Question

Let \[ A=\{1,2,3,4,\ldots,20\}. \] If three distinct elements are drawn from the set \(A\) at random, then the probability that the three numbers drawn are in increasing G.P. with common ratio as a non integral rational value is

• A. \(\dfrac{2}{255}\)
• B. \(\dfrac{3}{20}\)
• C. \(\dfrac{1}{380}\)
• D. \(\dfrac{2}{19}\)

Hint:

For G.P. problems involving integers, first generate all possible triples within the given range and then eliminate those having an integral common ratio.

Answer 1

The Correct Option is C

Step 1: Find the total number of selections. The total number of ways of choosing three distinct elements from \[ A=\{1,2,\ldots,20\} \] is \[ ^{20}C_3 = 1140. \]

Step 2: Find the favourable selections. Let the three numbers be \[ a,\;ar,\;ar^2, \] where \[ r>1 \] is a non-integral rational number. Checking all possible values within \(20\), the increasing G.P.s are \[ (1,2,4),\; (1,3,9),\; (2,4,8). \] Among these, \[ (1,2,4),\; (2,4,8) \] have integral common ratio. The only G.P. having a non-integral rational common ratio is \[ (4,6,9), \] whose common ratio is \[ \frac32. \] Hence, the number of favourable selections is \[ 1. \]

Step 3: Find the probability. Therefore, \[ P = \frac{1}{^{20}C_3} = \frac{1}{1140} = \frac{1}{380}. \] Hence, \[ \boxed{\frac{1}{380}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.