Answer to Question #147464 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

$\text{Total number of horizontal rectangles 1*2:}$

$24*3 =72\text{ multiplying the number in a horizontal row by the number of rows}$

$\text{Total number of vertical rectangles 2*1:}$

$2*25=50 \text{ multiplying the number in a vertical row by the number of rows}$

$72*50=3600 \text{ of all possible combinations of the arrangement}$

$\text{of a pair of rectangles 1*2 and 2*1}$

$\text{Rectangles 1 * 2 and 2 * 1 intersect in a square 2 * 2 with 2 * 2 = 4 options}$

$\text{The number of squares 2 * 2 on an area of 3 * 25 will be:}$

$2*24 =48$

$\text{Number of intersections of rectangles1*2 and 2*1:}$

$48*4 =192$

$\text{Intersection probability percentage:}$

$\frac{192}{3600}*100\approx5$

Answer: 5% Intersection probability