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STEP III 2007 question 14 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 14 solution (i) Consider the circle C of radius R (<=1) at the origin. The chance of a dart hitting C is In the case of the n-1 nearest darts, the chance of n-1 darts lying within a radius R is p(n darts lie within) + p(n-1 darts lie within) = Then the expected area is (ii) Now define S to be the square with sides 2M centered at the origin. The chance of a dart hitting S is (iii) Clearly greater, since the alllowed 'target' for the square dartboard includes the entirety of the 'target' for the circular one. For each dart, there's a chance Comment: At first, I took (iii) to be exclusively considering the square dartboard and comparing the area of the bounding circle to the area of the bounding square (which is a complete misread of the question - doh!). But it's quite an interesting question: it's clear the square is better for large n (since you will typically end up with at least one point near the corners of the dartboard, and bounding the corners gives a square of area 4 but a circle of area Solution by DFranklin. |
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![\displaystyle 2n(n-1)\pi\int_0^1 (R^{2n-3} -R^{2n-1})R^2 \text{ d} R = n(n-1)\pi \left[\frac{1}{n}-\frac{1}{n+1}\right] = \frac{n-1}{n+1}\pi](http://www.thestudentroom.co.uk/latexrender/pictures/88/88808c66636d5d5b49ac8b73a01b2a8e.png "\displaystyle 2n(n-1)\pi\int_0^1 (R^{2n-3} -R^{2n-1})R^2 \text{ d} R = n(n-1)\pi \left[\frac{1}{n}-\frac{1}{n+1}\right] = \frac{n-1}{n+1}\pi")
