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Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )

I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r

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- Thread starter user3
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Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )

I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r

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Dick

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Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )

I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r

That is just an opinion, and it's wrong. It's more likely to be seen near the ends. It's equally likely to appear anywhere on the semicircle descibing the dial. It's x coordinate is more likely to be near the ends, since the semicircle has more length near the ends than it does in the middle.

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Ok, could you please tell how to attack problems such as these effectively ?

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Ray Vickson

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Answer: 1/(pi r sin(theta) ) = 1/(pi r √(1-(x/r)^2 ) )

I can't see why the answer is not simply 1/2r. The shadow of the tip of the needle is equally likely to be seen at any point from -r to r

Work it out from first principles. Don't "guess". Probability---especially---is one of those subjects in which untrained intuition is often wrong.

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I Understand what he's doing. He took the same approach as the textbook by first finding rho of theta, but is there an approach where you don't have to do that? I would like to find rho of x directly.

- #6

HallsofIvy

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Assuming we are to take the 0 point at the center, just because that is simplest, then the horizontal line is the "near side" of a right triangle with angle [itex]\theta[/itex], the angle the needle makes with the horizontal, and hypotenuse of length r. The length of that horizontal is [itex]r cos(\theta)[/itex]. Further since that angle is "equally likely" to be anywhere between 0 and [itex]\pi[itex], its probability density is [itex]d\theta= (1/\pi)dt[/itex]. Now with [itex]x= r cos(\theta)[/itex], [itex]dx= -r sin(\theta)d\theta= -(r/\pi) sin(\theta) dt[/itex]. From [itex]d\theta= (1/\pi)dt[/itex] we get [itex]dx= -(r/\pi) sin(t/\pi)dt[/itex].

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Dick

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No, you can't! This statement that the needle is equally likely to come to rest at any angle is all that you know about the probability distribution. If you don't know that, then you don't know anything.

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Ray Vickson

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You want something that is not possible! What you know is that there is a relationship of the form ##X = h(\Theta)## between the random variable ##X## = horizontal location of needle's point and ##\Theta## = angle of needle. You are told the probability distribution of ##\Theta##, and you can figure out the function ##h##. From that you can deduce the probability distribution of the random variable ##X##. This is a standard "change of variable" problem in probability, and it is, basically, the only way to do the question.

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Intuition will only get you so far!

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Ray Vickson

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Probability is the area in which "intuition" has led more professional mathematicians astray than any other subject. Intuition can be dangerous until you have built up a solid encyclopedia of facts about the field---in other words, until you have 'trained' your intuition.

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