<p>Six points are there on a circle from which two triangles are drawn with no vertex common. Find the probability that none of the sides of the triangles intersect.</p>
Step-by-Step Solution
Key Concept: Non-intersecting triangles from 6 points on a circle requires that the 3 vertices of each triangle form a contiguous block when arranged cyclically, preventing any side crossings.
<p><strong>Step 1:</strong> Label 6 points on circle as 1,2,3,4,5,6 in order. We need to choose 3 for first triangle and remaining 3 for second triangle.</p><p><strong>Step 2:</strong> Total ways to partition 6 points into two groups of 3 = C(6,3)/2 = 20/2 = 10 (dividing by 2 since triangles are unordered).</p><p><strong>Step 3:</strong> For sides NOT to intersect, vertices of each triangle must be consecutive on the circle (forming a contiguous arc). For triangle with vertices at positions forming arc, the other 3 must also form a contiguous arc.</p><p><strong>Step 4:</strong> If triangle 1 has 3 consecutive vertices, triangle 2 automatically has the other 3 consecutive vertices. Count: starting positions for first consecutive block of 3: positions can start at 1,2,3,4,5, or 6 → 6 ways. But since triangles are unordered, we have 6/2 = 3 distinct configurations.</p><p><strong>Step 5:</strong> Actually, with unordered triangles: the valid non-intersecting pairs are exactly those where one triangle's vertices form a contiguous arc of 3. There are exactly 6 such configurations (by rotation), giving 6/2 = 3 unordered pairs.</p><p><strong>Step 6:</strong> Probability = 3/10</p><p>∴ Answer: B</p>
Correct Answer: B