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 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, we can follow these steps:

### Step 1: Understand the Set and Total Outcomes
The set consists of integers from 10 to 100, inclusive. The total number of elements in this set can be calculated as:
\[ \text{Total numbers} = 100 - 10 + 1 = 91 \]
The total number of ways to select 3 numbers from this set is given by the combination formula:
\[ \text{Total combinations} = \binom{91}{3} \]

### Step 2: Identify the Condition for GP
For three numbers \(a\), \(ar\), and \(ar^2\) to be in a GP, where \(r\) is the common ratio, we need \(r > 1\) and all three numbers must be within the range of 10 to 100.

### Step 3: Calculate Favorable Outcomes for Different Values of \(r\)
We will consider integral values of \(r\) starting from 2 and check how many valid GPs can be formed.

#### Case 1: \(r = 2\)
The numbers will be \(a\), \(2a\), and \(4a\).
- The maximum value for \(a\) such that \(4a \leq 100\) is \(a \leq 25\).
- Valid values for \(a\) are from 10 to 25, giving us \(25 - 10 + 1 = 16\) valid GPs.

#### Case 2: \(r = 3\)
The numbers will be \(a\), \(3a\), and \(9a\).
- The maximum value for \(a\) such that \(9a \leq 100\) is \(a \leq 11.11\), thus \(a\) can be 10 or 11.
- Valid values for \(a\) are 10 and 11, giving us \(2\) valid GPs.

#### Case 3: \(r = 4\)
The numbers will be \(a\), \(4a\), and \(16a\).
- The maximum value for \(a\) such that \(16a \leq 100\) is \(a \leq 6.25\), thus \(a\) can only be 10.
- No valid values for \(a\) are available.

#### Case 4: \(r = 5\)
The numbers will be \(a\), \(5a\), and \(25a\).
- The maximum value for \(a\) such that \(25a \leq 100\) is \(a \leq 4\), thus no valid values for \(a\).

Continuing this process for \(r = 6\) to \(r = 10\) will yield no additional valid GPs since the maximum values for \(a\) will be less than 10.

### Step 4: Total Favorable Outcomes
Adding the valid GPs from the cases:
- For \(r = 2\): 16 GPs
- For \(r = 3\): 2 GPs
- Total favorable outcomes = \(16 + 2 = 18\)

### Step 5: Calculate the Probability
The probability \(P\) that the three selected numbers form a GP with integral common ratio greater than 1 is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total combinations}} = \frac{18}{\binom{91}{3}} \]

### Step 6: Calculate \(\binom{91}{3}\)
\[ \binom{91}{3} = \frac{91 \times 90 \times 89}{3 \times 2 \times 1} = 123,410 \]
Thus, the probability becomes:
\[ P = \frac{18}{123,410} \]

### Final Answer
The probability that three randomly selected numbers from the set form a GP with an integral common ratio greater than 1 is:
\[ P = \frac{18}{123,410} \]