Probability
Counting with Constraints
Grade 12
Question:
<p>Three numbers are randomly selected from the set \(\{10, 11, 12, \ldots, 100\}\). Probability that they form a Geometric progression with integral common ratio greater than 1 is:</p>
<p>(a) \(\frac{15}{\binom{91}{3}}\)</p>
<p>(b) \(\frac{16}{\binom{91}{3}}\)</p>
<p>(c) \(\frac{17}{\binom{91}{3}}\)</p>
<p>(d) \(\frac{18}{\binom{91}{3}}\)</p>
Step-by-Step Solution
Key Concept: Count the number of geometric progressions with integral ratio \(r > 1\) where all three terms lie in the given range. Check each possible ratio separately.
<p>The set has 91 elements (from 10 to 100).</p><p>Total ways to select 3 numbers: \(\binom{91}{3}\)</p><p>For a GP with first term \(a\) and common ratio \(r > 1\) (integer), we need \(a, ar, ar^2\) all in \([10, 100]\).</p><p>For \(r = 2\): \(a \leq 25\) gives 16 GPs (\(a = 10, 11, \ldots, 25\))</p><p>For \(r = 3\): \(a \leq 11\) gives 2 GPs (\(a = 10, 11\))</p><p>Total favorable outcomes = 16 + 2 = 18</p><p>Probability = \(\frac{18}{\binom{91}{3}}\)</p>
Correct Answer: D