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澳洲代写-澳洲墨尔本代写数学作业HD作业范例paperPROBABILITY AND BUFFON'S NEEDLE

澳洲代写-澳洲墨尔本代写数学作业HD作业范例paperPROBABILITY AND BUFFON
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PROBABILITY AND BUFFON'S NEEDLE

 

Introduction

The probability theory is an important application of integration. The probability theory is at first focusing on dispersing cases like tossing a fair die, because they are calculable. However when it comes to problems like the Buffon’s Needle, the original theory does not apply anymore. This is when the integration is invited in. Since integration has a great property when dealing with continuous situations, it extended the range of probability applications and completed the basic theories of probability theory. And integrate the dispersing cases and the continuous cases into one formula: 

 

Buffon’s Needle

(1)When a segment randomly thrown onto a plane with marked horizontal lines with equal spacing L, it is certain that it will be placed between two of the horizontal lines. The probability of this event is 1. And the place of the segment between the two horizontal lines can be completely decided by y and, and the range of y and are respectively and . Thus, the sample space described all the possible conditions, and the variables in this sample space are y and .

In order to testify that if the uniform distribution  satisfies the condition , check if . In this case,, n=2, . And, , so it satisfies the condition .

(2)Through the figure below, it is clear that when the value of y, the distance from the midpoint to the closest horizontal line, is smaller than the value of , the segment crosses a horizontal line.

 

In light of this, the probability that the segment crosses a horizontal line can be calculated as below:

 

The in the equation above is . Since  can be any value in , and y can be any value in , and the probability of each point is identical. In addition, note that the range of y is , not , because when y>, this horizontal line is no longer the closest one to the segment anymore.

(3)This situation’s method actually goes the same as (2). We label the circle by the distance from the center to the closest horizontal line, x. It is clear that when , the circle crosses one of the horizontal line. So the probability of this event can be calculated as .

 

The Event with Zero Probability

(4)When A=1/2, .

When A is any finite set of numbers in [0, 1], . Since A is a finite set, the integral above is actually a finite group of 0, and =0.

(5)When the segment intersects a horizontal line along its entire length, we can describe this event by y=0 and =0. Then the formula for this event’ probability is .

That is to say, even though there are infinite horizontal lines in the plane, the event that the segment intersects one of them along its entire length seems to have no chance at all.

lready known, and we will get the value of .

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