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a point
is selected at random in a sphere. then find the probability that the point is closer to thecenter of the sphere than to its surface
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Assuming that all points are equally likely to be chosen, the probability of P being in any particular section of the sphere is equal to the ratio of the volume of that section to the volume of the entire sphere. If we assume the sphere has radius 1, then we want to find the probability that P is within 1/2 of the center, so the probability U that P is closer to the center than the surface is
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thanks bobbym and Dude
but for radius of 1unit, If we take a sphere of radius 1/2
than the point which is exactly on that sphere(1/2 radius sphere) . then these point are equal distance from center of sphere of radius 1 and also equal distance from surface
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We are equating volumes the surface of the inner sphere does not have a volume. It does not have any depth. It is two dimensional therefore no volume.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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bobbym i want more explanation
thanks
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What is the question?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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the line
We are equating volumes the surface of the inner sphere does not have a volume. It does not have any depth. It is two dimensional therefore no volume.
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Even though there is a possibility that the point P will land exactly at the halfway point between the center and the surface of the sphere, in a mathematical sense the probability of this happening is exactly 0. So there's no need to consider this possibility when answering this question.
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