Question

Suppose a number is chosen at random from the set {0,1,2,3,...,1721}. What is the probability that the number is a perfect cube?
Round your answer to 6 decimal places as needed.

Answer 1

Answer: 0.006388

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Explanation:

One way to go about this is to list out all the perfect cubes. A perfect cube is the result of taking any whole number and multiplying it by itself 3 times.

1 cubed = 1^3 = 1*1*1 = 1

2 cubed = 2^3 = 2*2*2 = 8

3 cubed = 3^3 = 3*3*3 = 27

4 cubed = 4^3 = 4*4*4 = 64

and so on. We stop once we reach 1721, or if we go over. Ignore any values larger than 1721. You'll find that 11^3 = 1331 and 12^3 = 1728. So we stop here and exclude 1728 as that is larger than 1721.

A quick way to see where we should stop is to apply the cube root to 1721 and we get

$\sqrt[3]{1721} = 1721^{1/3} \approx 11.98377$

The approximate result of 11.98377 tells us that 1721 is between the perfect cubes of 11^3 = 1331 and 12^3 = 1728

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So effectively, we have 11 perfect cubes in the set {0, 1, 2, 3, ..., 1721} and this is out of 1722 numbers in that same set. Note how I added 1 onto 1721 to get 1722. I'm adding an extra number because of the 0. If 0 wasn't part of the set, then we would have 1721 values total inside.

In summary: There are 11 values we want (11 perfect cubes) out of 1722 values total.

Divide 11 over 1722 to get

11/1722 = 0.00638792102207

which rounds to 0.006388