Answer to Question #146612 in Statistics and Probability for ankit

Question

A rectangle with height and width equal to 4 and 25 respectively, is drawn on a checkered paper. Bazil paints a random horizontal 1×3 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.

Answer 1

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:

$p=\frac{414}{4600}=0.09$