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Yes.
Hi bobbym,
is 9460912/124750080 the solution to post #374?
Hi Natkirky,
you're welcome.
Hi Natkirky,
Substitute x = t^2, after a step or two, you'll get
Hi magniack,
You can use a tree diagram, it'll be easier to visualize such problems.
The answer is 0.17050733
The exact answer is:
where p = 1/6 and q=5/6, the probability of getting 1, and binomial distribution is used.
Reasoning is 0 1's in first throw, then atleast 3 1's in the second, and similarly for 1 1's, 2 1's so on...
Everything agrees with the simulation.
*edit* I thought bob bundy's excel sheet was simulation, and posted the same solution. But I enjoyed solving, thanks.
Yes, you add special effect to math!
Hi SuperLynx,
Did you follow?
Oh, you were talking of breaking something inside you, and I thought you were talking of breaking something around you!
Better stick with math itself, we need you here.
Gee, that'll be nice.
Bye Howardroark...
I feel groggy if I sleep too much.
I'm hearing that for the first time!
Do something physical when you feel groggy, go out and turn a few somersaults!!
I do not need much sleep.
Whatever, but sleep well, atleast for 5 hours.
And yes there are 4 or 5 of us running around.
and take care not to bump against each other!!
Hi bobbym,
Thank you for all the beautiful questions. Solutions are nothing compared to it !
But why don't you sleep well? Do you love math so much that you can't sleep ?!
Or is there another bobbym?
That's nice! I'm from Mangalore.
Enjoy math, have fun
Hi Howardroark,
I'm from India, what about you?
Cool !
The problems are simply addictive!
Hi Howardroark,
You said it
Did you try some here: http://www.mathisfunforum.com/viewtopic.php?id=12832&p=7
Hi bobbym,
Hi SuperLynx,
Did you check the document : http://www.numbertheory.org/book/cha4.pdf
I don't know to explain better
Tell what you did not follow.
Hi Dragonshade,
You can check from the graph.
Since x>0, the symmetrical part for negativeve x is eliminated.
Hi SuperLynx,
You can do it by using the recursive definition for determinants.
E.g. consider a 3X3 determinant
That is, you take out the first element of the first row, delete the row and column where it lies, and then compute the determinant of the elements which remain.
Then take out the second element of the first row, prefix a minus sign, delete the row and column where it lies, and then compute the determinant of the elements which remain.
Then take out the third element of the first row, prefix a plus sign, delete the row and column where it lies, and then compute the determinant of the elements which remain.
Utimately, we end up in 2X2 determinant, which is easy to compute.
If there are more rows and columns, you compute it in the same way, with alternate plus and minus.
Go here to know some theorems and examples: http://www.numbertheory.org/book/cha4.pdf
Yes, I too could not find much information regarding this.
In fact I too did not think of this when I came across the square grids' problem!
But, since the no. of paths are less than the available paths in square grids', I guess we should subtract some value.
Maybe we will get more insight when we plot the total paths vs no. of open paths, the effect of rearranging open paths etc.
I'll leave it for now!
Hi bobbym,
At first I thought I could rearrange the longer paths to make the computation easy, but was disappointed to find out they give different answers...
Hi,
Good, good! You pose wonderful problems!!
Cool!
Am I right with the probability value also?