Answer to Question #142443 in Statistics and Probability for Peace College

Question

A rectangle with height and width equal to 4 and 25 respectively, is drawn on a checkered
paper. Bazil paints a random horizontal 1 x 3 rectangle, and Peter paints a random vertical 3 x 1
rectangle (each rectangle consists of 3 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

We assume that Bazil and Peter paint on the drawn rectangle. The picture below presents

rectangle $5\times6$.

As we can see, it is possible to paint 20 different horizontal $1\times3$ rectangles. In a similar way we conclude that on rectangle $4\times25$ it is possible to paint $23*4=92$ horizontal $1\times3$ rectangles and $25*2=50$ different verical $3\times1$ rectangles. We have $4600$ possible combinations in total. On $3\times3$ square it is possible to paint 9 different intersections of vertical and horizontal rectangles. We can consider $23*2=46$ different $3\times3$ squares on the rectangle. Thus, we obtain $46*9=414$ 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:$p=\frac{414}{4600}=0.09$.

Answer: 0.09