2019 Mock AMC 10B Problems/Problem 8

Revision as of 17:51, 11 November 2023 by Ir 55 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Solution 1:

the probability for each point will be the same if the points are equidistant from the origin because it will take the same amount of moves. There are 8 points inclusive of (4,3) --> (3,4),(-3,4),(-4,3),(3,-4),(4,-3), (-3,-4),(-4,-3)

Therefore the solution is 8