<p>If three numbers are selected from the set of the first 20 natural numbers, the probability that they are in GP, is</p>
<p>(a) \(\frac{4}{285}\)</p>
<p>(b) \(\frac{?}{285}\)</p>
<p>(c) \(\frac{11}{1140}\)</p>
<p>(d) \(\frac{1}{71}\)</p>
Step-by-Step Solution
Key Concept: Count all triples from {1, 2, ..., 20} that form a geometric progression, then divide by total ways to select 3 numbers.
<p><strong>Solution:</strong> Total number of ways of selecting 3 numbers from first 20 natural numbers: $n(S) = \binom{20}{3} = 1140$</p><p>Three numbers are in GP. The favourable cases are: 1, 2, 4; 1, 3, 9; 1, 4, 16; 2, 4, 8; 2, 6, 18; 3, 6, 12; 4, 8, 16; 5, 10, 20; 4, 6, 9; 8, 12, 18; 9, 12, 16</p><p>Number of favourable cases: $n(E) = 11$</p><p>Required probability: $P(E) = \frac{n(E)}{n(S)} = \frac{11}{1140}$</p>
Correct Answer: c