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 Random^Random (Posted on 2014-10-29)
Let A and B each be random real numbers chosen from the uniform interval (0,1).

Call Z the tenths place digit of AB.

Find the probability distribution of Z.

 See The Solution Submitted by Jer Rating: 5.0000 (1 votes)

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 towards an analytic solution | Comment 2 of 9 |
Start with the case of the digit 9. As all the Z values are below 1, what is sought here is the probability that Z > .9. Others can be found by subtracting this (and successibe probabilities) from the probability that Z > .8, etc.

P(Z>.9) = Integral{0 to 1} ln(.9)/ln(A) dA, treating A as the independent variable; use x if you prefer. Note that ln(.9) is negative, avoiding a singularity in the logarithmic integral function at A=1.

This turns out to involve the Logarithmic Integral function li(x). Note that ln(.9) is negative, avoiding a singularity in the logarithmic integral function at A=1.

An alternative, integrating over B, would involve the Exponential integral Ei(x):

Integral{0 to 1} e^(ln(.9)/B) dB.

 Posted by Charlie on 2014-10-29 14:35:18

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