If there are two natural numbers a, b such that their sum is 24, then the probability that `a.b ge 3/4` (maximum (a, b)) is

Options

A. `10/23`
B. `11/24`
C. `13/23`
D. `1/2`

Answer

To solve the problem, we need to find the probability that the product \( a \cdot b \) is greater than or equal to \( \frac{3}{4} \) of the maximum value of \( a \cdot b \) given that \( a + b = 24 \).

### Step-by-Step Solution:

1. **Identify the Maximum Value of \( a \cdot b \)**:
- Since \( a + b = 24 \), we can express \( b \) as \( b = 24 - a \).
- The product \( a \cdot b \) can be rewritten as:
\[ a \cdot b = a(24 - a) = 24a - a^2 \]
- This is a quadratic function in terms of \( a \). The maximum value occurs at the vertex of the parabola, which can be found using the formula \( a = -\frac{b}{2a} \) for a quadratic equation \( ax^2 + bx + c \).
- Here, \( a = -1 \) and \( b = 24 \):
\[ a = -\frac{24}{2 \cdot -1} = 12 \]
- Substituting \( a = 12 \) back into \( a \cdot b \):
\[ b = 24 - 12 = 12 \quad \Rightarrow \quad a \cdot b = 12 \cdot 12 = 144 \]

2. **Calculate \( \frac{3}{4} \) of the Maximum Value**:
- We need to find \( \frac{3}{4} \) of the maximum product:
\[ \frac{3}{4} \cdot 144 = 108 \]

3. **Set Up the Inequality**:
- We need to find the pairs \( (a, b) \) such that:
\[ a \cdot b \geq 108 \]
- Since \( b = 24 - a \), we substitute:
\[ a(24 - a) \geq 108 \quad \Rightarrow \quad 24a - a^2 \geq 108 \]
- Rearranging gives:
\[ a^2 - 24a + 108 \leq 0 \]

4. **Solve the Quadratic Inequality**:
- We can find the roots of the equation \( a^2 - 24a + 108 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{24 \pm \sqrt{24^2 - 4 \cdot 1 \cdot 108}}{2 \cdot 1} \]
\[ = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm \sqrt{144}}{2} = \frac{24 \pm 12}{2} \]
- This gives us the roots:
\[ a = \frac{36}{2} = 18 \quad \text{and} \quad a = \frac{12}{2} = 6 \]

5. **Determine the Valid Range for \( a \)**:
- The inequality \( a^2 - 24a + 108 \leq 0 \) holds between the roots:
\[ 6 \leq a \leq 18 \]

6. **Count the Favorable Cases**:
- The natural numbers \( a \) can take values from 6 to 18, inclusive. The possible values are:
\[ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 \]
- This gives us a total of \( 18 - 6 + 1 = 13 \) favorable cases.

7. **Calculate the Total Cases**:
- The total number of natural number pairs \( (a, b) \) such that \( a + b = 24 \) is from \( 1 \) to \( 23 \) (since \( a \) can be at most 23):
\[ 1, 2, 3, \ldots, 23 \]
- This gives us a total of 23 cases.

8. **Calculate the Probability**:
- The probability \( P \) is given by the ratio of favorable cases to total cases:
\[ P = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{13}{23} \]

### Final Answer:
The probability that \( a \cdot b \geq \frac{3}{4} \) of the maximum value of \( a \cdot b \) is \( \frac{13}{23} \).