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:D This site sucked up almost all my reading time last night/this morning :|
Thanks!
peterruse, you are right but I think that your solution will be more understandable if you simply use the definition of the uniform random variable instead of introducing sample space.Dropping the needles so that they all fall within the stick is equivalent to randomly picking \(n\) numbers (the left endpoints) each uniformly and independently drawn from \([0,1-h]\). Under this interpretation, the volume of the sample space is \((1-h)^n\). The volume of the region where the first needle is to the left of the second is to the left of the third ... is to the left of the \(n\)-th, is given by \(\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\), which comes out to \(\frac{(1-nh)^n}{n!}\). To account for the ordering we need to multiply this by \(n!\). The probability we want is then \(\frac{(1-nh)^n}{(1-h)^n}\)
peterruse, you are right but I think that your solution will be more understandable if you simply use the definition of the uniform random variable instead of introducing sample space.
The joint probability density function of X1, X2, ... ,Xn will be fX1 X2 .. Xn(x1,.. , xn) = fX1(x1)^n = \(\frac{1}{(1-h)^n}\) since each marginal density function fXi = \(\frac{1}{1-h}\) if 0<xi< 1-h and 0 otherwise. Thus, the probability will be
\(n!\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}\frac{1}{(1-h)^n}dx_ndx_{n-1}\cdots dx_1 = \frac{n!}{(1-h)^n} \int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\) = \(\frac{(1-nh)^n}{(1-h)^n}\)
Okay? And? You essentially repeated my solution...
Yes, I actually repeated your solution only changing its explanation. If possible explain why do you use 1 instead of \(\frac{1}{(1-h)^n}\) in the expression
\(\int_0^{1-nh}\int_{x_1+h}^{1-(n-1)h}\cdots\int_{x_{n-1}+h}^{1-h}1dx_ndx_{n-1}\cdots dx_1\)?