Question

Two numbers are selected at random (without replacement) from the first 50 natural numbers. The probability that their sum is even is:

• A. \( \frac{24}{50} \)
• B. \( \frac{24}{29} \)
• C. \( \frac{26}{49} \)
• D. \( \frac{1}{2} \)

Hint:

For probability of sums being even, consider the parity (odd/even) of numbers and compute favorable cases based on combinations.

Answer 1

The Correct Option is B

- The first 50 natural numbers are 1 to 50.
- A sum of two numbers is even if both are odd or both are even.
- Count odd and even numbers:
- Even numbers: 2, 4, ..., 50 → \( \frac{50}{2} = 25 \) even numbers.
- Odd numbers: 1, 3, ..., 49 → 25 odd numbers.
- Total ways to choose 2 numbers (without replacement): \[ \binom{50}{2} = \frac{50 \cdot 49}{2} = 1225 \]
- Favorable cases:
- Both even: Choose 2 even numbers: \( \binom{25}{2} = \frac{25 \cdot 24}{2} = 300 \)
- Both odd: Choose 2 odd numbers: \( \binom{25}{2} = 300 \)
- Total favorable cases: \( 300 + 300 = 600 \).
- Probability: \[ \frac{\text{Favorable cases}}{\text{Total cases}} = \frac{600}{1225} = \frac{600 \div 25}{1225 \div 25} = \frac{24}{49} \]