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Sprightly
Spy with right and l in
We can’t stop here, this is bat country - Hunter S Thompson RIP
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Correctamundo. Well done - quicker than I thought ...
Slogans aren't solutions.
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You have a 4” cube, and spray the outside with red paint.
You cut this big red cube into individual 1” cubes (the freshly exposed faces do not have any red paint on them).
All of these little cubes are placed into an urn, are thoroughly mixed, and one selected at random.
This chosen cube is then rolled like a die.
What is the probability that it lands with a red face up?
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A fish.
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...
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A bowl of petunias
I wanna be a eunuchs developer! Pass me a bread knife!
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Oh no, not again.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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[Insert your God's name here] will decide which face will come up. Hint: It will be the right, good and best one.
"It is easy to decipher extraterrestrial signals after deciphering Javascript and VB6 themselves.", ISanti[ ^]
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Yes indeed, but what is the probability that [Insert your God's name here] decides that the right, good and best face will come up?
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1
He will always decide it.
"It is easy to decipher extraterrestrial signals after deciphering Javascript and VB6 themselves.", ISanti[ ^]
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How did you come up with that number?
It's obviously 50%, either it lands on red or it doesn't
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I am a bit fuzzy on that.
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The simple answer would seem to be 25%:
- 8 inner cubes with no sides painted: 0%
- 24 face cubes with one side painted: (24 ÷ 64) × (1 ÷ 6) = 6.25%
- 24 edge cubes with two sides painted: (24 ÷ 64) × (2 ÷ 6) = 12.5%
- 8 corner cubes with three sides painted: (8 ÷ 64) × (3 ÷ 6) = 6.25%
- 6.25% + 6.25% + 12.5% = 25%
EDIT: No, a cube has six sides.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
modified 9-Feb-17 4:45am.
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Richard, a cube has 6 sides.
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And what if you pick one of the inner cubes without red?
You forgot to add the probability of picking one random out of the box.
M.D.V.
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
Rating helpful answers is nice, but saying thanks can be even nicer.
modified 9-Feb-17 5:26am.
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Nelek wrote: And what if you pick one of the inner cubes without red?
That's the first line: 8 inner cubes with no sides painted. Since there are no sides painted, the probability of the cube landing with a red side face-up given that you've picked one of these is 0%.
Nelek wrote: You forgot to add the probability of picking one random out of the box.
No I didn't. I might have initially cocked-up the number of sides a cube has, but I didn't forget "picking one at random" part. That's what the (24 ÷ 64) and (8 ÷ 64) parts are about.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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My bad... lack of coffee.
Somehow I thought that 25% was still too high and after the error of the faces of a cube... I thought you were missing coffee too
M.D.V.
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
Rating helpful answers is nice, but saying thanks can be even nicer.
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That's the first correct answer, with an explanation. But it's not the simplest.
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Jörgen Andersson wrote: That's the first correct answer, with an explanation.
But only after @CPallini reminded me how many sides a cube has.
And technically, his answer[^] was the first correct answer. He just forgot to explain it.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Richard Deeming wrote: He just forgot to explain it
He did indeed, so that's no win.
But since he gave you some important help I suppose you should share the honours.
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No explanation was required. However I used Richard's approach.
Now I see the simpler one (surface vs volume): The little red faces can be computed on the big red cube: 4x4x6. The odds of having a red face should be wighted by the total little faces, that is how many little cubes there are multiplied by 6: (4x4x4x6), hence 1/4.
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