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Category: | Probability |

Submitted By: | cms271828 |

Fun: | (2.23) |

Difficulty: | (3) |

Two squares each 17cm by 17cm are drawn randomly inside a square measuring 1m by 1m.

The edges of the two smaller squares are both parallel/perpendicular to the edges of the large square.

What is the exact probability the two smaller squares overlap (or touch)?

Since the centers of the two smaller squares each lie inside a square of side 83cm, we can simplify the problem to:

Find the probability that the horizontal and vertical distances between 2 points in a square of side 83 cm are less than or equal to 17 cm.

Considering the horizontal distances:

Let x and y be the horizontal distances of the two points from the left edge of the 83 by 83 square.

Clearly x and y can take any value from 0 to 83.

We can represent this space of possibilities using a grid where the lower left corner is (0,0), the horizontal axis represents x, and the vertical axis represents y.

We must work out what part of this space represents the two points being a distance of 17cm or less apart, we can do this by plotting a point where it is true.

Firstly assume y>x, then we must have y<=x+17.

This inequality is simply all the points on and below the straight line from (0,17) to (66,83).

Now assume, y<=x, then we must have y>=x-17.

This inequality is simply all the points on and above the straight line from (17,0) to (83,66).

Combining both inequalities gives a shaded area that lies between 2 right-angled triangles.

The smaller sides of each triangle will be 66 cm each. So the area of both triangles together is 66^2.

Hence the shaded area=83^2-66^2=2533

Hence the probability the distance between x and y is less than or equal to 17 is 2533/83^2=2533/6889.

Now Prob(any 2 points in 83 by 83 square having vertical AND horizontal distance <=17) =

Prob(horizontal distance between points is <=17) * Prob(vertical distance between points is <=17)

=2533/6889 * 2533/6889 (Due to symmetry)

=6,416,089/47,458,321

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