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 = 2.475 \%

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