Interesting prob: A point is chosen at random, with....

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Hello, I'm having trouble figuring out where to start, and it's been a LONG time since I've done a problem like this. Thanks for the help everyone! :)

"A point in the plane, both of whose rectangular coordinates are integers with absolute value less than or equal to four, is chosen at random, with all such points having an equal probability of being chosen. What is the probability that the distance from the point to the origin is at most two units? Give exact answers."
 
Re: Interesting word problem... very confused! :)-

Go here to get a bit of "remember when" :
http://www.themathpage.com/alg/rectangu ... inates.htm

We'd need chalk and blackboard to explain in detail!
Here's a few hints:

points (3,2) , (3,-2) , (-3,2), (-3,-2) are all same distance from origin;
so we can treat all 4 as being (3,2)

there are 25 such points:
(0,0) , (0,1) ... (0,4)
(1,0) , (1,1) ... (1,4)
...
...
(4,0) , (4,1) ... (4,4)

only 6 of those are 2 or less units from origin:
(0,0) , (0,1) , (0,2) , (1,0) , (1,1) , (2,0)

why not (2,1)?
(2,1) is sqrt(2^2 + 1^2) = sqrt(5) = 2.236... from origin

so probability is 6 / 25

Hope that helped you :idea:

Note: plus hope I didn't goof!
If so, Soroban et al will jump in :wink:
 
Hello, ArcainineFalls531!

Sorry, Denis, you over-counted the points on the axes.


A point in the plane, both of whose rectangular coordinates are integers
with absolute value less than or equal to four, is chosen at random,
with all such points having an equal probability of being chosen.
What is the probability that the distance from the point to the origin is at most two units?
Code:
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          *   *   *   *   o   *   *   *   *
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          *   *   *   o   o   o   *   *   *
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      - - * - * - o - o - o - o - o - * - * - -
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          *   *   *   o   o   o   *   *   *
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          *   *   *   *   o   *   *   *   *
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          *   *   *   *   *   *   *   *   *
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          *   *   *   *   *   *   *   *   *
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There are \(\displaystyle 81\) points in the sample space.

There are \(\displaystyle 13\) points which are at most 2 units from the origin.

Therefore, the probability is: \(\displaystyle \L\,\frac{13}{81}\)

 
Whoopsee doo; yer right: the ones on axes affect 2 quadrants each..
(1st mistake this year!)
Merci, Soroban.
 
That seems to make a lot of sense guys, thanks! But does the absolute value have anything to do with anything? It wouldn't just be the points in the first quadrant, would it?
 
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