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.
It is easier to count the number of integers from 1 to 150 that are perfect squares or perfect cubes. We see there are 12 perfect squares from 1 to 150, namely $1^{2}, 2^{2}, \ldots, 12^{2}$, and there are 5 perfect cubes, namely $1^{3}, \ldots, 5^{3}$. Then notice there are two repetitions, $1^{6} =1^2 = 1^3 = 1$ and $2^{6} = 8^2 = 4^3 = 64$. So there is a total of $12+5-2=15$ integers from 1 to 150 that are perfect squares or perfect cubes. Thus, we get $150-15=135$ integers from 1 to 150 that are neither perfect squares nor perfect cubes. So the probability that we select such a number is $\frac{135}{150} = \boxed{\frac{9}{10}}$.
\frac{9}{10}
Level 4
Counting & Probability
Problem: 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.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that a randomly selected number from 1 to 150 is neither a perfect square nor a perfect cube, we first need to determine the total number of outcomes. The total count of numbers from 1 to 150 is 150.

Next, we identify the perfect squares within this range. The perfect squares are the squares of integers from 1 to 12, since 12^2 = 144 and 13^2 = 169, which exceeds 150. The perfect squares are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. This gives us a total of 12 perfect squares.

Now, we identify the perfect cubes within the same range. The perfect cubes are the cubes of integers from 1 to 5, since 5^3 = 125 and 6^3 = 216, which exceeds 150. The perfect cubes are: 1, 8, 27, 64, 125. This gives us a total of 5 perfect cubes.

Next, we need to consider the overlap between perfect squares and perfect cubes, which are the perfect sixth powers. The perfect sixth powers within this range are the sixth powers of integers from 1 to 2, since 2^6 = 64 and 3^6 = 729, which exceeds 150. The perfect sixth powers are: 1 and 64. This gives us a total of 2 numbers that are both perfect squares and perfect cubes.

Using the principle of inclusion-exclusion, we calculate the total number of numbers that are either perfect squares or perfect cubes:

Total = (Number of perfect squares) + (Number of perfect cubes) - (Number of perfect sixth powers)
Total = 12 + 5 - 2 = 15.

Now, we find the count of numbers that are neither perfect squares nor perfect cubes by subtracting the total from the overall count:

Count of neither = Total numbers - Count of either
Count of neither = 150 - 15 = 135.

Finally, we calculate the probability that a randomly selected number is neither a perfect square nor a perfect cube:

Probability = Count of neither / Total numbers
Probability = 135 / 150.

To express this as a common fraction, we simplify:

135 and 150 share a common factor of 15. Dividing both by 15 gives:

135 / 15 = 9,
150 / 15 = 10.

Thus, the simplified probability is:

The final answer is 9/10.<|END_OF_LATENT|> The final answer is $\frac{9}{10}$.