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. **Count the Total Numbers in the Set**:
The total number of integers from 10 to 100 can be calculated as:
\[ 100 - 10 + 1 = 91 \]
Therefore, there are 91 numbers in total.

3. **Determine the Total Ways to Choose 3 Numbers**:
The total ways to select 3 numbers from 91 can be calculated using the combination formula:
\[ \binom{91}{3} = \frac{91 \times 90 \times 89}{3 \times 2 \times 1} = 125,970 \]

4. **Identify the Conditions for a Geometric Progression**:
For three numbers \( a, b, c \) to form a GP, they must satisfy the condition:
\[ b^2 = a \cdot c \]
where \( b \) is the middle term.

5. **Consider Integral Common Ratios Greater than 1**:
We will consider different integral common ratios \( r \) starting from \( r = 2 \).

6. **Calculate Favorable Cases for \( r = 2 \)**:
- If \( a = 10 \), then the terms are \( 10, 20, 40 \) (valid).
- If \( a = 11 \), then the terms are \( 11, 22, 44 \) (valid).
- Continue this process until \( a \) reaches a point where \( a \cdot r^2 \) exceeds 100.
- The valid sequences for \( r = 2 \) are:
- \( 10, 20, 40 \)
- \( 11, 22, 44 \)
- \( 12, 24, 48 \)
- \( 13, 26, 52 \)
- \( 14, 28, 56 \)
- \( 15, 30, 60 \)
- \( 16, 32, 64 \)
- \( 17, 34, 68 \)
- \( 18, 36, 72 \)
- \( 19, 38, 76 \)
- \( 20, 40, 80 \)
- \( 21, 42, 84 \)
- \( 22, 44, 88 \)
- \( 23, 46, 92 \)
- \( 24, 48, 96 \)
- \( 25, 50, 100 \)
- Total valid sequences for \( r = 2 \): 16.

7. **Calculate Favorable Cases for \( r = 3 \)**:
- If \( a = 10 \), then the terms are \( 10, 30, 90 \) (valid).
- If \( a = 11 \), then the terms are \( 11, 33, 99 \) (valid).
- If \( a = 12 \), then \( 12, 36, 108 \) (invalid).
- Total valid sequences for \( r = 3 \): 2.

8. **Calculate Total Favorable Cases**:
The total number of favorable cases for \( r = 2 \) and \( r = 3 \) is:
\[ 16 + 2 = 18 \]

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

### Final Answer:
The probability that three randomly selected numbers from the set {10, 11, 12, ..., 100} form a geometric progression with an integral common ratio greater than 1 is:
\[ \frac{18}{125970} \]