Question:

Integer n will be randomly selected from the integers 1 to 13, inclusive.
Column A: The probability that n will be even
Column B: The probability that n will be odd

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In any set of consecutive integers, the number of odd and even integers will either be equal or differ by one. If the total number of integers is odd (like 13 here), there will be one more of whichever number (odd or even) the set starts and ends with. This set starts and ends with an odd number, so there is one more odd number.
Updated On: Oct 1, 2025
  • The quantity in Column A is greater.
  • The quantity in Column B is greater.
  • The two quantities are equal.
  • The relationship cannot be determined from the information given.
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This problem requires us to calculate and compare two probabilities based on a given set of integers. The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Step 2: Detailed Explanation:
First, determine the total number of integers in the set. The set includes integers from 1 to 13, so there are 13 total outcomes.
Column A: Probability that n will be even
The even integers in the set are \{2, 4, 6, 8, 10, 12\}.
Number of even integers = 6.
The probability is:
\[ P(\text{even}) = \frac{\text{Number of even integers}}{\text{Total number of integers}} = \frac{6}{13} \]
Column B: Probability that n will be odd
The odd integers in the set are \{1, 3, 5, 7, 9, 11, 13\}.
Number of odd integers = 7.
The probability is:
\[ P(\text{odd}) = \frac{\text{Number of odd integers}}{\text{Total number of integers}} = \frac{7}{13} \]
Comparison:
We are comparing \(\frac{6}{13}\) (Column A) with \(\frac{7}{13}\) (Column B). Since the denominators are the same, we just compare the numerators.
Since \(6<7\), we have \(\frac{6}{13}<\frac{7}{13}\).
Step 3: Final Answer:
The quantity in Column B is greater.
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