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 solve the problem of determining the probability that the difference between two randomly selected numbers from the first 6 natural numbers is less than 3, we follow these steps: 

1. Identify the Sample Space:
   The first 6 natural numbers are: 1, 2, 3, 4, 5, 6. We are selecting 2 numbers without replacement, so the total number of ways to pick any 2 distinct numbers is given by \( \binom{6}{2} = 15 \). This is our sample space.
2. Calculate Favorable Outcomes:
   We need the difference between two selected numbers to be less than 3. Let's calculate the pairs:
   • Difference = 1: (2,1), (3,2), (4,3), (5,4), (6,5)
   • Difference = 2: (3,1), (4,2), (5,3), (6,4)
3. Calculate Probability:
   The probability is the ratio of the number of favorable outcomes to the total number of outcomes. Thus, the probability that the difference of the numbers is less than 3 is: \(\frac{9}{15} = \frac{3}{5}\).

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

Answer 2

Probability That the Difference Between Two Chosen Numbers Is Less Than 3

We are given a set of numbers: {1, 2, 3, 4, 5, 6}. The task is to find the probability that the absolute difference between two randomly chosen numbers from this set is less than 3.

Step 1: Total Number of Ways to Choose 2 Numbers

Since the order of selection doesn't matter, we use combinations to count the total number of possible unordered pairs:

\[ \binom{6}{2} = \frac{6 \times 5}{2} = 15 \]

Step 2: List All Favorable Pairs

We now list all unordered pairs where the absolute difference between the numbers is less than 3:

Valid pairs (difference < 3):

\[ (1,2),\ (1,3),\ (2,3),\ (2,4),\ (3,4),\ (3,5),\ (4,5),\ (4,6),\ (5,6) \]

Total favorable pairs = 9

Step 3: Calculate the Required Probability

Now we compute the probability using the ratio of favorable outcomes to total outcomes:

\[ P(\text{difference} < 3) = \frac{\text{Number of favorable pairs}}{\text{Total pairs}} = \frac{9}{15} = \frac{3}{5} \]

Final Answer:

The probability is \( \frac{3}{5} \).