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Point in the Square

by Andrew
(New York)










































Given a square with a side of 1 inch, let P be a random point inside the square. What is the probability that the side closest to point P is between 1/5 and 1/3 of an inch from P?

Comments for Point in the Square

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Dec 05, 2011
Point in the Square - Probability
by: Anonymous


Question:

by Andrew
(New York)

Given a square with a side of 1 inch, let P be a random point inside the square. What is the probability that the side closest to point P is between 1/5 and 1/3 of an inch from P?



Answer:

This is a problem in geometric probability.

The geometric probability is equal to the ratio area between 1/5 and 1/3 inches from the side of the square to the area of the square.


Open the following 3 links to view diagrams of this problem.


(1) If your browser is Firefox, click the following link to VIEW; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

Link showing square, sectors, and closest border:

http://www.solving-math-problems.com/images/square-01-sectors-closest-border-2011-12-04.png

Link showing square and measurements:

http://www.solving-math-problems.com/images/square-02-sectors-measurements-2011-12-04.png

Link showing square and geometric area of interest:

http://www.solving-math-problems.com/images/square-03-geometric-area-of-interest-2011-12-04.png


The area highlighted in red is the difference between the areas of the squares formed by the boarders


⅗ * ⅗ = 9/25 in²

⅓ * ⅓ = 1/9 in²


Area between 1/5 in and 1/3 inch from the border of the original large square:

9/25 in² - 1/9 in²

= 81/225 - 25/225

= 56/225 in²


Area of original large square:

1 in * 1 in = 1 in²


Probability that the side closest to point P is between 1/5 and 1/3 of an inch from P:

P = 56/225 in² / 1 in²

= 56/225

= 0.248889

= 24.9%


The final answer is: 24.9%





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Staff
www.solving-math-problems.com



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