I was reading this problem on calculating the probability of hitting a certain region on a dartboard. The number of hits the dart thrower will land at a certain radius R on the dartboard is proportional to e^(-R^2). The task is to take a certain portion of a ring (or annulus) on the dartboard and calculate the probability of hitting it. The full problem statement is at this link, if you could read the first two pages to see the parts relevant to my question, that would be nice: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-62-integrals-and-probability/MIT18_01SCF10_Ses62d.pdf(adsbygoogle = window.adsbygoogle || []).push({});

My question is: why is the probability a ratio of volumes and not areas? The pdf only partially addresses it. Given that the height of the graph ce^(r^2) is the number of hits, my intuitive feeling is that the area under the curve is the number of hits for that range of radii. Could someone explain why it's a volume?

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# Dartboards and Normal Distribution: Probability of Hitting

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