Perfect cubes

Suppose a number is chosen at random from the set (0,1,2,3,. .. ,202).
What is the probability that the number is a perfect cube?

Result

p =  2.475 %

Solution:

c1=13 c2=23=8 c3=33=27 c4=43=64 c5=53=125 c6=63=216  0<c1...c5<202  p=100⋅ 5202−0=250101≐2.4752≐2.475%c_{1}=1^3 \ \\ c_{2}=2^3=8 \ \\ c_{3}=3^3=27 \ \\ c_{4}=4^3=64 \ \\ c_{5}=5^3=125 \ \\ c_{6}=6^3=216 \ \\ \ \\ 0<c_{1} ... c_{5}<202 \ \\ \ \\ p=100 \cdot \ \dfrac{ 5 }{ 202-0 }=\dfrac{ 250 }{ 101 } \doteq 2.4752 \doteq 2.475 \%

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