Explanation:
Okay, so I need to find the sum of the positive divisors of 500, which is called A, and then find the sum of the distinct prime divisors of A. Let me start by recalling how to calculate the sum of divisors of a number.

First, I think the formula for the sum of divisors involves the prime factorization of the number. If I can find the prime factors of 500, then I can use the formula for the sum of divisors. Let me try to factorize 500.

Starting with the smallest prime, 2. 500 divided by 2 is 250. Then 250 divided by 2 is 125. So that's two 2s. Now 125 is 5 cubed, right? Because 5*5=25, 25*5=125. So the prime factorization of 500 is 2² * 5³. Let me check that: 2 squared is 4, 5 cubed is 125, 4*125 is indeed 500. Okay, so that's correct.

Now, the formula for the sum of divisors function, often denoted as σ(n), is if n = p₁^a₁ * p₂^a₂ * ... * p_k^a_k, then σ(n) = (p₁^0 + p₁^1 + ... + p₁^a₁) * (p₂^0 + p₂^1 + ... + p₂^a₂) * ... * (p_k^0 + p_k^1 + ... + p_k^a_k).

So applying that to 500, which is 2² * 5³, the sum of divisors σ(500) should be (1 + 2 + 2²) multiplied by (1 + 5 + 5² + 5³). Let me compute each part step by step.

First, for the 2² part: 1 + 2 + 4. Let me add those up: 1 + 2 is 3, plus 4 is 7.

Then for the 5³ part: 1 + 5 + 25 + 125. Let me compute that: 1 + 5 is 6, 6 + 25 is 31, 31 + 125 is 156.

So multiplying those two results together: 7 * 156. Let me compute that.

Hmm, 7 * 150 is 1050, and 7 * 6 is 42, so adding those gives 1050 + 42 = 1092. Therefore, A is 1092.

Wait, let me verify that again. Maybe I made a mistake in the multiplication. Let me do 156 * 7 another way. 156 * 7: 100*7=700, 50*7=350, 6*7=42. So 700 + 350 = 1050, plus 42 is 1092. Yeah, that's correct. So A = 1092.

Now, I need to find the sum of the distinct prime divisors of A. So first, I need to find the prime factors of 1092, then take the distinct ones and add them up.

So let me factorize 1092. Let's start with the smallest primes.

First, check divisibility by 2. 1092 is even, so divide by 2: 1092 / 2 = 546.

Again, 546 is even, divide by 2 again: 546 / 2 = 273.

Now, 273 is not even. Let's check divisibility by 3. The sum of digits of 273 is 2 + 7 + 3 = 12, which is divisible by 3. So divide by 3: 273 / 3 = 91.

Now, 91. Let me check divisibility by 7: 7 * 13 is 91. So 91 factors into 7 and 13.

So compiling all the factors: 1092 = 2² * 3 * 7 * 13.

Wait, let me check that again. Starting with 1092:

Divide by 2: 1092 / 2 = 546

Divide by 2 again: 546 / 2 = 273

Divide by 3: 273 / 3 = 91

Divide by 7: 91 / 7 = 13

And 13 is prime.

So yes, prime factors are 2, 3, 7, 13. The exponents are 2² * 3¹ * 7¹ * 13¹.

So the distinct prime divisors are 2, 3, 7, 13.

Therefore, the sum is 2 + 3 + 7 + 13. Let me compute that.

2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25.

So the sum of the distinct prime divisors of A is 25.

Wait a second, let me check if I did everything correctly. Let me verify the prime factors of 1092 again.

Alternatively, maybe I can check the multiplication steps again.

Original A was calculated as σ(500) = 1092. Let me check if that's correct.

Sum of divisors of 500. The divisors of 500 can be listed, but that would take time. Alternatively, since we have the formula, maybe I can check another way.

The formula is (1 + 2 + 4) * (1 + 5 + 25 + 125).

1 + 2 + 4 is 7, as before.

1 + 5 + 25 + 125: 1 + 5 is 6, 6 + 25 is 31, 31 + 125 is 156.

7 * 156: As before, 7*150 + 7*6 = 1050 + 42 = 1092. So that seems correct.

So A is indeed 1092.

Now factorizing 1092, as above: 2² * 3 * 7 * 13.

So primes are 2, 3, 7, 13. Their sum is 2 + 3 + 7 + 13 = 25.

Wait, 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25. Yes, that's correct.

But let me just double-check the factorization of 1092 by another method.

Let me divide 1092 by 2: 1092 / 2 = 546.

Divide by 2 again: 546 / 2 = 273.

Now 273: divide by 3: 273 / 3 = 91.

Divide 91 by 7: 91 / 7 = 13.

So yes, that gives the same result.

Alternatively, check if 1092 is divisible by other primes. For example, 5? Last digit is 2, so no. 11? Let's check 11. 1 - 0 + 9 - 2 = 1 - 0 + 9 - 2 = 8, which is not divisible by 11. So no. 13? Well, 13 is already a factor.

Alternatively, maybe I can check if 1092 has any other prime factors. Let me check if 13 is prime, which it is, and 7 is prime, 3 is prime, 2 is prime. So the prime factors are indeed 2, 3, 7, 13.

So their sum is 25.

Therefore, the answer should be 25.

But just to be thorough, let me check if there is any chance I made a mistake in the initial sum of divisors.

Original number is 500. Let me list all the divisors and sum them up.

But that might take some time. Let me try.

The prime factors are 2² * 5³. So the divisors are generated by exponents of 2 from 0 to 2 and exponents of 5 from 0 to 3.

So:

For 2^0 (which is 1):

1 * 5^0 = 1

1 * 5^1 = 5

1 * 5^2 = 25

1 * 5^3 = 125

For 2^1 (which is 2):

2 * 5^0 = 2

2 * 5^1 = 10

2 * 5^2 = 50

2 * 5^3 = 250

For 2^2 (which is 4):

4 * 5^0 = 4

4 * 5^1 = 20

4 * 5^2 = 100

4 * 5^3 = 500

So all divisors are: 1, 5, 25, 125, 2, 10, 50, 250, 4, 20, 100, 500.

Let me count them: 12 divisors. Now let me sum them up.

First, group them for easier addition:

Group 1: 1 + 5 + 25 + 125

That's 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156

Group 2: 2 + 10 + 50 + 250

2 + 10 = 12, 12 + 50 = 62, 62 + 250 = 312

Group 3: 4 + 20 + 100 + 500

4 + 20 = 24, 24 + 100 = 124, 124 + 500 = 624

Total sum: 156 + 312 + 624

Compute 156 + 312 first: 156 + 312 = 468

Then 468 + 624: 468 + 600 = 1068, 1068 + 24 = 1092

Yes, that's correct. So the sum of divisors is indeed 1092. Therefore, A is 1092.

So my previous calculation of the prime factors of 1092 is correct. Therefore, the distinct primes are 2, 3, 7, 13, sum to 25.

Is there any chance I missed a prime factor? Let me check 1092 divided by 2, 3, 7, 13.

Let me check if there's another prime factor. Let me try dividing 1092 by 13: 1092 / 13. Let me compute 13*80 = 1040, 1092 - 1040 = 52, 52 /13=4, so total is 80 + 4 = 84. So 13*84=1092.

Then 84 factors into 2² * 3 * 7, as before.

Therefore, 1092 = 13 * 84 = 13 * 2² * 3 * 7. Therefore, the prime factors are indeed 2, 3, 7, 13.

Therefore, the answer is 2 + 3 + 7 + 13 = 25.

Therefore, the sum of the distinct prime divisors of A is 25.

I think that's solid. I double-checked all the steps. The key was ensuring that the sum of divisors was calculated correctly, which I did by both the formula and by listing the divisors. Then factorizing 1092 by division and verifying the prime factors.

So I feel confident that the answer is 25.

**Final Answer**
The sum of the distinct prime divisors of $ A $ is \boxed{25}.

To solve the problem, we are asked to compute the sum of the *distinct prime divisors* of a number $ A $, where $ A $ is defined as the sum of the positive divisors of 500.

---

### Step 1: Prime Factorization of 500

We begin by finding the prime factorization of 500:

$$
500 = 2^2 \cdot 5^3
$$

---

### Step 2: Use the Divisor Sum Formula

The function that computes the sum of all positive divisors of a number is given by the **sigma function** $ \sigma(n) $, which for $ n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k} $ is:

$$
\sigma(n) = \prod_{i=1}^k \left(1 + p_i + p_i^2 + \cdots + p_i^{a_i}\right)
$$

Applying this to $ n = 2^2 \cdot 5^3 $:

$$
\sigma(500) = (1 + 2 + 2^2)(1 + 5 + 5^2 + 5^3)
$$

Compute each part:

- $ 1 + 2 + 4 = 7 $
- $ 1 + 5 + 25 + 125 = 156 $

Now multiply:

$$
A = \sigma(500) = 7 \cdot 156 = 1092
$$

---

### Step 3: Prime Factorization of 1092

Now we need to find the **prime factorization** of 1092, and then sum the **distinct** prime factors.

Start by dividing:

- $ 1092 \div 2 = 546 $
- $ 546 \div 2 = 273 $
- $ 273 \div 3 = 91 $
- $ 91 \div 7 = 13 $

So, the prime factorization of 1092 is:

$$
1092 = 2^2 \cdot 3 \cdot 7 \cdot 13
$$

These are all primes, and the distinct ones are: $ 2, 3, 7, 13 $

---

### Step 4: Sum of the Distinct Prime Divisors

Finally, compute the sum of these primes:

$$
2 + 3 + 7 + 13 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the positive divisors of 500, which is called A, and then find the sum of the distinct prime divisors of A. Let me start by recalling how to calculate the sum of divisors of a number.

First, I think the formula for the sum of divisors involves the prime factorization of the number. If I can find the prime factors of 500, then I can use the formula for the sum of divisors. Let me try to factorize 500.

Starting with the smallest prime, 2. 500 divided by 2 is 250. Then 250 divided by 2 is 125. So that's two 2s. Now 125 is 5 cubed, right? Because 5*5=25, 25*5=125. So the prime factorization of 500 is 2² * 5³. Let me check that: 2 squared is 4, 5 cubed is 125, 4*125 is indeed 500. Okay, so that's correct.

Now, the formula for the sum of divisors function, often denoted as σ(n), is if n = p₁^a₁ * p₂^a₂ * ... * p_k^a_k, then σ(n) = (p₁^0 + p₁^1 + ... + p₁^a₁) * (p₂^0 + p₂^1 + ... + p₂^a₂) * ... * (p_k^0 + p_k^1 + ... + p_k^a_k).

So applying that to 500, which is 2² * 5³, the sum of divisors σ(500) should be (1 + 2 + 2²) multiplied by (1 + 5 + 5² + 5³). Let me compute each part step by step.

First, for the 2² part: 1 + 2 + 4. Let me add those up: 1 + 2 is 3, plus 4 is 7.

Then for the 5³ part: 1 + 5 + 25 + 125. Let me compute that: 1 + 5 is 6, 6 + 25 is 31, 31 + 125 is 156.

So multiplying those two results together: 7 * 156. Let me compute that.

Hmm, 7 * 150 is 1050, and 7 * 6 is 42, so adding those gives 1050 + 42 = 1092. Therefore, A is 1092.

Wait, let me verify that again. Maybe I made a mistake in the multiplication. Let me do 156 * 7 another way. 156 * 7: 100*7=700, 50*7=350, 6*7=42. So 700 + 350 = 1050, plus 42 is 1092. Yeah, that's correct. So A = 1092.

Now, I need to find the sum of the distinct prime divisors of A. So first, I need to find the prime factors of 1092, then take the distinct ones and add them up.

So let me factorize 1092. Let's start with the smallest primes.

First, check divisibility by 2. 1092 is even, so divide by 2: 1092 / 2 = 546.

Again, 546 is even, divide by 2 again: 546 / 2 = 273.

Now, 273 is not even. Let's check divisibility by 3. The sum of digits of 273 is 2 + 7 + 3 = 12, which is divisible by 3. So divide by 3: 273 / 3 = 91.

Now, 91. Let me check divisibility by 7: 7 * 13 is 91. So 91 factors into 7 and 13.

So compiling all the factors: 1092 = 2² * 3 * 7 * 13.

Wait, let me check that again. Starting with 1092:

Divide by 2: 1092 / 2 = 546

Divide by 2 again: 546 / 2 = 273

Divide by 3: 273 / 3 = 91

Divide by 7: 91 / 7 = 13

And 13 is prime.

So yes, prime factors are 2, 3, 7, 13. The exponents are 2² * 3¹ * 7¹ * 13¹.

So the distinct prime divisors are 2, 3, 7, 13.

Therefore, the sum is 2 + 3 + 7 + 13. Let me compute that.

2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25.

So the sum of the distinct prime divisors of A is 25.

Wait a second, let me check if I did everything correctly. Let me verify the prime factors of 1092 again.

Alternatively, maybe I can check the multiplication steps again.

Original A was calculated as σ(500) = 1092. Let me check if that's correct.

Sum of divisors of 500. The divisors of 500 can be listed, but that would take time. Alternatively, since we have the formula, maybe I can check another way.

The formula is (1 + 2 + 4) * (1 + 5 + 25 + 125).

1 + 2 + 4 is 7, as before.

1 + 5 + 25 + 125: 1 + 5 is 6, 6 + 25 is 31, 31 + 125 is 156.

7 * 156: As before, 7*150 + 7*6 = 1050 + 42 = 1092. So that seems correct.

So A is indeed 1092.

Now factorizing 1092, as above: 2² * 3 * 7 * 13.

So primes are 2, 3, 7, 13. Their sum is 2 + 3 + 7 + 13 = 25.

Wait, 2 + 3 is 5, 5 + 7 is 12, 12 + 13 is 25. Yes, that's correct.

But let me just double-check the factorization of 1092 by another method.

Let me divide 1092 by 2: 1092 / 2 = 546.

Divide by 2 again: 546 / 2 = 273.

Now 273: divide by 3: 273 / 3 = 91.

Divide 91 by 7: 91 / 7 = 13.

So yes, that gives the same result.

Alternatively, check if 1092 is divisible by other primes. For example, 5? Last digit is 2, so no. 11? Let's check 11. 1 - 0 + 9 - 2 = 1 - 0 + 9 - 2 = 8, which is not divisible by 11. So no. 13? Well, 13 is already a factor.

Alternatively, maybe I can check if 1092 has any other prime factors. Let me check if 13 is prime, which it is, and 7 is prime, 3 is prime, 2 is prime. So the prime factors are indeed 2, 3, 7, 13.

So their sum is 25.

Therefore, the answer should be 25.

But just to be thorough, let me check if there is any chance I made a mistake in the initial sum of divisors.

Original number is 500. Let me list all the divisors and sum them up.

But that might take some time. Let me try.

The prime factors are 2² * 5³. So the divisors are generated by exponents of 2 from 0 to 2 and exponents of 5 from 0 to 3.

So:

For 2^0 (which is 1):

1 * 5^0 = 1

1 * 5^1 = 5

1 * 5^2 = 25

1 * 5^3 = 125

For 2^1 (which is 2):

2 * 5^0 = 2

2 * 5^1 = 10

2 * 5^2 = 50

2 * 5^3 = 250

For 2^2 (which is 4):

4 * 5^0 = 4

4 * 5^1 = 20

4 * 5^2 = 100

4 * 5^3 = 500

So all divisors are: 1, 5, 25, 125, 2, 10, 50, 250, 4, 20, 100, 500.

Let me count them: 12 divisors. Now let me sum them up.

First, group them for easier addition:

Group 1: 1 + 5 + 25 + 125

That's 1 + 5 = 6, 6 + 25 = 31, 31 + 125 = 156

Group 2: 2 + 10 + 50 + 250

2 + 10 = 12, 12 + 50 = 62, 62 + 250 = 312

Group 3: 4 + 20 + 100 + 500

4 + 20 = 24, 24 + 100 = 124, 124 + 500 = 624

Total sum: 156 + 312 + 624

Compute 156 + 312 first: 156 + 312 = 468

Then 468 + 624: 468 + 600 = 1068, 1068 + 24 = 1092

Yes, that's correct. So the sum of divisors is indeed 1092. Therefore, A is 1092.

So my previous calculation of the prime factors of 1092 is correct. Therefore, the distinct primes are 2, 3, 7, 13, sum to 25.

Is there any chance I missed a prime factor? Let me check 1092 divided by 2, 3, 7, 13.

Let me check if there's another prime factor. Let me try dividing 1092 by 13: 1092 / 13. Let me compute 13*80 = 1040, 1092 - 1040 = 52, 52 /13=4, so total is 80 + 4 = 84. So 13*84=1092.

Then 84 factors into 2² * 3 * 7, as before.

Therefore, 1092 = 13 * 84 = 13 * 2² * 3 * 7. Therefore, the prime factors are indeed 2, 3, 7, 13.

Therefore, the answer is 2 + 3 + 7 + 13 = 25.

Therefore, the sum of the distinct prime divisors of A is 25.

I think that's solid. I double-checked all the steps. The key was ensuring that the sum of divisors was calculated correctly, which I did by both the formula and by listing the divisors. Then factorizing 1092 by division and verifying the prime factors.

So I feel confident that the answer is 25.

**Final Answer**
The sum of the distinct prime divisors of $ A $ is \boxed{25}.

To solve the problem, we are asked to compute the sum of the *distinct prime divisors* of a number $ A $, where $ A $ is defined as the sum of the positive divisors of 500.

---

### Step 1: Prime Factorization of 500

We begin by finding the prime factorization of 500:

$$
500 = 2^2 \cdot 5^3
$$

---

### Step 2: Use the Divisor Sum Formula

The function that computes the sum of all positive divisors of a number is given by the **sigma function** $ \sigma(n) $, which for $ n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k} $ is:

$$
\sigma(n) = \prod_{i=1}^k \left(1 + p_i + p_i^2 + \cdots + p_i^{a_i}\right)
$$

Applying this to $ n = 2^2 \cdot 5^3 $:

$$
\sigma(500) = (1 + 2 + 2^2)(1 + 5 + 5^2 + 5^3)
$$

Compute each part:

- $ 1 + 2 + 4 = 7 $
- $ 1 + 5 + 25 + 125 = 156 $

Now multiply:

$$
A = \sigma(500) = 7 \cdot 156 = 1092
$$

---

### Step 3: Prime Factorization of 1092

Now we need to find the **prime factorization** of 1092, and then sum the **distinct** prime factors.

Start by dividing:

- $ 1092 \div 2 = 546 $
- $ 546 \div 2 = 273 $
- $ 273 \div 3 = 91 $
- $ 91 \div 7 = 13 $

So, the prime factorization of 1092 is:

$$
1092 = 2^2 \cdot 3 \cdot 7 \cdot 13
$$

These are all primes, and the distinct ones are: $ 2, 3, 7, 13 $

---

### Step 4: Sum of the Distinct Prime Divisors

Finally, compute the sum of these primes:

$$
2 + 3 + 7 + 13 = 25
$$

---

### Final Answer

$$
\boxed{25}
$$

Solution:
First, we find $A$. The prime factorization of $500$ is $2^2 \cdot 5^3$. Therefore,  $$A=(1+2+2^2)(1+5+5^2+5^3)=(7)(156).$$To see why $(1+2+2^2)(1+5+5^2+5^3)$ equals the sum of the divisors of 500, note that if you distribute (without simplifying), you get 12 terms, with each divisor of $2^2\cdot 5^3$ appearing exactly once.

Now we prime factorize $7 \cdot 156 = 7 \cdot 2^2 \cdot 3 \cdot 13$. The sum of the prime divisors of $A$ is $2+3+7+13=\boxed{25}$.


Explanation:
Okay, so I need to find the probability that three lengths—1, x, and y—can form a triangle, where x and y are randomly selected from the interval (0, 3). Hmm, let me recall the triangle inequality theorem. For three sides to form a triangle, the sum of any two sides must be greater than the third side. That means I need to check all three combinations:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

Since x and y are both between 0 and 3, I need to figure out the region in the coordinate plane where these inequalities hold, and then calculate the area of that region divided by the total area of the possible region for x and y (which is a square from (0,0) to (3,3)).

First, let me note the total area. Since x and y are each between 0 and 3, the total area is 3*3 = 9. So the probability will be the area of the region satisfying the triangle inequalities divided by 9.

Now, let me think about each inequality one by one.

Starting with the third inequality: x + y > 1. Since x and y are both positive numbers between 0 and 3, this inequality will be satisfied except for the region near (0,0). The area where x + y ≤ 1 is a small triangle with vertices at (0,0), (0,1), and (1,0). The area of this triangle is (1*1)/2 = 0.5. So the area where x + y > 1 is total area 9 minus 0.5, which is 8.5. But wait, but I also have the other inequalities to consider. The first two inequalities might restrict the region further.

Wait, but actually, the triangle inequalities are three separate conditions. So even though x + y > 1 is necessary, we also need 1 + x > y and 1 + y > x. So, maybe I need to consider the intersection of all three inequalities? Let me think.

Alternatively, maybe I can think of all three conditions together. Let me write down all three inequalities again:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

But since x and y are positive, maybe some of these can be simplified or considered in terms of each other?

For example, the first inequality 1 + x > y can be rewritten as y < 1 + x. Similarly, the second inequality is y > x - 1. The third is y > 1 - x. Hmm, so combining these, maybe I can find the region where all three inequalities hold.

But I need to visualize this. Let me try to sketch mentally the square [0,3] x [0,3]. The inequalities define regions within this square. Let me consider each inequality:

Starting with the third inequality x + y > 1. As I mentioned, this excludes the area under the line x + y = 1, which is a small triangle near the origin. However, the other two inequalities add more constraints.

The first inequality y < 1 + x is a region below the line y = 1 + x. Similarly, the second inequality y > x - 1 is the region above the line y = x - 1.

So, combining all three inequalities, the feasible region is the set of points (x, y) in [0,3]^2 where:

y > max(x - 1, 1 - x) and y < 1 + x

But also, since x and y are between 0 and 3, we have to consider the boundaries.

Alternatively, maybe breaking it down into cases based on the values of x and y?

Alternatively, maybe I can find the area where all three inequalities hold by considering the overlap of these regions. Let me try to sketch this mentally.

First, note that the inequalities y < 1 + x and y > x - 1. Let me think about the region between these two lines. The lines y = x + 1 and y = x - 1 are parallel? Wait, no, they have the same slope (1) and are parallel. The distance between them is 2 units vertically. However, since we are in the square [0,3]^2, the area between them will depend on where they intersect the square.

But actually, the line y = x + 1 intersects the square where? Let's see, if x is between 0 and 3, then y would go from 1 (when x=0) up to 4 (when x=3). But since our square only goes up to y=3, so the line y = x + 1 intersects the square at (0,1) and (2,3). Similarly, the line y = x - 1 intersects the square at (1,0) and (3,2). Therefore, between those lines, the region between them is kind of a band around the line y = x. However, we also have the third inequality x + y > 1. But since x and y are positive, x + y > 1 is automatically true except near the origin. However, maybe the combination of all three inequalities is the intersection of y < 1 + x, y > x - 1, and x + y > 1.

Wait, but perhaps I can find the area where all three inequalities hold. Let me consider that the three inequalities:

1. y < 1 + x

2. y > x - 1

3. y > 1 - x

But the third inequality is y > 1 - x. So if I combine inequalities 2 and 3, since y must be greater than both x - 1 and 1 - x. So, depending on the value of x, which one is larger?

Let me consider different ranges of x. For example, when x is between 0 and 1, 1 - x is greater than x - 1 (since 1 - x > x -1 when 1 +1 > x + x => 2 > 2x => x < 1). Therefore, for x < 1, the lower bound for y is 1 - x. For x > 1, the lower bound is x - 1. At x = 1, both are equal to 0. So, the region where y > max(1 - x, x - 1) is equivalent to y > |x - 1|.

Therefore, combining with the upper bound y < 1 + x, the region is between y = |x - 1| and y = 1 + x.

Therefore, the area where all three inequalities hold is the area between the lines y = |x - 1| and y = 1 + x, within the square [0,3]x[0,3].

But also, we need to make sure that y is between 0 and 3. But since we are already in the square [0,3]^2, the values of y are already bounded. However, we need to check if the lines y = |x - 1| and y = 1 + x intersect the boundaries of the square.

Let me try to visualize this. For x between 0 and 1, the lower boundary is y = 1 - x, which goes from (0,1) down to (1,0). The upper boundary is y = 1 + x, which goes from (0,1) up to (1,2). Therefore, for x between 0 and 1, the vertical slice between these two lines is from y = 1 - x to y = 1 + x.

Similarly, for x between 1 and 3, the lower boundary is y = x - 1, which goes from (1,0) up to (3,2). The upper boundary is still y = 1 + x, which goes from (0,1) up to (3,4), but since our square only goes up to y=3, the upper boundary intersects the square at (2,3) when x=2 (since 1 + x = 3 when x=2) and then continues to (3,4) which is outside. Therefore, for x between 1 and 2, the upper bound is y = 1 + x (from y=2 up to y=3), and for x between 2 and 3, the upper bound is y=3 instead of y = 1 + x because 1 + x would be greater than 3. So we need to split the integral into different ranges.

Therefore, to compute the area, I can split the integration into three parts:

1. x from 0 to 1
2. x from 1 to 2
3. x from 2 to 3

Let me check that.

First, for x in [0,1], upper bound is 1 + x, lower bound is 1 - x. So the vertical slice length is (1 + x) - (1 - x) = 2x.

But wait, no. Wait, the vertical slice is from y = lower bound to upper bound. So the length is (upper - lower). For x between 0 and 1, upper bound is 1 + x, lower bound is 1 - x. Therefore, vertical slice length is (1 + x) - (1 - x) = 2x. Therefore, integrating 2x from 0 to 1 gives the area for this interval.

For x between 1 and 2, the lower bound is y = x - 1, upper bound is y = 1 + x. The vertical slice length is (1 + x) - (x - 1) = 2. Therefore, integrating 2 from 1 to 2 gives 2*(2 -1) = 2*1 = 2.

For x between 2 and 3, the lower bound is still y = x -1, but upper bound is y = 3, since 1 + x would be more than 3. Therefore, vertical slice length is 3 - (x - 1) = 4 - x. Therefore, integrating (4 - x) from x=2 to x=3.

Therefore, total area would be:

Integral from 0 to1 of 2x dx + Integral from 1 to2 of 2 dx + Integral from 2 to3 of (4 - x) dx

Let me compute each integral.

First integral: ∫₀¹ 2x dx = [x²]₀¹ = 1 - 0 = 1

Second integral: ∫₁² 2 dx = 2*(2 - 1) = 2*1 = 2

Third integral: ∫₂³ (4 - x) dx. Let's compute that:

Antiderivative of 4 - x is 4x - (1/2)x². Evaluated from 2 to 3.

At x=3: 4*3 - 0.5*(9) = 12 - 4.5 = 7.5

At x=2: 4*2 - 0.5*(4) = 8 - 2 = 6

Subtracting: 7.5 - 6 = 1.5

Therefore, third integral is 1.5

Adding them up: 1 + 2 + 1.5 = 4.5

Therefore, the total area where all three inequalities hold is 4.5.

But wait, is that right? Let me check again.

But hold on, isn't the total area of the square 9, and we're saying that the area where triangle inequalities hold is 4.5? Then the probability would be 4.5 / 9 = 0.5. But that seems too simple. Let me check if I considered all constraints correctly.

Wait, but actually, I think there is a mistake here. Because the region I calculated (4.5) is the area where all three inequalities hold. But maybe I also need to consider that x and y are between 0 and 3. Wait, but no, since we already considered the square [0,3]^2. However, I need to check whether the inequalities x + y > 1 is already incorporated into the other inequalities?

Wait, no. Wait, the three inequalities are:

1. 1 + x > y => y < 1 + x

2. 1 + y > x => y > x - 1

3. x + y > 1 => y > 1 - x

But in my calculation, I considered combining 2 and 3 as y > max(x - 1, 1 - x), which is equivalent to y > |x -1|. Then combined with y < 1 + x. Therefore, the area between y = |x - 1| and y = 1 + x. But does this automatically include the condition x + y > 1? Wait, if we have y > |x - 1|, then for x between 0 and 1, y > 1 - x. Since 1 - x is positive, and since y is positive, then x + y is certainly greater than x + (1 - x) = 1. Therefore, x + y > 1 is automatically satisfied in that case. Similarly, for x between 1 and 3, y > x - 1. Then x + y > x + (x - 1) = 2x -1. When x ≥ 1, 2x -1 ≥ 1, so x + y ≥ 1. Therefore, in both cases, the third inequality x + y > 1 is automatically satisfied if we have y > |x -1|. Therefore, the region we computed (between y = |x -1| and y = 1 + x) automatically satisfies all three triangle inequalities.

Therefore, the area is indeed 4.5. Therefore, the probability would be 4.5 / 9 = 0.5. But that seems too straightforward. Let me check with some test points.

For example, suppose x = 0.5, y = 0.5. Then sides are 1, 0.5, 0.5. But 0.5 + 0.5 = 1, which is not greater than 1. Therefore, such a point should not be included. However, according to our region, y > |x - 1|. For x = 0.5, |x -1| = 0.5, so y must be greater than 0.5. Therefore, y = 0.5 is not included. Therefore, the region we calculated excludes points where y = |x -1|. So the area we calculated is correct in that sense.

Another example: x = 0.5, y = 1. Then 1, 0.5, 1. The sum of the two smaller sides is 0.5 + 1 = 1.5 > 1. So that's okay. So the region seems okay.

Wait, but let me check another point. Let me take x = 0.5, y = 0.6. Then 0.6 > |0.5 -1| = 0.5. So it is in the region. The sides are 1, 0.5, 0.6. The sum of the two smaller sides is 0.5 + 0.6 = 1.1 > 1. Okay. Similarly, for x = 2, y = 3. The sides are 1, 2, 3. But 1 + 2 = 3, which is not greater than 3. Therefore, this should not be in the region. However, according to our region, for x = 2, y must be greater than |2 -1| = 1, and less than 1 + 2 = 3. Therefore, y is in (1, 3). However, at y = 3, it's excluded. So the upper bound is y = 3. So when x = 2, y approaches 3, but does not include y = 3. So the point (2,3) is excluded, which is correct because 1 + 2 = 3 is not greater than 3. Therefore, our calculation seems correct.

But then the area is 4.5. Let me just verify the integrals again.

First integral from 0 to 1: ∫2x dx = 1. That's correct.

Second integral from 1 to 2: 2*(2 - 1) = 2. Correct.

Third integral from 2 to 3: ∫(4 - x) dx = [4x - x²/2] from 2 to 3. At 3: 12 - 9/2 = 12 - 4.5 = 7.5. At 2: 8 - 4/2 = 8 - 2 = 6. 7.5 - 6 = 1.5. Correct. Total area 1 + 2 + 1.5 = 4.5. So 4.5 / 9 = 0.5. Therefore, probability is 1/2. But is this correct?

Wait, but let me think again. If x and y are selected from (0,3), but the problem says selected at random from the interval (0,3). Wait, the interval is open, but since the probability of selecting exactly 0 or 3 is zero in continuous distribution, we can treat it as [0,3] for area calculation.

But let me think again if there are other constraints or if I missed something.

Wait, but suppose we take x and y both approaching 3. For example, x = 3 and y = 3. Then sides are 1, 3, 3. Then 1 + 3 > 3 is true (4 > 3), 3 + 3 > 1 is true, and 1 + 3 > 3 is true. So that's okay. So the point (3,3) is included. However, our calculation for x=3, y would have to be greater than |3 -1| = 2, and less than 1 + 3 = 4, but since y is up to 3, then y is between 2 and 3. However, the integral for x from 2 to 3 is ∫(4 - x) dx, which for x=3 gives 4 - 3 = 1. So at x=3, the vertical slice is 1, which corresponds to y between 2 and 3. So that's correct. Therefore, (3,3) is included, since y=3 is included in the upper bound. Wait, but in the integral, we integrated up to 3, and the upper bound was y=3. So the area is correct.

But then, according to this, the probability is 0.5. However, I have a feeling that the answer might be different. Maybe I made a mistake in considering the triangle inequalities?

Wait, let me think again. The three conditions are:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

But when we combine them, we have to make sure that all three hold. However, in our calculation, we considered the area between y = |x -1| and y = 1 + x. Is that sufficient?

Wait, let's check another point. Let me take x = 0.5, y = 0.6. Then 1 + x = 1.5 > 0.6; 1 + y = 1.6 > 0.5; x + y = 1.1 > 1. All good.

Take another point: x = 0.5, y = 0.5. But that's excluded since y needs to be greater than |x - 1| = 0.5. So the point y = 0.5 is on the boundary. Since the original problem says "selected at random", which is a continuous distribution, the probability of y being exactly 0.5 is zero. So that's okay.

Another point: x = 1.5, y = 0.5. Wait, but according to our region, for x = 1.5, the lower bound is |1.5 - 1| = 0.5. Therefore, y must be greater than 0.5, and less than 1 + 1.5 = 2.5. So y = 0.5 is excluded. But if we take y = 0.5, then check the triangle inequalities:

1 + x = 1 + 1.5 = 2.5 > y = 0.5 ✔

1 + y = 1 + 0.5 = 1.5 > x = 1.5? No, 1.5 is not greater than 1.5. It's equal. Therefore, the triangle inequality is not satisfied. Therefore, the point (1.5, 0.5) is excluded, which is consistent with our region.

But if we take y = 0.6, which is greater than 0.5, then 1 + y = 1.6 > 1.5. Therefore, it's okay. Therefore, our region is correct.

Therefore, the area seems correct. So probability is 4.5 / 9 = 0.5. So 1/2. Therefore, the answer is 1/2? But let me check another approach.

Alternative approach: Let's consider all three inequalities.

First, the triangle inequalities:

For sides 1, x, y, the three conditions are:

1. 1 + x > y

2. 1 + y > x

3. x + y > 1

Note that x and y are in (0,3). Let me consider these inequalities.

Let me rearrange inequalities 1 and 2:

From 1: y < 1 + x

From 2: y > x - 1

From 3: y > 1 - x

Therefore, combining 2 and 3, y > max(x - 1, 1 - x). So as before. Therefore, the region is between y = |x - 1| and y = 1 + x. Then integrating over x from 0 to 3.

But then, in the previous calculation, that gave area 4.5. Therefore, probability 0.5.

But maybe there's an error in assuming all three inequalities are satisfied? Let me think if there is another constraint.

Wait, but for example, if x is very small, say approaching 0, then y must be greater than |x -1| = 1 - x (since x is small). So y > 1 - x. But since x is approaching 0, y must be greater than 1. However, then, the other inequality y < 1 + x. If x approaches 0, then y < 1 + x approaches 1. Therefore, y must be greater than 1 and less than 1, which is impossible. Therefore, in this case, there is no solution? Wait, but this contradicts with our earlier integral. Wait, but when x approaches 0, for example, take x = 0.1. Then y must be greater than |0.1 -1| = 0.9, and less than 1 + 0.1 = 1.1. So y is between 0.9 and 1.1. Therefore, for x=0.1, there is a region. So when x approaches 0, the vertical slice is between y = 1 - x and y = 1 + x. So as x approaches 0, the length of the vertical slice approaches 2x, which goes to 0. Therefore, that is correct.

Moreover, at x = 0, the vertical slice length is 0. Therefore, the area is indeed 1. So that seems okay.

But let me think of another way. Suppose we imagine the square [0,3]x[0,3]. The region where the triangle inequalities hold is the set of points (x,y) such that all three inequalities hold. Let me try to visualize this.

First, the region where x + y > 1 is the square minus the triangle near (0,0). Then, the regions where 1 + x > y and 1 + y > x are also regions within the square. The intersection of these three regions is what we need.

Alternatively, maybe using geometric probability.

Alternatively, think of the problem as follows: for x and y in (0, 3), what's the probability that 1, x, y satisfy triangle inequalities. Which is equivalent to:

max(1, x, y) < 1 + min(1, x, y)

Because for three sides a, b, c, the triangle inequality is equivalent to the sum of the two smaller sides being greater than the largest side. Therefore, if we let a, b, c be 1, x, y. Then, the maximum of the three must be less than the sum of the other two.

Therefore, we can consider cases based on which of 1, x, y is the largest.

Case 1: 1 is the largest side. Then, we need 1 < x + y.

Case 2: x is the largest side. Then, we need x < 1 + y.

Case 3: y is the largest side. Then, we need y < 1 + x.

However, since x and y are symmetric, Cases 2 and 3 will have the same area. Therefore, we can compute Case 1, then double the area for Cases 2 and 3, but we have to be careful about overlapping.

Wait, actually, it's better to compute all cases.

But maybe this approach can help verify.

Case 1: 1 is the largest side. That is, x ≤ 1 and y ≤ 1. Then, we need x + y > 1. So this is the region where x and y are both between 0 and 1, and x + y > 1. The area here is the area of the square [0,1]x[0,1] minus the area where x + y ≤ 1. The area of [0,1]^2 is 1. The area where x + y ≤1 is 0.5, so the area for Case 1 is 1 - 0.5 = 0.5.

Case 2: x is the largest side. That is, x ≥ 1 and x ≥ y. Then, we need x < 1 + y. Which simplifies to y > x - 1. But since x ≥ 1 and x ≥ y, we have y ≤ x. Also, since x ≥ 1 and y ≤ x, and y > x - 1. So the region is x between 1 and 3, y between max(x - 1, 0) and x. However, since y must be between 0 and 3. But also, since we are in the case where x is the largest side, y ≤ x. So for x from 1 to 3, y is between max(x - 1, 0) and x. But since x ≥ 1, max(x -1, 0) = x -1.

Therefore, the area for Case 2 is ∫ from x=1 to 3 [x - (x - 1)] dx = ∫ from 1 to 3 1 dx = 3 - 1 = 2.

Wait, but hold on. Wait, that can't be right. Because for each x, the upper bound of y is x, and the lower bound is x - 1. Therefore, the length is x - (x -1 ) = 1. Therefore, integrating 1 from x=1 to 3 gives area 2. However, we need to check if y is within [0, 3]. Since x is up to 3, y is up to x=3, so it's okay. Therefore, the area for Case 2 is 2.

But wait, this seems different from the previous approach. Because in the previous approach, we had total area 4.5, but with Case 1 being 0.5 and Case 2 and Case 3 contributing 2 each? Wait, let me check.

Wait, Case 3: y is the largest side. Similarly, y ≥ 1 and y ≥ x. Then, we need y < 1 + x. Which is similar to Case 2 but with x and y swapped. Therefore, the area for Case 3 is also 2.

But then total area would be Case1 + Case2 + Case3 = 0.5 + 2 + 2 = 4.5, which matches our previous result. Therefore, the total area is 4.5. Therefore, probability is 4.5 / 9 = 0.5. Therefore, 1/2. Therefore, the probability is 1/2.

But let me check Case 2 again. For Case 2: x is the largest side. So x ≥ max(1, y). Therefore, for x between 1 and 3, and y between x - 1 and x. But when x is between 1 and 2, y ranges from x - 1 (which is between 0 and 1) to x (which is between 1 and 2). When x is between 2 and 3, y ranges from x -1 (which is between 1 and 2) to x (which is between 2 and 3). Therefore, integrating over x from 1 to 3, the vertical slice length is always 1. Therefore, area is 2. Similarly for Case 3.

But then, why does this approach give the same result as before? Therefore, seems that 0.5 is correct. However, I need to check if there's any overlap between the cases. For example, when x=1 and y=1, which case does it belong to? In Case 1, since all sides are equal (1,1,1), so 1 is the largest side. Also, x and y are equal to 1. However, in Case 2, x=1 and y=1: since x is the largest side, which is true. But the problem is that the cases are not mutually exclusive. However, since in Case 1, x ≤ 1 and y ≤ 1. In Case 2, x ≥ 1 and x ≥ y. In Case 3, y ≥ 1 and y ≥ x. Therefore, these cases are actually mutually exclusive. Because if x ≤1 and y ≤1, then neither x nor y is the largest unless they are both equal to 1. Similarly, if x is the largest, then x ≥1 and y ≤ x. If y is the largest, then y ≥1 and x ≤ y. Therefore, there is no overlap between Case 1 and Cases 2 and 3, nor between Case 2 and Case 3. Therefore, the total area is indeed 0.5 + 2 + 2 = 4.5.

Therefore, the probability is 4.5 / 9 = 0.5. Therefore, 1/2.

But let me check with another method. For instance, using coordinates.

The total area is 9. The area where triangle inequalities hold is 4.5. So probability is 1/2.

But let me think about when x and y are selected from (0, 3). Let me think of the three conditions again.

But maybe I can think of the problem as follows. Since the triangle inequalities are equivalent to all sides being less than the sum of the other two. Which is the same as:

- x + y > 1

- x + 1 > y => y < x + 1

- y + 1 > x => x < y + 1

Therefore, combining these, the region is defined by:

x < y + 1

y < x + 1

x + y > 1

Which is equivalent to the region between the lines y = x + 1, y = x - 1, and x + y = 1.

But graphically, in the square [0,3]x[0,3], the area where these conditions are satisfied is the area between the lines y = x + 1, y = x -1, and above the line x + y =1. Which as we calculated earlier is 4.5. Hence, probability is 0.5. So perhaps this is correct.

But I'm still a bit uncertain because I might have missed some constraints. Let me try to compute the area using another method. Let's consider that the problem is symmetric with respect to x and y. Therefore, the area where the triangle inequalities hold should be the same as the area where all three inequalities hold.

Alternatively, if we consider all possible x and y, the probability that they satisfy triangle inequalities is equal to the probability that x + y > 1, and |x - y| < 1. Wait, because:

If we have 1, x, y, then the triangle inequalities are equivalent to:

x + y > 1 and |x - y| < 1.

Wait, is that correct?

Wait, let me think. The triangle inequalities are:

1. x + y > 1

2. x + 1 > y => x - y > -1 => y - x < 1

3. y + 1 > x => y - x > -1 => same as above.

Therefore, combining inequalities 2 and 3 gives: |x - y| < 1. Therefore, the triangle inequalities are equivalent to:

x + y > 1 and |x - y| < 1.

Therefore, the region we need is the intersection of x + y > 1 and |x - y| < 1.

So, if I can compute this area, that would be easier.

Let me try this approach.

First, |x - y| < 1 is equivalent to y -1 < x < y +1. So for each y, x is between y -1 and y +1. However, x must also be in (0, 3), and similarly for y.

But also, x + y > 1.

Therefore, the region is the set of (x, y) such that |x - y| < 1 and x + y > 1.

Therefore, the area can be calculated by integrating over the square [0,3]^2 the region where both conditions hold.

Alternatively, since this is symmetric, maybe compute the area where |x - y| < 1 and x + y > 1.

But how?

Alternatively, since |x - y| < 1 defines a band around the line y = x, between y = x -1 and y = x + 1. The intersection with x + y > 1 is the region in this band above the line x + y =1.

Therefore, we can compute this area.

But perhaps again, breaking into cases based on the value of x and y.

Alternatively, note that the area where |x - y| < 1 is the area between the lines y = x + 1 and y = x -1. Which in the square [0,3]^2, is a band around the diagonal. The area of this band can be calculated as follows:

The area between y = x + 1 and y = x -1 in the square.

But since the square is from (0,0) to (3,3), the area of the band |x - y| < 1 is total area minus the areas of the regions where |x - y| ≥ 1.

Which would be the two triangles where y ≥ x + 1 and y ≤ x - 1.

The area where y ≥ x + 1 is a triangle with vertices at (0,1), (0,3), (2,3). Similarly, the area where y ≤ x -1 is a triangle with vertices at (1,0), (3,0), (3,2). Both triangles have base and height of 2, area is 0.5*2*2 = 2 each. So total area of these regions is 2 + 2 = 4. Therefore, the area of the band |x - y| < 1 is total area 9 - 4 = 5.

But we also need to intersect this band with the region x + y > 1. However, the band |x - y| < 1 already includes regions where x + y might be less than 1. For example, near the origin, where x and y are small, and |x - y| < 1. But x + y could be less than 1. Therefore, we need to find the area of the intersection between |x - y| < 1 and x + y > 1.

Therefore, the area we want is equal to the area of the band |x - y| < 1 minus the area of the part of the band where x + y ≤ 1.

So, compute area_band = 5, area_part = area of band where x + y ≤ 1. Then desired area = 5 - area_part.

What is area_part?

The region where |x - y| < 1 and x + y ≤ 1.

This region is the intersection of the band |x - y| < 1 and the region below x + y = 1.

Let me visualize this. The line x + y = 1 intersects the band |x - y| < 1.

To find the area where both |x - y| < 1 and x + y ≤ 1, let's find the overlapping region.

First, the region x + y ≤ 1 is a triangle with vertices at (0,0), (0,1), (1,0).

Within this triangle, we need to find the region where |x - y| < 1. However, since x and y are both between 0 and 1, and x + y ≤ 1, then |x - y| ≤ 1. Because the maximum difference between x and y occurs when one is 0 and the other is 1, which gives |x - y| = 1. Therefore, within the triangle x + y ≤ 1, the condition |x - y| < 1 is automatically satisfied except on the boundary where |x - y| = 1. But since it's a continuous distribution, the area on the boundary is zero. Therefore, the area_part = area of the triangle x + y ≤1 = 0.5.

Therefore, desired area = 5 - 0.5 = 4.5. Which matches our previous result!

Therefore, the area is indeed 4.5, and the probability is 4.5 / 9 = 0.5.

Therefore, the probability is 1/2.

Wait, but now I feel more confident. So why did I think it was too simple? Maybe because I didn't expect such a clean answer. However, all methods point to 4.5 / 9 = 1/2. Therefore, the probability is 1/2.

Therefore, the answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a triangle with side lengths 1, $ x $, and $ y $ can exist when $ x $ and $ y $ are selected randomly from the interval $ (0,3) $, we must rely on the **triangle inequality**. For three sides $ a $, $ b $, and $ c $ to form a triangle, the following must all hold:

- $ a + b > c $
- $ a + c > b $
- $ b + c > a $

Here, $ a = 1 $, $ b = x $, and $ c = y $, so the triangle condition becomes:

- $ x + y > 1 $
- $ 1 + x > y \Rightarrow y < x + 1 $
- $ 1 + y > x \Rightarrow x < y + 1 $

These three inequalities together define a region in the $ xy $-plane within the square $ [0,3] \times [0,3] $. The sample space is this square, which has area:

$$
\text{Total area} = 3 \times 3 = 9
$$

Our goal is to find the **area of the region** that satisfies all three inequalities, then divide by 9 to get the probability.

---

### Step 1: Geometric Interpretation

The three inequalities define a region of interest. Note that:

- $ x + y > 1 $: the area **above** the line $ x + y = 1 $
- $ x < y + 1 \Rightarrow y - x < 1 $: **below** the line $ y = x + 1 $
- $ y < x + 1 \Rightarrow x - y < 1 $: **above** the line $ y = x - 1 $

Thus, the region is:

$$
\text{Region} = \{ (x, y) \in [0,3]^2 \mid x + y > 1, \ |x - y| < 1 \}
$$

Which is the **intersection** of the area **above** the line $ x + y = 1 $ and the area **between** the lines $ y = x - 1 $ and $ y = x + 1 $.

This region lies **between the two diagonal lines** and **above the line $ x + y = 1 $**.

---

### Step 2: Compute the Desired Area

We can find the desired area more systematically. Consider the square $ [0,3] \times [0,3] $, with total area 9.

Let’s define the region $ A $ where $ |x - y| < 1 $, which is the area **between** the lines $ y = x + 1 $ and $ y = x - 1 $. The square is divided by these lines into regions where $ |x - y| < 1 $ and $ |x - y| \ge 1 $.

- The region where $ |x - y| \ge 1 $ consists of two triangles:
  - One above $ y = x + 1 $, with vertices at $ (0,1), (0,3), (2,3) $
  - One below $ y = x - 1 $, with vertices at $ (1,0), (3,0), (3,2) $

Each of these triangles has area:

$$
\text{Area of triangle} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 2 = 2
$$

So total area outside the region $ |x - y| < 1 $ is:

$$
2 + 2 = 4
$$

Thus, the area where $ |x - y| < 1 $ is:

$$
9 - 4 = 5
$$

Now, from this band, we must remove the part where $ x + y \leq 1 $, since this violates the triangle inequality.

- The region where $ x + y \leq 1 $ is a triangle with vertices at $ (0,0), (0,1), (1,0) $, so:

$$
\text{Area} = \frac{1}{2} \cdot 1 \cdot 1 = 0.5
$$

So the area where all triangle inequalities are satisfied is:

$$
5 - 0.5 = 4.5
$$

---

### Step 3: Compute the Probability

The total area of the sample space is 9. Hence, the probability is:

$$
\text{Probability} = \frac{\text{Desired Area}}{\text{Total Area}} = \frac{4.5}{9} = \frac{1}{2}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that three lengths—1, x, and y—can form a triangle, where x and y are randomly selected from the interval (0, 3). Hmm, let me recall the triangle inequality theorem. For three sides to form a triangle, the sum of any two sides must be greater than the third side. That means I need to check all three combinations:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

Since x and y are both between 0 and 3, I need to figure out the region in the coordinate plane where these inequalities hold, and then calculate the area of that region divided by the total area of the possible region for x and y (which is a square from (0,0) to (3,3)).

First, let me note the total area. Since x and y are each between 0 and 3, the total area is 3*3 = 9. So the probability will be the area of the region satisfying the triangle inequalities divided by 9.

Now, let me think about each inequality one by one.

Starting with the third inequality: x + y > 1. Since x and y are both positive numbers between 0 and 3, this inequality will be satisfied except for the region near (0,0). The area where x + y ≤ 1 is a small triangle with vertices at (0,0), (0,1), and (1,0). The area of this triangle is (1*1)/2 = 0.5. So the area where x + y > 1 is total area 9 minus 0.5, which is 8.5. But wait, but I also have the other inequalities to consider. The first two inequalities might restrict the region further.

Wait, but actually, the triangle inequalities are three separate conditions. So even though x + y > 1 is necessary, we also need 1 + x > y and 1 + y > x. So, maybe I need to consider the intersection of all three inequalities? Let me think.

Alternatively, maybe I can think of all three conditions together. Let me write down all three inequalities again:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

But since x and y are positive, maybe some of these can be simplified or considered in terms of each other?

For example, the first inequality 1 + x > y can be rewritten as y < 1 + x. Similarly, the second inequality is y > x - 1. The third is y > 1 - x. Hmm, so combining these, maybe I can find the region where all three inequalities hold.

But I need to visualize this. Let me try to sketch mentally the square [0,3] x [0,3]. The inequalities define regions within this square. Let me consider each inequality:

Starting with the third inequality x + y > 1. As I mentioned, this excludes the area under the line x + y = 1, which is a small triangle near the origin. However, the other two inequalities add more constraints.

The first inequality y < 1 + x is a region below the line y = 1 + x. Similarly, the second inequality y > x - 1 is the region above the line y = x - 1.

So, combining all three inequalities, the feasible region is the set of points (x, y) in [0,3]^2 where:

y > max(x - 1, 1 - x) and y < 1 + x

But also, since x and y are between 0 and 3, we have to consider the boundaries.

Alternatively, maybe breaking it down into cases based on the values of x and y?

Alternatively, maybe I can find the area where all three inequalities hold by considering the overlap of these regions. Let me try to sketch this mentally.

First, note that the inequalities y < 1 + x and y > x - 1. Let me think about the region between these two lines. The lines y = x + 1 and y = x - 1 are parallel? Wait, no, they have the same slope (1) and are parallel. The distance between them is 2 units vertically. However, since we are in the square [0,3]^2, the area between them will depend on where they intersect the square.

But actually, the line y = x + 1 intersects the square where? Let's see, if x is between 0 and 3, then y would go from 1 (when x=0) up to 4 (when x=3). But since our square only goes up to y=3, so the line y = x + 1 intersects the square at (0,1) and (2,3). Similarly, the line y = x - 1 intersects the square at (1,0) and (3,2). Therefore, between those lines, the region between them is kind of a band around the line y = x. However, we also have the third inequality x + y > 1. But since x and y are positive, x + y > 1 is automatically true except near the origin. However, maybe the combination of all three inequalities is the intersection of y < 1 + x, y > x - 1, and x + y > 1.

Wait, but perhaps I can find the area where all three inequalities hold. Let me consider that the three inequalities:

1. y < 1 + x

2. y > x - 1

3. y > 1 - x

But the third inequality is y > 1 - x. So if I combine inequalities 2 and 3, since y must be greater than both x - 1 and 1 - x. So, depending on the value of x, which one is larger?

Let me consider different ranges of x. For example, when x is between 0 and 1, 1 - x is greater than x - 1 (since 1 - x > x -1 when 1 +1 > x + x => 2 > 2x => x < 1). Therefore, for x < 1, the lower bound for y is 1 - x. For x > 1, the lower bound is x - 1. At x = 1, both are equal to 0. So, the region where y > max(1 - x, x - 1) is equivalent to y > |x - 1|.

Therefore, combining with the upper bound y < 1 + x, the region is between y = |x - 1| and y = 1 + x.

Therefore, the area where all three inequalities hold is the area between the lines y = |x - 1| and y = 1 + x, within the square [0,3]x[0,3].

But also, we need to make sure that y is between 0 and 3. But since we are already in the square [0,3]^2, the values of y are already bounded. However, we need to check if the lines y = |x - 1| and y = 1 + x intersect the boundaries of the square.

Let me try to visualize this. For x between 0 and 1, the lower boundary is y = 1 - x, which goes from (0,1) down to (1,0). The upper boundary is y = 1 + x, which goes from (0,1) up to (1,2). Therefore, for x between 0 and 1, the vertical slice between these two lines is from y = 1 - x to y = 1 + x.

Similarly, for x between 1 and 3, the lower boundary is y = x - 1, which goes from (1,0) up to (3,2). The upper boundary is still y = 1 + x, which goes from (0,1) up to (3,4), but since our square only goes up to y=3, the upper boundary intersects the square at (2,3) when x=2 (since 1 + x = 3 when x=2) and then continues to (3,4) which is outside. Therefore, for x between 1 and 2, the upper bound is y = 1 + x (from y=2 up to y=3), and for x between 2 and 3, the upper bound is y=3 instead of y = 1 + x because 1 + x would be greater than 3. So we need to split the integral into different ranges.

Therefore, to compute the area, I can split the integration into three parts:

1. x from 0 to 1
2. x from 1 to 2
3. x from 2 to 3

Let me check that.

First, for x in [0,1], upper bound is 1 + x, lower bound is 1 - x. So the vertical slice length is (1 + x) - (1 - x) = 2x.

But wait, no. Wait, the vertical slice is from y = lower bound to upper bound. So the length is (upper - lower). For x between 0 and 1, upper bound is 1 + x, lower bound is 1 - x. Therefore, vertical slice length is (1 + x) - (1 - x) = 2x. Therefore, integrating 2x from 0 to 1 gives the area for this interval.

For x between 1 and 2, the lower bound is y = x - 1, upper bound is y = 1 + x. The vertical slice length is (1 + x) - (x - 1) = 2. Therefore, integrating 2 from 1 to 2 gives 2*(2 -1) = 2*1 = 2.

For x between 2 and 3, the lower bound is still y = x -1, but upper bound is y = 3, since 1 + x would be more than 3. Therefore, vertical slice length is 3 - (x - 1) = 4 - x. Therefore, integrating (4 - x) from x=2 to x=3.

Therefore, total area would be:

Integral from 0 to1 of 2x dx + Integral from 1 to2 of 2 dx + Integral from 2 to3 of (4 - x) dx

Let me compute each integral.

First integral: ∫₀¹ 2x dx = [x²]₀¹ = 1 - 0 = 1

Second integral: ∫₁² 2 dx = 2*(2 - 1) = 2*1 = 2

Third integral: ∫₂³ (4 - x) dx. Let's compute that:

Antiderivative of 4 - x is 4x - (1/2)x². Evaluated from 2 to 3.

At x=3: 4*3 - 0.5*(9) = 12 - 4.5 = 7.5

At x=2: 4*2 - 0.5*(4) = 8 - 2 = 6

Subtracting: 7.5 - 6 = 1.5

Therefore, third integral is 1.5

Adding them up: 1 + 2 + 1.5 = 4.5

Therefore, the total area where all three inequalities hold is 4.5.

But wait, is that right? Let me check again.

But hold on, isn't the total area of the square 9, and we're saying that the area where triangle inequalities hold is 4.5? Then the probability would be 4.5 / 9 = 0.5. But that seems too simple. Let me check if I considered all constraints correctly.

Wait, but actually, I think there is a mistake here. Because the region I calculated (4.5) is the area where all three inequalities hold. But maybe I also need to consider that x and y are between 0 and 3. Wait, but no, since we already considered the square [0,3]^2. However, I need to check whether the inequalities x + y > 1 is already incorporated into the other inequalities?

Wait, no. Wait, the three inequalities are:

1. 1 + x > y => y < 1 + x

2. 1 + y > x => y > x - 1

3. x + y > 1 => y > 1 - x

But in my calculation, I considered combining 2 and 3 as y > max(x - 1, 1 - x), which is equivalent to y > |x -1|. Then combined with y < 1 + x. Therefore, the area between y = |x - 1| and y = 1 + x. But does this automatically include the condition x + y > 1? Wait, if we have y > |x - 1|, then for x between 0 and 1, y > 1 - x. Since 1 - x is positive, and since y is positive, then x + y is certainly greater than x + (1 - x) = 1. Therefore, x + y > 1 is automatically satisfied in that case. Similarly, for x between 1 and 3, y > x - 1. Then x + y > x + (x - 1) = 2x -1. When x ≥ 1, 2x -1 ≥ 1, so x + y ≥ 1. Therefore, in both cases, the third inequality x + y > 1 is automatically satisfied if we have y > |x -1|. Therefore, the region we computed (between y = |x -1| and y = 1 + x) automatically satisfies all three triangle inequalities.

Therefore, the area is indeed 4.5. Therefore, the probability would be 4.5 / 9 = 0.5. But that seems too straightforward. Let me check with some test points.

For example, suppose x = 0.5, y = 0.5. Then sides are 1, 0.5, 0.5. But 0.5 + 0.5 = 1, which is not greater than 1. Therefore, such a point should not be included. However, according to our region, y > |x - 1|. For x = 0.5, |x -1| = 0.5, so y must be greater than 0.5. Therefore, y = 0.5 is not included. Therefore, the region we calculated excludes points where y = |x -1|. So the area we calculated is correct in that sense.

Another example: x = 0.5, y = 1. Then 1, 0.5, 1. The sum of the two smaller sides is 0.5 + 1 = 1.5 > 1. So that's okay. So the region seems okay.

Wait, but let me check another point. Let me take x = 0.5, y = 0.6. Then 0.6 > |0.5 -1| = 0.5. So it is in the region. The sides are 1, 0.5, 0.6. The sum of the two smaller sides is 0.5 + 0.6 = 1.1 > 1. Okay. Similarly, for x = 2, y = 3. The sides are 1, 2, 3. But 1 + 2 = 3, which is not greater than 3. Therefore, this should not be in the region. However, according to our region, for x = 2, y must be greater than |2 -1| = 1, and less than 1 + 2 = 3. Therefore, y is in (1, 3). However, at y = 3, it's excluded. So the upper bound is y = 3. So when x = 2, y approaches 3, but does not include y = 3. So the point (2,3) is excluded, which is correct because 1 + 2 = 3 is not greater than 3. Therefore, our calculation seems correct.

But then the area is 4.5. Let me just verify the integrals again.

First integral from 0 to 1: ∫2x dx = 1. That's correct.

Second integral from 1 to 2: 2*(2 - 1) = 2. Correct.

Third integral from 2 to 3: ∫(4 - x) dx = [4x - x²/2] from 2 to 3. At 3: 12 - 9/2 = 12 - 4.5 = 7.5. At 2: 8 - 4/2 = 8 - 2 = 6. 7.5 - 6 = 1.5. Correct. Total area 1 + 2 + 1.5 = 4.5. So 4.5 / 9 = 0.5. Therefore, probability is 1/2. But is this correct?

Wait, but let me think again. If x and y are selected from (0,3), but the problem says selected at random from the interval (0,3). Wait, the interval is open, but since the probability of selecting exactly 0 or 3 is zero in continuous distribution, we can treat it as [0,3] for area calculation.

But let me think again if there are other constraints or if I missed something.

Wait, but suppose we take x and y both approaching 3. For example, x = 3 and y = 3. Then sides are 1, 3, 3. Then 1 + 3 > 3 is true (4 > 3), 3 + 3 > 1 is true, and 1 + 3 > 3 is true. So that's okay. So the point (3,3) is included. However, our calculation for x=3, y would have to be greater than |3 -1| = 2, and less than 1 + 3 = 4, but since y is up to 3, then y is between 2 and 3. However, the integral for x from 2 to 3 is ∫(4 - x) dx, which for x=3 gives 4 - 3 = 1. So at x=3, the vertical slice is 1, which corresponds to y between 2 and 3. So that's correct. Therefore, (3,3) is included, since y=3 is included in the upper bound. Wait, but in the integral, we integrated up to 3, and the upper bound was y=3. So the area is correct.

But then, according to this, the probability is 0.5. However, I have a feeling that the answer might be different. Maybe I made a mistake in considering the triangle inequalities?

Wait, let me think again. The three conditions are:

1. 1 + x > y
2. 1 + y > x
3. x + y > 1

But when we combine them, we have to make sure that all three hold. However, in our calculation, we considered the area between y = |x -1| and y = 1 + x. Is that sufficient?

Wait, let's check another point. Let me take x = 0.5, y = 0.6. Then 1 + x = 1.5 > 0.6; 1 + y = 1.6 > 0.5; x + y = 1.1 > 1. All good.

Take another point: x = 0.5, y = 0.5. But that's excluded since y needs to be greater than |x - 1| = 0.5. So the point y = 0.5 is on the boundary. Since the original problem says "selected at random", which is a continuous distribution, the probability of y being exactly 0.5 is zero. So that's okay.

Another point: x = 1.5, y = 0.5. Wait, but according to our region, for x = 1.5, the lower bound is |1.5 - 1| = 0.5. Therefore, y must be greater than 0.5, and less than 1 + 1.5 = 2.5. So y = 0.5 is excluded. But if we take y = 0.5, then check the triangle inequalities:

1 + x = 1 + 1.5 = 2.5 > y = 0.5 ✔

1 + y = 1 + 0.5 = 1.5 > x = 1.5? No, 1.5 is not greater than 1.5. It's equal. Therefore, the triangle inequality is not satisfied. Therefore, the point (1.5, 0.5) is excluded, which is consistent with our region.

But if we take y = 0.6, which is greater than 0.5, then 1 + y = 1.6 > 1.5. Therefore, it's okay. Therefore, our region is correct.

Therefore, the area seems correct. So probability is 4.5 / 9 = 0.5. So 1/2. Therefore, the answer is 1/2? But let me check another approach.

Alternative approach: Let's consider all three inequalities.

First, the triangle inequalities:

For sides 1, x, y, the three conditions are:

1. 1 + x > y

2. 1 + y > x

3. x + y > 1

Note that x and y are in (0,3). Let me consider these inequalities.

Let me rearrange inequalities 1 and 2:

From 1: y < 1 + x

From 2: y > x - 1

From 3: y > 1 - x

Therefore, combining 2 and 3, y > max(x - 1, 1 - x). So as before. Therefore, the region is between y = |x - 1| and y = 1 + x. Then integrating over x from 0 to 3.

But then, in the previous calculation, that gave area 4.5. Therefore, probability 0.5.

But maybe there's an error in assuming all three inequalities are satisfied? Let me think if there is another constraint.

Wait, but for example, if x is very small, say approaching 0, then y must be greater than |x -1| = 1 - x (since x is small). So y > 1 - x. But since x is approaching 0, y must be greater than 1. However, then, the other inequality y < 1 + x. If x approaches 0, then y < 1 + x approaches 1. Therefore, y must be greater than 1 and less than 1, which is impossible. Therefore, in this case, there is no solution? Wait, but this contradicts with our earlier integral. Wait, but when x approaches 0, for example, take x = 0.1. Then y must be greater than |0.1 -1| = 0.9, and less than 1 + 0.1 = 1.1. So y is between 0.9 and 1.1. Therefore, for x=0.1, there is a region. So when x approaches 0, the vertical slice is between y = 1 - x and y = 1 + x. So as x approaches 0, the length of the vertical slice approaches 2x, which goes to 0. Therefore, that is correct.

Moreover, at x = 0, the vertical slice length is 0. Therefore, the area is indeed 1. So that seems okay.

But let me think of another way. Suppose we imagine the square [0,3]x[0,3]. The region where the triangle inequalities hold is the set of points (x,y) such that all three inequalities hold. Let me try to visualize this.

First, the region where x + y > 1 is the square minus the triangle near (0,0). Then, the regions where 1 + x > y and 1 + y > x are also regions within the square. The intersection of these three regions is what we need.

Alternatively, maybe using geometric probability.

Alternatively, think of the problem as follows: for x and y in (0, 3), what's the probability that 1, x, y satisfy triangle inequalities. Which is equivalent to:

max(1, x, y) < 1 + min(1, x, y)

Because for three sides a, b, c, the triangle inequality is equivalent to the sum of the two smaller sides being greater than the largest side. Therefore, if we let a, b, c be 1, x, y. Then, the maximum of the three must be less than the sum of the other two.

Therefore, we can consider cases based on which of 1, x, y is the largest.

Case 1: 1 is the largest side. Then, we need 1 < x + y.

Case 2: x is the largest side. Then, we need x < 1 + y.

Case 3: y is the largest side. Then, we need y < 1 + x.

However, since x and y are symmetric, Cases 2 and 3 will have the same area. Therefore, we can compute Case 1, then double the area for Cases 2 and 3, but we have to be careful about overlapping.

Wait, actually, it's better to compute all cases.

But maybe this approach can help verify.

Case 1: 1 is the largest side. That is, x ≤ 1 and y ≤ 1. Then, we need x + y > 1. So this is the region where x and y are both between 0 and 1, and x + y > 1. The area here is the area of the square [0,1]x[0,1] minus the area where x + y ≤ 1. The area of [0,1]^2 is 1. The area where x + y ≤1 is 0.5, so the area for Case 1 is 1 - 0.5 = 0.5.

Case 2: x is the largest side. That is, x ≥ 1 and x ≥ y. Then, we need x < 1 + y. Which simplifies to y > x - 1. But since x ≥ 1 and x ≥ y, we have y ≤ x. Also, since x ≥ 1 and y ≤ x, and y > x - 1. So the region is x between 1 and 3, y between max(x - 1, 0) and x. However, since y must be between 0 and 3. But also, since we are in the case where x is the largest side, y ≤ x. So for x from 1 to 3, y is between max(x - 1, 0) and x. But since x ≥ 1, max(x -1, 0) = x -1.

Therefore, the area for Case 2 is ∫ from x=1 to 3 [x - (x - 1)] dx = ∫ from 1 to 3 1 dx = 3 - 1 = 2.

Wait, but hold on. Wait, that can't be right. Because for each x, the upper bound of y is x, and the lower bound is x - 1. Therefore, the length is x - (x -1 ) = 1. Therefore, integrating 1 from x=1 to 3 gives area 2. However, we need to check if y is within [0, 3]. Since x is up to 3, y is up to x=3, so it's okay. Therefore, the area for Case 2 is 2.

But wait, this seems different from the previous approach. Because in the previous approach, we had total area 4.5, but with Case 1 being 0.5 and Case 2 and Case 3 contributing 2 each? Wait, let me check.

Wait, Case 3: y is the largest side. Similarly, y ≥ 1 and y ≥ x. Then, we need y < 1 + x. Which is similar to Case 2 but with x and y swapped. Therefore, the area for Case 3 is also 2.

But then total area would be Case1 + Case2 + Case3 = 0.5 + 2 + 2 = 4.5, which matches our previous result. Therefore, the total area is 4.5. Therefore, probability is 4.5 / 9 = 0.5. Therefore, 1/2. Therefore, the probability is 1/2.

But let me check Case 2 again. For Case 2: x is the largest side. So x ≥ max(1, y). Therefore, for x between 1 and 3, and y between x - 1 and x. But when x is between 1 and 2, y ranges from x - 1 (which is between 0 and 1) to x (which is between 1 and 2). When x is between 2 and 3, y ranges from x -1 (which is between 1 and 2) to x (which is between 2 and 3). Therefore, integrating over x from 1 to 3, the vertical slice length is always 1. Therefore, area is 2. Similarly for Case 3.

But then, why does this approach give the same result as before? Therefore, seems that 0.5 is correct. However, I need to check if there's any overlap between the cases. For example, when x=1 and y=1, which case does it belong to? In Case 1, since all sides are equal (1,1,1), so 1 is the largest side. Also, x and y are equal to 1. However, in Case 2, x=1 and y=1: since x is the largest side, which is true. But the problem is that the cases are not mutually exclusive. However, since in Case 1, x ≤ 1 and y ≤ 1. In Case 2, x ≥ 1 and x ≥ y. In Case 3, y ≥ 1 and y ≥ x. Therefore, these cases are actually mutually exclusive. Because if x ≤1 and y ≤1, then neither x nor y is the largest unless they are both equal to 1. Similarly, if x is the largest, then x ≥1 and y ≤ x. If y is the largest, then y ≥1 and x ≤ y. Therefore, there is no overlap between Case 1 and Cases 2 and 3, nor between Case 2 and Case 3. Therefore, the total area is indeed 0.5 + 2 + 2 = 4.5.

Therefore, the probability is 4.5 / 9 = 0.5. Therefore, 1/2.

But let me check with another method. For instance, using coordinates.

The total area is 9. The area where triangle inequalities hold is 4.5. So probability is 1/2.

But let me think about when x and y are selected from (0, 3). Let me think of the three conditions again.

But maybe I can think of the problem as follows. Since the triangle inequalities are equivalent to all sides being less than the sum of the other two. Which is the same as:

- x + y > 1

- x + 1 > y => y < x + 1

- y + 1 > x => x < y + 1

Therefore, combining these, the region is defined by:

x < y + 1

y < x + 1

x + y > 1

Which is equivalent to the region between the lines y = x + 1, y = x - 1, and x + y = 1.

But graphically, in the square [0,3]x[0,3], the area where these conditions are satisfied is the area between the lines y = x + 1, y = x -1, and above the line x + y =1. Which as we calculated earlier is 4.5. Hence, probability is 0.5. So perhaps this is correct.

But I'm still a bit uncertain because I might have missed some constraints. Let me try to compute the area using another method. Let's consider that the problem is symmetric with respect to x and y. Therefore, the area where the triangle inequalities hold should be the same as the area where all three inequalities hold.

Alternatively, if we consider all possible x and y, the probability that they satisfy triangle inequalities is equal to the probability that x + y > 1, and |x - y| < 1. Wait, because:

If we have 1, x, y, then the triangle inequalities are equivalent to:

x + y > 1 and |x - y| < 1.

Wait, is that correct?

Wait, let me think. The triangle inequalities are:

1. x + y > 1

2. x + 1 > y => x - y > -1 => y - x < 1

3. y + 1 > x => y - x > -1 => same as above.

Therefore, combining inequalities 2 and 3 gives: |x - y| < 1. Therefore, the triangle inequalities are equivalent to:

x + y > 1 and |x - y| < 1.

Therefore, the region we need is the intersection of x + y > 1 and |x - y| < 1.

So, if I can compute this area, that would be easier.

Let me try this approach.

First, |x - y| < 1 is equivalent to y -1 < x < y +1. So for each y, x is between y -1 and y +1. However, x must also be in (0, 3), and similarly for y.

But also, x + y > 1.

Therefore, the region is the set of (x, y) such that |x - y| < 1 and x + y > 1.

Therefore, the area can be calculated by integrating over the square [0,3]^2 the region where both conditions hold.

Alternatively, since this is symmetric, maybe compute the area where |x - y| < 1 and x + y > 1.

But how?

Alternatively, since |x - y| < 1 defines a band around the line y = x, between y = x -1 and y = x + 1. The intersection with x + y > 1 is the region in this band above the line x + y =1.

Therefore, we can compute this area.

But perhaps again, breaking into cases based on the value of x and y.

Alternatively, note that the area where |x - y| < 1 is the area between the lines y = x + 1 and y = x -1. Which in the square [0,3]^2, is a band around the diagonal. The area of this band can be calculated as follows:

The area between y = x + 1 and y = x -1 in the square.

But since the square is from (0,0) to (3,3), the area of the band |x - y| < 1 is total area minus the areas of the regions where |x - y| ≥ 1.

Which would be the two triangles where y ≥ x + 1 and y ≤ x - 1.

The area where y ≥ x + 1 is a triangle with vertices at (0,1), (0,3), (2,3). Similarly, the area where y ≤ x -1 is a triangle with vertices at (1,0), (3,0), (3,2). Both triangles have base and height of 2, area is 0.5*2*2 = 2 each. So total area of these regions is 2 + 2 = 4. Therefore, the area of the band |x - y| < 1 is total area 9 - 4 = 5.

But we also need to intersect this band with the region x + y > 1. However, the band |x - y| < 1 already includes regions where x + y might be less than 1. For example, near the origin, where x and y are small, and |x - y| < 1. But x + y could be less than 1. Therefore, we need to find the area of the intersection between |x - y| < 1 and x + y > 1.

Therefore, the area we want is equal to the area of the band |x - y| < 1 minus the area of the part of the band where x + y ≤ 1.

So, compute area_band = 5, area_part = area of band where x + y ≤ 1. Then desired area = 5 - area_part.

What is area_part?

The region where |x - y| < 1 and x + y ≤ 1.

This region is the intersection of the band |x - y| < 1 and the region below x + y = 1.

Let me visualize this. The line x + y = 1 intersects the band |x - y| < 1.

To find the area where both |x - y| < 1 and x + y ≤ 1, let's find the overlapping region.

First, the region x + y ≤ 1 is a triangle with vertices at (0,0), (0,1), (1,0).

Within this triangle, we need to find the region where |x - y| < 1. However, since x and y are both between 0 and 1, and x + y ≤ 1, then |x - y| ≤ 1. Because the maximum difference between x and y occurs when one is 0 and the other is 1, which gives |x - y| = 1. Therefore, within the triangle x + y ≤ 1, the condition |x - y| < 1 is automatically satisfied except on the boundary where |x - y| = 1. But since it's a continuous distribution, the area on the boundary is zero. Therefore, the area_part = area of the triangle x + y ≤1 = 0.5.

Therefore, desired area = 5 - 0.5 = 4.5. Which matches our previous result!

Therefore, the area is indeed 4.5, and the probability is 4.5 / 9 = 0.5.

Therefore, the probability is 1/2.

Wait, but now I feel more confident. So why did I think it was too simple? Maybe because I didn't expect such a clean answer. However, all methods point to 4.5 / 9 = 1/2. Therefore, the probability is 1/2.

Therefore, the answer is \boxed{\dfrac{1}{2}}.

**Final Answer**
\boxed{\dfrac{1}{2}}

To determine the probability that a triangle with side lengths 1, $ x $, and $ y $ can exist when $ x $ and $ y $ are selected randomly from the interval $ (0,3) $, we must rely on the **triangle inequality**. For three sides $ a $, $ b $, and $ c $ to form a triangle, the following must all hold:

- $ a + b > c $
- $ a + c > b $
- $ b + c > a $

Here, $ a = 1 $, $ b = x $, and $ c = y $, so the triangle condition becomes:

- $ x + y > 1 $
- $ 1 + x > y \Rightarrow y < x + 1 $
- $ 1 + y > x \Rightarrow x < y + 1 $

These three inequalities together define a region in the $ xy $-plane within the square $ [0,3] \times [0,3] $. The sample space is this square, which has area:

$$
\text{Total area} = 3 \times 3 = 9
$$

Our goal is to find the **area of the region** that satisfies all three inequalities, then divide by 9 to get the probability.

---

### Step 1: Geometric Interpretation

The three inequalities define a region of interest. Note that:

- $ x + y > 1 $: the area **above** the line $ x + y = 1 $
- $ x < y + 1 \Rightarrow y - x < 1 $: **below** the line $ y = x + 1 $
- $ y < x + 1 \Rightarrow x - y < 1 $: **above** the line $ y = x - 1 $

Thus, the region is:

$$
\text{Region} = \{ (x, y) \in [0,3]^2 \mid x + y > 1, \ |x - y| < 1 \}
$$

Which is the **intersection** of the area **above** the line $ x + y = 1 $ and the area **between** the lines $ y = x - 1 $ and $ y = x + 1 $.

This region lies **between the two diagonal lines** and **above the line $ x + y = 1 $**.

---

### Step 2: Compute the Desired Area

We can find the desired area more systematically. Consider the square $ [0,3] \times [0,3] $, with total area 9.

Let’s define the region $ A $ where $ |x - y| < 1 $, which is the area **between** the lines $ y = x + 1 $ and $ y = x - 1 $. The square is divided by these lines into regions where $ |x - y| < 1 $ and $ |x - y| \ge 1 $.

- The region where $ |x - y| \ge 1 $ consists of two triangles:
  - One above $ y = x + 1 $, with vertices at $ (0,1), (0,3), (2,3) $
  - One below $ y = x - 1 $, with vertices at $ (1,0), (3,0), (3,2) $

Each of these triangles has area:

$$
\text{Area of triangle} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 2 = 2
$$

So total area outside the region $ |x - y| < 1 $ is:

$$
2 + 2 = 4
$$

Thus, the area where $ |x - y| < 1 $ is:

$$
9 - 4 = 5
$$

Now, from this band, we must remove the part where $ x + y \leq 1 $, since this violates the triangle inequality.

- The region where $ x + y \leq 1 $ is a triangle with vertices at $ (0,0), (0,1), (1,0) $, so:

$$
\text{Area} = \frac{1}{2} \cdot 1 \cdot 1 = 0.5
$$

So the area where all triangle inequalities are satisfied is:

$$
5 - 0.5 = 4.5
$$

---

### Step 3: Compute the Probability

The total area of the sample space is 9. Hence, the probability is:

$$
\text{Probability} = \frac{\text{Desired Area}}{\text{Total Area}} = \frac{4.5}{9} = \frac{1}{2}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{2}}
$$

Solution:
If a triangle with sides of length 1, $x$, and $y$ exists, the triangle inequality must be satisfied, which states that $x+y>1$, $1+x>y$, and $1+y>x$.  We can draw a plane with $x$ and $y$ axes and shade in the area where all of these inequalities are satisfied.

[asy]
draw((0,0)--(3,0)--(3,3)--(0,3));
draw((0,0)--(0,3));
label("$x$",(3,0),S);
label("$y$",(0,3),W);
fill((1,0)--(3,2)--(3,3)--(2,3)--(0,1)--cycle,gray(.7));
draw((1,-.1)--(1,.1));
draw((2,-.1)--(2,.1));
draw((.1,1)--(-.1,1));
draw((.1,2)--(-.1,2));

draw((1,0)--(0,1));
draw((1,0)--(3,2));
draw((0,1)--(2,3));

[/asy]

The total area of the square is $3^2=9$.  The area of the unshaded region is $2^2+1/2=9/2$.  Thus, the shaded area is $9/2$ and the probability that such a triangle exists is $(9/2)/9=\boxed{\frac{1}{2}}$.