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:

c1=13 c2=23=8 c3=33=27 c4=43=64 c5=53=125 c6=63=216  0<c1c5<202  p=100 52020=250101=2.4752%



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