Question

If three numbers are randomly selected from the set \( \{1,2,3,\dots,50\} \), then the probability that they are in arithmetic progression is: 

• A. \( \frac{3}{50} \)
• B. \( \frac{3}{98} \)
• C. \( \frac{3}{49} \)
• D. \( \frac{3}{25} \)

Hint:

To count sequences forming an arithmetic progression, consider choosing the middle term first and then counting valid common differences while ensuring values remain within the given set.

Answer 1

The Correct Option is B

Step 1: Total Ways to Select Three Numbers 
The given set is: \[ \{1,2,3,\dots,50\}. \] The number of ways to choose any three numbers from this set is: \[ \text{Total selections} = \binom{50}{3} = \frac{50!}{3!(50-3)!} = \frac{50 \times 49 \times 48}{6} = 19600. \] 

Step 2: Counting Selections that Form an Arithmetic Progression 
For three numbers to be in arithmetic progression, they must be of the form: \[ a-d, a, a+d. \] where \( a \) is the middle term, and \( d \) is the common difference. The constraints are: - \( a-d \geq 1 \), - \( a+d \leq 50 \). Rewriting: \[ 1 \leq a - d, \quad a + d \leq 50. \] The number of valid sequences is determined by choosing \( a \) and \( d \) such that they remain within the set boundaries. Counting possible values of \( (a,d) \): - The middle term \( a \) can be any value from 2 to 49. - The common difference \( d \) can range from 1 to \( \min(a-1, 50-a) \). Summing up all valid pairs: \[ 1 + 2 + \dots + 24 = \frac{24 \times 25}{2} = 300. \]

Step 3: Computing Probability 
\[ P(\text{AP}) = \frac{\text{favorable cases}}{\text{total cases}} = \frac{300}{19600}. \] Simplifying: \[ P(\text{AP}) = \frac{3}{98}. \] 

Step 4: Conclusion 
Thus, the final probability that three randomly selected numbers form an arithmetic progression is: \[ \boxed{\frac{3}{98}}. \]