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?

Correct answer:

p =  2.4752 %

Step-by-step explanation:

$$\begin{gathered} c_1=1^3 \\ {c}_2=2^3=8 \\ {c}_3=3^3=27 \\ {c}_4=4^3=64 \\ {c}_5=5^3=125 \\ {c}_6=6^3=216 \\ 0<c_1\dots c_5<202 \\ p=100\cdot {\displaystyle \frac{5}{202-0}}={\displaystyle \frac{250}{101}}\mathrm{\%}=2.4752\mathrm{\%} \end{gathered}$$

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