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Science And Math, Odds, Mathematics, Probability, Maths, Dice My Recent Posts

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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