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A real number x is chosen at random between 0 and 1. Find the probability that the first nonzero digit in the decimal expansion of \(\sqrt{x}\) is 3.

 Mar 16, 2020
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0.3^2 = 0.09

0.4^2 = 0.16

 

so 

sqrt(0.16) = 0.4

sqrt(0.09) = 0.3

So 

\(0.09\le x <0.16\)

 

0.16-0.09 = 0.07

probability is 7%

 

Actually this is not quite right because I have assumed that the first non zero is in the tenths place value but it could be in some other place value.  So the answer is slightly higher than this. If you want the exact answer you'd need to set up a series  with all possible answers and look at it.

 Mar 16, 2020
edited by Melody  Mar 16, 2020

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