`x_(1) , x_(2) , x_(3)`… . `x_(50)` are fifty real numbers such that `x_(r) lt x_(r + 1)` for r = 1 ,2 , 3 …….. , 49 . Five numbers out of these are picked up at random . The probability that the five numbers have `x_(20)` as the middle number is :

Options

A. `(""^(20) C_(2) xx ""^(30) C_(2))/(""^(20) C_(5))`
B. `(""^(30)C_(2) xx ""^(19) C_(2))/(""^(50) C_(5))`
C. `(""^(19) C_(2) xx ""^(30) C_(3))/(""^(50) C_(5))`
D. None of these

Answer

To solve the problem, we need to find the probability that when we randomly select 5 numbers from the set of 50 real numbers \( x_1, x_2, \ldots, x_{50} \) (where \( x_r < x_{r+1} \) for \( r = 1, 2, \ldots, 49 \)), the middle number of those 5 numbers is \( x_{20} \).

### Step-by-Step Solution:

1. **Understanding the Middle Number**:
- When we select 5 numbers, the middle number will be the 3rd number when arranged in increasing order. For \( x_{20} \) to be the middle number, we need 2 numbers less than \( x_{20} \) and 2 numbers greater than \( x_{20} \).

2. **Identifying the Numbers**:
- The numbers less than \( x_{20} \) are \( x_1, x_2, \ldots, x_{19} \) (total of 19 numbers).
- The numbers greater than \( x_{20} \) are \( x_{21}, x_{22}, \ldots, x_{50} \) (total of 30 numbers).

3. **Choosing the Numbers**:
- We need to choose 2 numbers from the 19 numbers less than \( x_{20} \).
- We also need to choose 2 numbers from the 30 numbers greater than \( x_{20} \).

4. **Calculating the Combinations**:
- The number of ways to choose 2 numbers from the 19 numbers is given by \( \binom{19}{2} \).
- The number of ways to choose 2 numbers from the 30 numbers is given by \( \binom{30}{2} \).

5. **Total Ways to Choose 5 Numbers**:
- The total number of ways to choose any 5 numbers from the 50 numbers is given by \( \binom{50}{5} \).

6. **Calculating the Probability**:
- The probability \( P \) that the middle number is \( x_{20} \) is given by the ratio of the favorable outcomes to the total outcomes:
\[ P = \frac{\binom{19}{2} \cdot \binom{30}{2}}{\binom{50}{5}} \]

### Final Expression:
Thus, the required probability that the middle number is \( x_{20} \) is:
\[ P = \frac{\binom{19}{2} \cdot \binom{30}{2}}{\binom{50}{5}} \]