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Topic: probability to cut the marble (Read 937 times) |
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inexorable
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probability to cut the marble
« on: Jan 8th, 2006, 5:17am » |
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Suppose a round marble of radius 1 inch is randomly situated in some cake batter that is placed into a pan and baked into a cake of radius 14 inches. If the cake is cut into 12 congruent pieces via six cuts along diameters, what is the probability that one of the cuts strikes the marble?
Ps:- i guess this has n't been posted before , as i could not find on search
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towr
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Re: probability to cut the marble
« Reply #1 on: Jan 8th, 2006, 7:34am » |
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Provided I didn't make any mistakes.
24* [1/2 a / ([pi]*132) + phi/(2 [pi])]
where a = sqrt(168) - 1/tan([pi]/12) and phi = asin(1/13)
about a 50.3 % chance
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| « Last Edit: Jan 8th, 2006, 7:36am by towr » |
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inexorable
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Re: probability to cut the marble
« Reply #2 on: Jan 8th, 2006, 12:06pm » |
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Yes it is correct. can you please explain the procedure  ?
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towr
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Re: probability to cut the marble
« Reply #3 on: Jan 8th, 2006, 12:49pm » |
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| hidden: | I first drew it. The point is to figure out where you can place the center of the marble in the pan, such that it can be struck. First off the center of the marble need to be 1" from the edge of the pan. Which accounts for the 13 in the answer. The marble can only be in a 13" radius of the centre.
Then it needs to be with 1 inch of the cuts. Due to symmetries you only need to look at half a wedge, so that's 1/24 (accounting for the multiplication by 24)
The area where the marble has to be can be split in two areas: A triangle (1/2 * a * 1) , and a thin wedge (with angle phi).
A drawing would really make this clearer, but I don't have a good program to make one. |
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| « Last Edit: Jan 8th, 2006, 12:59pm by towr » |
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