$2005 \times 2005$ checkerboard with rows shaded, probability random rectangle includes a shaded square? 
A $2005 \times 2005$ square consists of $(2005)^2$ unit squares. The middle square of each row is shaded. If a rectangle (of any size) is chosen at random, what is the probability that the rectangle includes a shaded square?

EDIT: Okay let me explain my thought process. The probability is going to be$${{\text{total }\#\text{ of rectangles with a shaded square}}\over{\text{total }\#\text{ of rectangles}}}$$Let's first calculate the total number of rectangles with a shaded square.

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*Number of rectangles with unit height passing through the center shaded column and with ending squares on both sides of the column: $1002(2005)(1002)$

*Number of rectangles of height at least two passing through the center shaded column with their vertical sides not directly on the center: ${{1002(2005)(2004)(1002)}\over2}$ (we divide by two since two squares not in the same row or column determine a rectangle and form one of the diagonals, but there's also the other diagonal pair and so we divide by $2$ to avoid overcounting)

*Number of rectangles with unit height with an ending square lying on the center shaded column (and not a $1 \times 1$ square): $2(2005)(1002)$

*Number of rectangles of height at least two with one of its vertical sides on the center shaded column: ${{2(2005)(2004)(1002)}\over2}$

*Number of vertical unit width rectangles on the center shaded column with height at least $2$: ${{2005(2004)}\over2}$

*Number of unit squares on the center shaded column: $2005$
Adding that all up, the total number of rectangles with a shaded square is $2023099189135$.
Okay, next let's calculate the total number of rectangles:

*

*Number of rectangles of either unit height or width (and not a $1 \times 1$ square): ${{2(2005)(2005)(2004)}\over2}$

*Number of $1 \times 1$ squares: $2005(2005)$

*Number of rectangles of at least both height $2$ and width $2$: ${{2005(2005)(2004)(2004)}\over{2(2)}}$
Adding that all up, the total number of rectangles is $4044181330225$.
Finally, plugging into Wolfram Alpha, I end up getting:$${{2023099189135}\over{4044181330225}} = {{1003}\over{2005}}$$
Okay, that was a lot of work, and realistically I don't think I would've been able to evaluate all that by hand quickly.

*

*Am I correct?

*If so, is there an easier i.e. less laborious way to solve this problem?

 A: I think a good first observation is to reduce the dimension.
In other words, to select a rectangle you have to select each of the borders (top,down,left,right), but for our problem we just care about the left and right borders.
there are $\binom{2005}{2}+2005$ ways to select the left and right borders, and there are $1003\times 1003$ ways to do it so that one is in the left side (including the center) and the other is in the right side (including the center), which is what we want.
Hence the answer is $\frac{1003^2}{\frac{2005\cdot 2004}{2} + 2005}= \frac{1003^2}{\frac{2005\cdot 2006}{2}}= \frac{1003}{2005}$
A: For such questions, it might be easier for you to consider the question as a pattern and finding the formula/general trend. You will notice that most of the time, it will simplify to the point you won't need to deal with ridiculously large numbers. (In addition, if ever a scenario appears where u need to deal with a similar question but with different numbers, say a 3007 $\times$ 3007 checkerboard for example, you won't need to go through all the calculations again.)
In this case, we can see the question as a board with $n$ by $n$ unit squares, where $n$ must be an odd number (so that we can have a middle row).
The total number of rectangles that can be formed = no. of 1$\times$1 squares + no. of $k \times m$ rectangles + no. of 1$\times k$ rectangles (where $k$ and $m$ ≥ 2)
= $n^2$ + $n\choose2$$^2$ +  $2$$n$$n\choose2$
= $n^2$ + $n^2(n-1)^2\over4$ + $2$$n$$n(n-1)\over2$
= $n^2\over 4$$(4+(n-1)^2+4(n-1))$
= $n^2\over 4$$((n-1)+2)^2$
= $n^2\over 4$$(n+1)^2$
The total number of rectangles with a shaded square = no. of 1$\times$1 squares + no. of rectangles entirely on centre row + no. of rectangles whose border do not lie on center row + no. of rectangles whose border lie on the centre rectangle
= $n$ + $n\choose2$ + ($n-1\over2$)$^2$[$n\choose2$+$n$] + $(n-1)$[$n\choose2$+$n$]
= $n$ + $n(n-1)\over2$ + $(n-1)^2\over4$[$n(n-1)\over2$+$n$] + $(n-1)$[$n(n-1)\over2$+$n$]
= $n(n+1)\over2$ + $(n-1)^2\over4$[$n(n+1)\over2$] + $(n-1)$[$n(n+1)\over2$]
= $n(n+1)\over8$$(4+(n-1)^2+4(n-1))$ [we know the 2nd part is$(n+1)^2$ from earlier calculations]
= $n(n+1)^3\over8$
So now we know that the probability, regardless of the value of $n$, is going to be {$n(n+1)^3\over8$} $\div$ {$n^2\over 4$$(n+1)^2$} which gives us $n+1\over 2n$.
Now we just need to substitute $n$ = 2005 to get $2006\over4010$ = $1003\over2005$. So your answer is correct.
disclaimer: I am entirely convinced that there is an even easier way to get the final probability as $n+1\over 2n$. Unfortunately at this point of time, I am unable to think of how to do so. Sorry about that. :(
