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The numbers from 1 to 150, inclusive, are placed in a bag and a number is randomly selected from the bag. What is the probability it is neither a perfect square nor a perfect cube? Express your answer as a common fraction.

 

 

I tried to do this by finding all the perfect squares and perfect cubes over all the numbers: 8/75 and then subtract it from 1, but I got 67/75 which was wrong.

ANotSmartPerson  Dec 6, 2018
 #1
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Perfect squares =(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144) 
Perfect Cubes = (1, 8, 27, 64, 125)                                                     
Perfect squares + Perfect cubes = 12 + 5  - 2 = 15
150 - 15 = 135 - which are neither Perfect square or Perfect cubes.
So, the probability is = 135 / 150 = 9 / 10

Guest Dec 6, 2018
edited by Guest  Dec 6, 2018
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you fixed it :D

Rom  Dec 6, 2018
edited by Rom  Dec 6, 2018
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I sure did! Forgot about 1 being counted twice!!.

Guest Dec 6, 2018
 #2
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there's probably a number or two in there that are both perfect squares and perfect cubes

 

\(P[\text{# is neither perfect cube or square}] = 1 - P[\text{it is}] = \\ 1 \\ -P[\text{# is a perfect cube}]\\ -P[\text{# is a perfect square}] \\ +P[\text{# is both a perfect cube and perfect square}]\)

 

\(P[\text{perfect cube}]=\dfrac{5}{150}=\dfrac{1}{30}\\ P[\text{perfect square}] = \dfrac{12}{150} = \dfrac{2}{25}\\ P[\text{both}] = \dfrac{2}{150}\)

 

\(P[\text{# is neither cube or square}] = 1 - \dfrac{1}{30}-\dfrac{2}{25}+\dfrac{2}{150}= \dfrac{9}{10}\)

Rom  Dec 6, 2018

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