Probability
Geometric/Classical Probability
Grade 12
Question:
<p>In a \(n\)-sided regular polygon, the probability that the two diagonal chosen at random will intersect inside the polygon is</p>
<p>(1) \(\dfrac{{}^{2n}C_2}{{}^{({}^nC_2 - n)}C_2}\)</p>
<p>(2) \(\dfrac{{}^{n(n-1)}C_2}{{}^{({}^nC_2 - n)}C_2}\)</p>
<p>(3) \(\dfrac{{}^nC_4}{{}^{({}^nC_2 - n)}C_2}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: For two diagonals to intersect inside a convex polygon, they must connect four distinct vertices forming a convex quadrilateral. Any 4 vertices of a regular polygon automatically form a unique pair of intersecting diagonals, so the probability equals the ratio of ways to choose 4 vertices to total ways to choose 2 diagonals.
<p><strong>Step 1:</strong> Count total diagonals in an n-sided polygon.</p><p>Total diagonals = $\binom{n}{2} - n = \frac{n(n-3)}{2}$</p><p><strong>Step 2:</strong> Count ways to choose 2 diagonals.</p><p>Total ways = $\binom{\frac{n(n-3)}{2}}{2} = \frac{n(n-3)(n(n-3)-2)}{8}$</p><p><strong>Step 3:</strong> Two diagonals intersect inside the polygon if and only if their four endpoints are all distinct and form a convex quadrilateral.</p><p>This occurs for every selection of 4 vertices from n vertices, giving exactly one pair of intersecting diagonals.</p><p>Intersecting diagonal pairs = $\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}$</p><p><strong>Step 4:</strong> Calculate probability.</p><p>$$P = \frac{\binom{n}{4}}{\binom{\frac{n(n-3)}{2}}{2}} = \frac{\frac{n(n-1)(n-2)(n-3)}{24}}{\frac{n(n-3)(n(n-3)-2)}{8}}$$</p><p>$$= \frac{n(n-1)(n-2)(n-3)}{24} \cdot \frac{8}{n(n-3)(n^2-3n-2)}$$</p><p>$$= \frac{(n-1)(n-2)}{3(n^2-3n-2)} = \boxed{\frac{(n-1)(n-2)}{3(n-4)(n+1)}}$$</p><p>Or equivalently: $\binom{n}{4}/\binom{\frac{n(n-3)}{2}}{2}$</p><p>∴ Answer: C</p>
Correct Answer: C