- #1

cyberdeathreaper

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The needle on a broken car spedometer is free to swing, and bounces perfectly off the pins at either end, so that if you give it a flick it is equally likely to come to rest at any angle between 0 and [itex] \pi [/itex].

Consider the x-coordinate of the needle point - that is, the "shadow," or "projection," of the needle on the horizontal line.

What is the probability density [itex] \rho(x) [/itex] if r is the length of the needle?

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Conceptually, I'm trying to clarify the following:

In a previous question, they asked to find [itex] \rho(\theta) [/itex], which I found as simply [itex] \frac{1}_{\pi} [/itex]. This implies the theta-derived answer moves the x coordinate from r to -r as [itex] \theta [/itex] goes from 0 to [itex] \pi [/itex].

In calculating the probability for this question, I assumed x moved in the same fashion (ie, x goes from r at [itex] \theta = 0 [/itex] to -r at [itex] \theta = \pi [/itex]). This reverse movement gives a probability density that is always negative.

My question is, is this possible? The total probability does equal 1 if I move from right to left, though I'm not sure if this is permissible.

Likewise, could I simply change the sign from negative to positive for the probability density, only reasoning that if x moves left to right, the signs should be reversed?