<![CDATA[Math Is Fun Forum / Probability and Statistics]]> 2011-03-01T06:00:15Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=2985 <![CDATA[Re: Probability and Statistics]]> Hi G-man;

Welcome to the forum. Which question are you answering?

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https://www.mathisfunforum.com/profile.php?id=33790 2011-03-01T06:00:15Z https://www.mathisfunforum.com/viewtopic.php?pid=166562#p166562
<![CDATA[Re: Probability and Statistics]]> If I'm not wrong.
Answer to question in first post.

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https://www.mathisfunforum.com/profile.php?id=102287 2011-03-01T05:42:16Z https://www.mathisfunforum.com/viewtopic.php?pid=166561#p166561
<![CDATA[Re: Probability and Statistics]]> Hi maudish;

Using the ordinary generating function for the rook polynomials on a r x c board.

The series can be eliminated because we only seek the coefficient of x^2 ( 2 rooks). Also since the board is square r = c. So.

The above is the number of 2 non attacking rooks on a c x c chessboard.

From a combinatorical argument and playing much spot the pattern.

We can solve for n and clean up:

Where n is the number of ways 2 rooks can be positioned on the white squares of a c x c chessboard when c is even. The above will generate the table given in the previous post, i.e.

c = 2 then n = 0
c = 4 then n = 8
c = 6 then n = 36
c = 8 then n = 96
c = 10 then n = 200
.
.
.

This is not a proof, just a conjecture. I have tested it for c = 16 by direct count. I suppose it might be proven by induction but the correct method is by partitioning the chessboard with it's forbidden black squares into disjoint boards and then using the rook polynomials to prove it. When I do that I will post it.

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https://www.mathisfunforum.com/profile.php?id=33790 2009-10-13T02:59:42Z https://www.mathisfunforum.com/viewtopic.php?pid=121594#p121594
<![CDATA[Re: Probability and Statistics]]> Hi maudish;

I believe the sequence looks like this:

For various n x n boards:
2 x 2 =0,
3 x 3 =4
4 x 4 =8,
5 x 5 =22
6 x 6 =36
7 x 7 = ?,
8 x 8 = 96

All by actual count. If i find a formula for n x n will post it.

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https://www.mathisfunforum.com/profile.php?id=33790 2009-10-11T22:46:43Z https://www.mathisfunforum.com/viewtopic.php?pid=121492#p121492
In how many ways you can choose 2 white squares on a chessboard such that they are either in the same row or same column?

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https://www.mathisfunforum.com/profile.php?id=47885 2009-10-11T21:35:44Z https://www.mathisfunforum.com/viewtopic.php?pid=121485#p121485
<![CDATA[Re: Probability and Statistics]]> Excellent, mathsyperson!

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https://www.mathisfunforum.com/profile.php?id=682 2006-03-10T04:13:39Z https://www.mathisfunforum.com/viewtopic.php?pid=30946#p30946
<![CDATA[Re: Probability and Statistics]]> Oh, silly me. I hate it when I make stupid mistakes from not reading the question properly. Ah well.

The probability of it raining and of them winning is 0.4*0.3 = 0.12.
The probability of it not raining and of them winning is 0.6*0.55 = 0.33.

Therefore, the probability of rain if they won is 0.12/(0.12+0.33) = 0.26666... = 27% (nearest %)

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https://www.mathisfunforum.com/profile.php?id=641 2006-03-09T17:01:25Z https://www.mathisfunforum.com/viewtopic.php?pid=30906#p30906
<![CDATA[Re: Probability and Statistics]]> PS # 4

The probability of rain was 40%. If it rained, the Redskins had a 30% chance of winning, if it did not rain, they had a 55% chance of winning. Given that the Redskins won, what is the probability that it rained?

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https://www.mathisfunforum.com/profile.php?id=682 2006-03-09T04:17:05Z https://www.mathisfunforum.com/viewtopic.php?pid=30849#p30849
<![CDATA[Re: Probability and Statistics]]> ganesh wrote:

and the probability that he will not pass in mathematics is 5/9.

mathsyperson, did you notice that?
The probability that he will pass in mathematics is, therefore, 4/9.
Hence, the probability of passing both the subjects would be
2/3 + 4/9 - 4/5 = 14/45.

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https://www.mathisfunforum.com/profile.php?id=682 2006-03-03T04:43:36Z https://www.mathisfunforum.com/viewtopic.php?pid=29718#p29718
<![CDATA[Re: Probability and Statistics]]> P(AnB) = P(A) + P(B) - P(AuB).

Therefore, P(Passing both) = 2/3 + 5/9 - 4/5 = 19/45.

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https://www.mathisfunforum.com/profile.php?id=641 2006-03-02T16:31:26Z https://www.mathisfunforum.com/viewtopic.php?pid=29650#p29650
<![CDATA[Re: Probability and Statistics]]> PS # 3

The probabitlity that a student will pass in Statistics examination is 2/3 and the probability that he will not pass in mathematics is 5/9. The probability that he will pass in atleast one of the examinations is 4/5. Find the probability of his passing in both the examinations.

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https://www.mathisfunforum.com/profile.php?id=682 2006-03-02T04:19:11Z https://www.mathisfunforum.com/viewtopic.php?pid=29535#p29535
<![CDATA[Re: Probability and Statistics]]> Excellent! Well done, mathsyperson! ]]>
https://www.mathisfunforum.com/profile.php?id=682 2006-03-01T15:58:37Z https://www.mathisfunforum.com/viewtopic.php?pid=29435#p29435
<![CDATA[Re: Probability and Statistics]]> Ooh, interesting. Let's see... a chessboard has 4 corner squares, 24 edge squares and 36 centre squares.

On the corner squares, the chance of the other square being next to it is 2/63.
On the edge squares, the chance is 3/63 and on the centres, the chance is 4/63.

Combine this with the chances of the first square being each of the types and we get (4/64*2/63) + (24/64*3/63) + (36/64*4/63) = 224/4032 = 1/18.

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https://www.mathisfunforum.com/profile.php?id=641 2006-03-01T15:47:18Z https://www.mathisfunforum.com/viewtopic.php?pid=29430#p29430
<![CDATA[Re: Probability and Statistics]]> PS # 2

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

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https://www.mathisfunforum.com/profile.php?id=682 2006-03-01T08:08:07Z https://www.mathisfunforum.com/viewtopic.php?pid=29387#p29387
<![CDATA[Re: Probability and Statistics]]> Very good, mathsyperson. You're correct. ]]>
https://www.mathisfunforum.com/profile.php?id=682 2006-02-28T16:41:45Z https://www.mathisfunforum.com/viewtopic.php?pid=29215#p29215