Question

Two numbers are selected at random (without replacement) from the first 6 natural numbers. What is the probability that the difference of the numbers is less than 3?

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{5}{15} \)

Hint:

\textbf{Tip:} Always count unordered combinations to avoid double-counting in probability of selection problems.

Answer 1

The Correct Option is C

To determine the probability that the difference between two distinct numbers randomly selected from the set {1, 2, 3, 4, 5, 6} is less than 3, we proceed as follows:

1. Define the Sample Space:
   The set of the first 6 natural numbers is {1, 2, 3, 4, 5, 6}. The number of ways to select 2 distinct numbers from this set (without regard to order) is given by the combination formula \( \binom{6}{2} \), which equals 15. This represents the total possible outcomes.
2. Identify Favorable Outcomes:
   We seek pairs where the absolute difference between the two selected numbers is less than 3. These pairs are:
   
   • Absolute difference of 1: (2,1), (3,2), (4,3), (5,4), (6,5)
   • Absolute difference of 2: (3,1), (4,2), (5,3), (6,4)
   There are 9 such favorable outcomes.
3. Calculate Probability:
   Probability is calculated as the ratio of favorable outcomes to the total number of outcomes. The probability is therefore \( \frac{9}{15} \), which simplifies to \( \frac{3}{5} \).

Consequently, the probability that the difference between the two randomly selected numbers is less than 3 is \( \frac{3}{5} \).