Question

A probability question. sorry for bad drawing,squares are actually equal

Answer 1

Count the number of single square rectangles, 2 square rectangles (horizontal and vertical), 3 square, etc. (there are 90). Now count the number of 1x1 squares, 2x2 squares, and 3x3 squares (there are 26). The probability of selecting a square is 26/90, or 13/45.

There may be a formula to determine the number of rectangles and squares. I'll leave that up to you!

Answer 2

thank you very much.
C(10,2) gives the number of rectangles and i just counted squares.

Answer 3

You're welcome, but how do you get C(10,2)? The problem with the rectangle selection is that not all combinations are rectangles.

Answer 4

I am sorry it is actually C(4,2)*C(6,2). you select 2 of horizontal and 2 of vertical lanes. it always results in a rectangle. then you just count the number of squares .

Answer 5

Hmm...your deleted response is actually very clever but it doesn't quite work. A rectangle is going to have two vertical sides that are equal in length and horizontally aligned, so if you select 2 of the possible single length sides in the same row, or 2 of the possible double length sides in the same row, or 2 of the possible triple length sides in the same row, then you cover all possible combinations. So the number of rectangle choices would be:

\[3C(6,2) + 2C(6,2) + C(6,2) = (3 + 2 + 1)C(6,2) = 6C(6,2) = 6x15 = 90\]
In this particular case it happens to be the same as C(4,2)*C(6,2), but it's a coincidence. If the grid was one unit higher, it wouldn't be the same.

Answer 6

Just for the sake of completion (since I've already spent way too much time on this), the number of squares can also be calculated:

\[(3 \times 5) + (2 \times 4) + (1 \times 3) = 15 + 8 + 3 = 26\]

Both of these formulas can be generalized for a grid of any size.