Answer to Question #147517 in Statistics and Probability for Amir

Question

A rectangle with height and width equal to 3 and 25 respectively, is drawn on a checkered paper. Bazil paints a random horizontal 1×2 rectangle, and Peter paints a random vertical 2×1rectangle (each rectangle consists of 2 sells). Find the probability that at least one of the cells is painted twice. Express the answer in percent, and round to the nearest integer.

Answer 1

As we can see, it is possible to paint 25 different horizontal $1\times2$ rectangles. In a similar way we conclude that on rectangle $3\times 19$ it is possible to paint $3 \cdot 24 =72$ horizontal $1\times2$ rectangles and $2 \cdot 25=50$ different vertical $2\times1$ rectangles. We have $3600$ possible combinations in total.

On $2\times2$ square it is possible to paint 4 different intersections of vertical and horizontal rectangles. We can consider $2 \cdot 24=48$ different $2\times2$ squares on the rectangle. Thus, we obtain $48 \cdot 4=192$ different combination, when two rectangles intersect each other. I.e., there will be cells painted twice.

Thus, the probability that one cell is painted twice is: $p=\frac{192}{3600}=0.05 = 5\%$