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/sci/ - Science & Math

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>> No.9410960 [View]
File: 22 KB, 371x394, 1a0efdf01b8dd2dbd1ca358b283eb6da891a0a01_00.jpg [View same] [iqdb] [saucenao] [google]
9410960

does sitting like this icreases my iq?

>> No.8397836 [View]
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8397836

>>8394067
>>8394243
>>8396875
I'll give this a shot.

In order to draw a (non-degenerate) triangle, we require each of its three vertices to be unique, that is, no vertex can be picked more than once. Since we don't care about the order in which the vertices are picked, the number of ways to choose 3 elements out of a possible 16 is given by the binomial coefficient
[eqn]{{16}\choose{3}} = \frac{16!}{3!(16-3)!} = \frac{14\cdot15\cdot16}{2\cdot3} = 560[/eqn]
so there are 560 possible distinct combinations to try. To guarantee success, we'll have to try all of them.

In nine minutes, there are [math]9\cdot60 = 540[/math] seconds. Since Chris can draw one triangle per second, he'll be able to try 540 out of the 560 combinations, which means he's not guaranteed to be able to save Dana. He is fairly likely to do so, though.

Fun sidenotes: The number of ways to choose 3 vertices from the 3 "correct" dots and 0 vertices from the 16-3 = 13 "incorrect" ones is given by
[eqn]{{3}\choose{3}} {{13}\choose{0}} = 1[/eqn]
Thus, the probability of succeeding on any one try is [math]\frac{1}{560}[/math], as is to be expected. However, this method of determining the probability shows that this exercise can be described by a hypergeometric distribution of the form
[eqn]p_x(k) = \frac{ {{v}\choose{k}} {{s}\choose{n-k}} }{{{v+s}\choose{n}}}[/eqn]

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