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Ok so...

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Thread replies: 18
Thread images: 3

File: 1.jpg (78KB, 852x480px) Image search: [Google]
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If you were to take a 3 cans of alphabetti spaghetti with exactly the same letters in each one and 200 letters in each can and pour them on to a plate. What are the chances, if picked at random, that you pick a 'Z'?
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>>8112977
1/26
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>>8112979
Well, that tricked nobody.
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>>8112977
You could make the problem more complicated by adding a distribution for the letters and a probability that they're broken.
Even worse, we could include a probability of particular letters being indistinguishable from others if they're broken in a certain way.
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>>8112977
how many ways can i permute the alphabet if A becomes W and B does NOT become T?
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>>8112977

>alphabetti spaghetti

ABSOLUTELY DISGUSTING.jpg
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>>8112982
There might still be a 300 post thread in this.
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1/13

The Z is also a N
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>>8112977
Insufficient data. Cannot compute.

We don't know if all letters are present. Each can could be filled with 200 A's.
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1/(26/3)

you can turn an n sideways or flip an s and squish it into a z shape
>>
50/50 you either do or you don't. Jeez it's like literally no one ever took a stats course on this board.
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>>8112992
25!-24!
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Impossible to solve.

There is no way to solve due to the absence of information that states how many letters are in each can, or if some show up in each can, repeat multiple times per can, or simply don't show up at all.
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>>8112977
>>8112979
but with 200 in each one don't you have to worry about whether or not they have z's or some other letter? You don't know the Z density.

(26^600-25^600)/(26^600) would be the probability that there's at least one z on the plate. From there I'm not sure how to calculate the probability of actually picking a z with so many permutations. (Obviously order wouldn't matter, but it still seems like it would be tedious to calculate.)
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File: Screenshot (39).png (13KB, 472x157px) Image search: [Google]
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>>8113801
Same guy. I think it's right, but it's too hard for me to think about to check and go over it again. And I don't know how to type in equations so i just screenshot it.
>>
0.999999999...
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You know that the cans have exactly the same letters in each one. That letter is [math]Z[/math]. Therefore [math]P(Z)=1[/math]
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>>8113729
you could also take a W and bite one of the ends off
Thread posts: 18
Thread images: 3


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