.299) + (1 \times 1 \times 0.108) + (1 \times 2 \times 0.041)\)
- \(+ (2 \times 0 \times 0.054) + (2 \times 1 \times 0.014) + (2 \times 2 \times 0.035)\)

Simplifying:

\[
E(XY) = 0 + 0 + 0 + 0 + 0.108 + 0.082 + 0 + 0.028 + 0.14 = 0.358
\]

---

### Summary of Answers:
1. Probability that the product is evaluated as qualified in both rounds: **0.198**.
2. Probability that the product is qualified in the second round given that it is not qualified in the first round: **0.69**.
3. \(E(XY)\): **0.358**.

---

Would you like any further details or clarifications? 

Here are some related questions:
1. How would the probabilities change if a new joint probability table was given?
2. What if the qualifications were based on different thresholds?
3. How would you compute \(E(X^2)\) or \(E(Y^2)\) based on the same table?
4. How does conditional probability work in more complex joint distributions?
5. Can we visualize the probabilities using charts or graphs?

**Tip**: Always verify that the probabilities in a joint distribution sum to 1 as a consistency check!

In a product quality inspection, products are tested in two rounds. A product is qualified if it has a rating of 1 or 2, and it is not qualified if it has a rating of 0. The following is the joint probability distribution:

1. If a product is randomly chosen, find the probability that this product is evaluated as qualified in both rounds.
2. If a product is randomly chosen, find the probability that this product is evaluated as qualified in the second round given that it is not qualified in the first round.
3. Compute E(XY).

Learn how to solve a probability problem involving product quality inspection with ratings across two rounds. The solution covers the computation of joint probabilities, conditional probabilities, and expected value (E(XY)). This example is perfect for high school or college-level students studying probability and statistics.

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