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(C) Let $S = \{1, 2, 3, \ldots, 1000\}$. The total number of outcomes is $n(S) = 1000$. Let $A$ be the set of perfect cubes in $S$. Since $10^3 = 1000

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$\frac{19}{500}$

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(C) Let $S = \{1, 2, 3, \ldots, 1000\}$. The total number of outcomes is $n(S) = 1000$. Let $A$ be the set of perfect cubes in $S$. Since $10^3 = 1000$,$A = \{1^3, 2^3, \ldots, 10^3\}$,so $n(A) = 10$. Let $B$ be the set of natural numbers having an odd number of divisors. $A$ number has an odd number of divisors if and only if it is a perfect square. The largest perfect square $\le 1000$ is $31^2 = 961$. Thus,$B = \{1^2, 2^2, \ldots, 31^2\}$,so $n(B) = 31$. The intersection $A \cap B$ consists of numbers that are both perfect cubes and perfect squares,i.e.,perfect sixth powers. These are $1^6 = 1$,$2^6 = 64$,and $3^6 = 729$. Thus,$n(A \cap B) = 3$. Using the inclusion-exclusion principle,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 10 + 31 - 3 = 38$. The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{38}{1000} = \frac{19}{500}$.

$A$ number is selected at random from the set $\{1, 2, 3, 4, \ldots, 1000\}$. What is the probability of getting a number which is a perfect cube or a natural number having an odd number of divisors?

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