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Question:

An 8in by 8in area is covered with with 1in square tiles. Two of the 64 tiles turn out to by faulty.?

What is the probability that the two faulty tiles share an edge?

Answer:

Break down the titles into 3 categories: - Corner tiles (there are 4) - Outer tiles that aren't corner tiles (there are 6*4=24) - The rest, on the inside (there are 6*6 = 36) If a tile is on a corner, there are only 2 possible other squares in the grid that it can share an edge with. If the tile is in the second category, there are 3. And if the tile is somewhere in the middle, there are 4 neighboring tiles it could share an edge with. The total probability that two random tiles share an edge is: ( P(tile is a corner tile)*(2/63) + P(tile is on end)*(3/63) + P(tile is in middle)*(4/63) ) / 2. We have to divide by 2 because we're counting each pair twice. So this is: ((4/64)(2/63) + (24/64)(3/63) + (36/64)(4/63)) / 2 = (8 + 72 + 144) / 2(64*63) = 112/4032 = 0.027777... EDIT: Right after I first posted this I noticed that the user above me got the same answer unsing a different method. I like his method better, but it would seem to be that the total number of unique pairs on the grid is not 64*63, but 64*63/2 because there are two different ways you can pick tiles to make the same pair (depends on which order you pick them in). So maybe the answer is 112 / 2016 and I missed something in my reasoning!
The number of ways you can select a position for the first tile is 64. The number of ways to position the second tile is then 63. So, the total number of ways to position 2 tiles is 64*63= 4032. But, since it doesn't matter if tile #1 and #2 are switched, you have to divide this by 2 (at least, if the tiles are all the same). That gives you 2016 different ways. The number of ways you can place a pair of tiles (so that they share an edge) is just the number of ways you can tile the pair horizontally plus the number of ways you can do it vertically, which for a square will be the same. So, twice the number of ways to tile the square horizontally is 2* (4+3)*8. The 2 is because we are taking twice the number, the 4 is the number of ways you can do it starting at one edge, the 3 is the number of ways starting one tile in, and the 8 is the number of horizontal rows. The total is 112. So, there are 112 of the 2016 total ways where the tiles share an edge. This makes the final probability 112/2016.

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