A rectangle with height and width equal to 4 and 25 respectively, is drawn on a checkered paper. Bazil paints a random horizontal 1×3 rectangle, and Peter paints a random vertical 3×1 rectangle (each rectangle consists of 3 sells). Find the probability that at least one of the cells ispainted twice. Express the answer in percent, and round to the nearest integer.
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Expert's answer
2020-12-02T19:57:06-0500
Formula of count of horizontal restangles = c*(n-2), of vertical restangles = n*2. We conclude that on rectangle it is possible to paint horizontal rectangles and different vertical rectangles. We can consider "(25-2)*4=92" different "3\\times3" squares and 25*2=50 vertical rectangles "3\\times1" rectangles. We have 92*50 = 4600 combinations. On "3\\times3" square it is possible to paint 9 different intersections of vertical and horizontal rectangles. We can consider "(25-2)*2=46" different "3\\times3" squares on the rectangle. Then, we obtain "46*3*3=414" different combination, when two rectangles intersect each other. Then, we can get probability of to be cells painted twice.Totally, the probability that one cell is painted twice is:
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