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Suppose a number is chosen at random from the set {0,1,2,3,...,497}.

What is the probability that the number is a perfect cube?

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Answer: 4/249

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Step-by-step explanation:

Cube values in the set {0,1,2,3,...} until we reach a result larger than 497

  • 0^3 = 0
  • 1^3 = 1
  • 2^3 = 8
  • 3^3 = 27
  • 4^3 = 64
  • 5^3 = 125
  • 6^3 = 216
  • 7^3 = 343
  • 8^3 = 512

We stop here because 512 is larger than 497.

Or you could note that
\sqrt[3]{497} = 497^(1/3) \approx 7.921 helping us see that we stop at 8.

The list of nonnegative perfect cubes less than 497 is {0,1,8,27,64,125,216,343}

There are 8 items in that set out of 498 items in the set {0,1,2,3,...,497}

So the probability of getting a perfect cube is 8/498 = 4/249

User Stephen Swensen
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