<p>Mr. A lives at origin on the Cartesian plane and has his office at (4, 5). His friend lives at (2, 3) on the same plane. Mr. A can go to his office traveling one block at a time either in the +y or +x direction. If all possible paths are equally likely then the probability that Mr. A passed his friends house is (shortest path for any event must be considered)</p>
Step-by-Step Solution
Key Concept: The probability equals the ratio of paths passing through the friend's house to total shortest paths. Use the combination formula C(m+n, m) to count lattice paths: total paths from origin to office = C(9,4), and paths through friend's house = C(5,2) × C(4,2).
<p><strong>Step 1:</strong> Find total shortest paths from origin (0,0) to office (4,5). Moving right 4 times and up 5 times requires 9 total moves. Total paths = C(9,4) = C(9,5) = 126</p><p><strong>Step 2:</strong> Find paths passing through friend's house at (2,3). These are shortest paths that go (0,0)→(2,3)→(4,5). From (0,0) to (2,3): need 2 right, 3 up moves = C(5,2) = 10 paths. From (2,3) to (4,5): need 2 right, 2 up moves = C(4,2) = 6 paths.</p><p><strong>Step 3:</strong> Paths through friend's house = C(5,2) × C(4,2) = 10 × 6 = 60</p><p><strong>Step 4:</strong> Probability = 60/126 = 10/21</p><p>∴ Answer: 10/21</p>
Correct Answer: 2