Answer to Question #147464 in Statistics and Probability for Amir

Question #147464
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.
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Expert's answer
2020-12-01T02:14:14-0500

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

243=72 multiplying the number in a horizontal row by the number of rows24*3 =72\text{ multiplying the number in a horizontal row by the number of rows}

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

225=50 multiplying the number in a vertical row by the number of rows2*25=50 \text{ multiplying the number in a vertical row by the number of rows}

7250=3600 of all possible combinations of the arrangement72*50=3600 \text{ of all possible combinations of the arrangement}

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

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

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

224=482*24 =48

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

484=19248*4 =192

Intersection probability percentage:\text{Intersection probability percentage:}

19236001005\frac{192}{3600}*100\approx5

Answer: 5% Intersection probability





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