Member since July 29, 2011

w0lfshad3

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w0lfshad3
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Mathematics, Probability, Maths, Science And Math, Odds, Dice
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Tue 29 May, 2012 01:15 am - All you needed was: http://en.wikipedia.org/wiki/Science Also, if you meditate upon the basis of knowledge, what is identity, that is, an entity and it's unique id, or what makes it... (view)
Wed 7 Sep, 2011 09:07 am - This is a conditional probability problem. ABCDE;E AAA;EEE was calculated as 4/45 experiment sais 0.0659658333 3/45 is very close: 0.0(6) Anyone knows how to calculate this probability... (view)
Wed 7 Sep, 2011 09:06 am - This is a conditional probability problem. I experimented on matching all numbers between ABCD;EE and ABCD;EE Hypothetic (calculated) probability: 1/1260 Real probability: ~1.75/1260... (view)
Sat 3 Sep, 2011 05:02 pm - Here there's many matches of 3. Because it's a match of 6/6 it means that there are C(6,3)=20 matches of 3... I think (view)
Sat 3 Sep, 2011 04:25 pm - 1. is (1/252)*(1/5)=1/1260 2. is hard; take 2 simplified, it's still hard: What is the probability of the player matching 3 numbers from a 10 number lottery (0...9) w/o repetition, given... (view)
Fri 2 Sep, 2011 12:15 pm - Is this the answer to 1? C(5,5)C(10-5,0)/C(10,5)=1/252 + (1/252)*(1/5)=6/1260=1/210 or just (1/252)*(1/5)=1/1260? And this to 2? C(5,4)C(10-5,0)/C(10,4)=(5/210)*(2/5)=2/210? (view)
Fri 2 Sep, 2011 12:01 pm - Problem 1 data: A lottery of 10 numbers (0...9) w/o repetition drawing sets of 6. Statistical data: A. The statistical probability of a lottery drawing 6 distinct numbers is 19.66% E.g.:... (view)
Fri 2 Sep, 2011 11:04 am - Well, that wiki bit was inspiring, even if it that formula I posted is wrong for this problem because it's for no replacement balls. I should've posted this one instead perhaps: The... (view)
Fri 2 Sep, 2011 08:16 am - 1. What is the probability of the player matching 3 numbers from a 10 number lottery (0...9) w/o repetition, given the conditions: - player and lottery pick 6 numbers - player and lottery always... (view)
Fri 2 Sep, 2011 03:16 am - Actually the answer is the hypergeometric distribution: [kCx] [N-kCn-x] / [NCn] 5 units (6 out of which 2 are repeated which counts as one distinct), out of a space of 10 units, matching 4... (view)
 
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