Question

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?

Answer 1

Answer: 4/249

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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