Three numbers are randomly selected from the set {10, 11, 12, ………, 100}. Probability that they form a Geometric progression with integral common ratio greater than 1 is :

Options

A. `(15)/(""^(91)C_(3))`
B. `(16)/(""^(91)C_(3))`
C. `(17)/(""^(91)C_(3))`
D. `(18)/(""^(91)C_(3))`

Answer

To solve the problem, we need to find the probability that three randomly selected numbers from the set {10, 11, 12, ..., 100} form a geometric progression (GP) with an integral common ratio greater than 1.

### Step-by-Step Solution:

1. **Identify the Set of Numbers**:
The set of numbers is {10, 11, 12, ..., 100}. This set contains all integers from 10 to 100 inclusive.

2. **Determine the Total Number of Elements**:
The total number of elements in this set can be calculated as:
\[ \text{Total numbers} = 100 - 10 + 1 = 91 \]

3. **Form of a Geometric Progression**:
A geometric progression with first term \( a \) and common ratio \( r \) will have the terms:
\[ a, ar, ar^2 \]
We need to ensure that all three terms are within the set {10, 11, ..., 100}.

4. **Constraints on \( a \)**:
The first term \( a \) must satisfy:
\[ a \geq 10 \]
Additionally, since \( ar^2 \) must be less than or equal to 100, we have:
\[ ar^2 \leq 100 \implies a \leq \frac{100}{r^2} \]

5. **Finding Suitable Values for \( r \)**:
The common ratio \( r \) must be an integer greater than 1. We will consider values of \( r = 2, 3, \ldots \) until \( r^2 > 10 \) (since \( a \) must be at least 10).

- For \( r = 2 \):
\[ a \leq \frac{100}{2^2} = \frac{100}{4} = 25 \]
Thus, \( a \) can take values from 10 to 25, which gives us:
\[ 25 - 10 + 1 = 16 \text{ values} \]

- For \( r = 3 \):
\[ a \leq \frac{100}{3^2} = \frac{100}{9} \approx 11.11 \]
Thus, \( a \) can take values from 10 to 11, which gives us:
\[ 11 - 10 + 1 = 2 \text{ values} \]

- For \( r \geq 4 \):
\[ a \leq \frac{100}{r^2} < 10 \text{ (not possible)} \]
Therefore, we stop here.

6. **Total Favorable Outcomes**:
The total number of favorable outcomes is:
\[ 16 \text{ (for } r = 2\text{)} + 2 \text{ (for } r = 3\text{)} = 18 \]

7. **Calculate the Total Ways to Choose 3 Numbers**:
The total number of ways to choose 3 numbers from 91 is given by the combination:
\[ \binom{91}{3} = \frac{91 \times 90 \times 89}{3 \times 2 \times 1} = 125970 \]

8. **Calculate the Probability**:
The probability \( P \) that the three selected numbers form a GP with an integral common ratio greater than 1 is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{18}{\binom{91}{3}} = \frac{18}{125970} \]

### Final Answer:
Thus, the probability that the three numbers form a geometric progression with an integral common ratio greater than 1 is:
\[ \frac{18}{125970} \]