Question:
If a number is chosen at random from the set
\[
\{1,2,3,\ldots,100\},
\]
then the probability that the chosen number is a perfect cube is:
If a number is chosen at random from the set
\[
\{1,2,3,\ldots,100\},
\]
then the probability that the chosen number is a perfect cube is:
Show Hint
To count perfect cubes up to a number \(N\), find the largest integer \(k\) such that \(k^3\leq N\). The number of perfect cubes is then \(k\).
Updated On: Jun 26, 2026
- \(\frac{1}{25}\)
- \(\frac{1}{2}\)
- \(\frac{1}{10}\)
- \(\frac{9}{13}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find all perfect cubes between 1 and 100.
A perfect cube is a number of the form \[ n^3. \] The perfect cubes not exceeding 100 are: \[ 1^3=1,\quad 2^3=8,\quad 3^3=27,\quad 4^3=64. \] Since \[ 5^3=125\gt 100, \] there are only \(4\) perfect cubes in the given set.
Step 2: Determine the total number of possible outcomes.
The set \[ \{1,2,3,\ldots,100\} \] contains \[ 100 \] numbers.
Step 3: Apply the probability formula.
Probability is given by \[ P(E)=\frac{\text{Number of favourable outcomes}} {\text{Total number of outcomes}}. \] Therefore, \[ P(\text{perfect cube}) = \frac{4}{100} = \frac{1}{25}. \]
Step 4: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{1}{25}}. \]
A perfect cube is a number of the form \[ n^3. \] The perfect cubes not exceeding 100 are: \[ 1^3=1,\quad 2^3=8,\quad 3^3=27,\quad 4^3=64. \] Since \[ 5^3=125\gt 100, \] there are only \(4\) perfect cubes in the given set.
Step 2: Determine the total number of possible outcomes.
The set \[ \{1,2,3,\ldots,100\} \] contains \[ 100 \] numbers.
Step 3: Apply the probability formula.
Probability is given by \[ P(E)=\frac{\text{Number of favourable outcomes}} {\text{Total number of outcomes}}. \] Therefore, \[ P(\text{perfect cube}) = \frac{4}{100} = \frac{1}{25}. \]
Step 4: Final conclusion.
Hence, the required probability is \[ \boxed{\frac{1}{25}}. \]
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