Statement-1: A natural x is chosen at random from the first 100 natural numbers. The probability that
`((x-10)(x-50))/(x-30) lt 0` is 0.28
Statement-2 : For any event `A, 0 le P(A) le 1`.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze the inequality given and determine the probability that it holds true for a natural number \( x \) chosen from the first 100 natural numbers.

### Step 1: Analyze the inequality

We start with the inequality:

\[ \frac{(x-10)(x-50)}{(x-30)} < 0 \]

This inequality can be broken down into two parts: the numerator \((x-10)(x-50)\) and the denominator \((x-30)\).

### Step 2: Identify the critical points

The critical points occur where the expression is equal to zero or undefined. These points are:

- From the numerator: \( x - 10 = 0 \) gives \( x = 10 \) and \( x - 50 = 0 \) gives \( x = 50 \).
- From the denominator: \( x - 30 = 0 \) gives \( x = 30 \).

Thus, the critical points are \( x = 10, 30, 50 \).

### Step 3: Determine the intervals

The critical points divide the number line into intervals. We will test the sign of the expression in each interval:

1. \( (-\infty, 10) \)
2. \( (10, 30) \)
3. \( (30, 50) \)
4. \( (50, \infty) \)

### Step 4: Test each interval

- **Interval 1: \( (-\infty, 10) \)** (e.g., \( x = 0 \)):
\[ \frac{(0-10)(0-50)}{(0-30)} = \frac{(-10)(-50)}{-30} = \frac{500}{-30} < 0 \quad \text{(True)} \]

- **Interval 2: \( (10, 30) \)** (e.g., \( x = 20 \)):
\[ \frac{(20-10)(20-50)}{(20-30)} = \frac{(10)(-30)}{-10} = \frac{-300}{-10} > 0 \quad \text{(False)} \]

- **Interval 3: \( (30, 50) \)** (e.g., \( x = 40 \)):
\[ \frac{(40-10)(40-50)}{(40-30)} = \frac{(30)(-10)}{10} = \frac{-300}{10} < 0 \quad \text{(True)} \]

- **Interval 4: \( (50, \infty) \)** (e.g., \( x = 60 \)):
\[ \frac{(60-10)(60-50)}{(60-30)} = \frac{(50)(10)}{30} > 0 \quad \text{(False)} \]

### Step 5: Collect valid intervals

From our tests, the expression is negative in the intervals:

- \( (-\infty, 10) \)
- \( (30, 50) \)

### Step 6: Count the valid natural numbers

Now we find the natural numbers in these intervals within the range of the first 100 natural numbers:

1. For \( (-\infty, 10) \): The valid natural numbers are \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \) (total of 9 numbers).
2. For \( (30, 50) \): The valid natural numbers are \( 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49 \) (total of 19 numbers).

### Step 7: Calculate the total valid numbers

Total valid numbers = \( 9 + 19 = 28 \).

### Step 8: Calculate the probability

The probability \( P \) that the inequality holds is given by:

\[ P = \frac{\text{Number of valid outcomes}}{\text{Total outcomes}} = \frac{28}{100} = 0.28 \]

### Conclusion

Thus, Statement 1 is true, and since Statement 2 is also true (as it states that the probability of any event lies between 0 and 1), both statements are true.