Question:medium

A group of 200 people are chosen randomly at a conference. Later, it was found that 120 of them like blue pens and 140 like green pens while 70 like both green and blue pens. Determine the number of people thus chosen who like neither green nor blue pens?

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Use inclusion-exclusion: $n(A \cup (b) = n((a) + n((b) - n(A \cap (b)$.
Updated On: Jun 15, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
This problem can be solved using Set Theory and Venn Diagrams. We need to find the number of elements outside the union of two sets.
Step 2: Key Formula or Approach:
The Principle of Inclusion-Exclusion for two sets A and B is: \[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \] Where: - \(n(A \cup B)\) = number of people who like at least one of the pens. - \(n(A)\) = number of people who like blue pens. - \(n(B)\) = number of people who like green pens. - \(n(A \cap B)\) = number of people who like both. People liking neither = Total people - \(n(A \cup B)\).
Step 3: Detailed Explanation:
1. Let \(U\) be the universal set of all people at the conference, so \(n(U) = 200\). 2. Let \(B\) be the set of people who like blue pens, so \(n(B) = 120\). 3. Let \(G\) be the set of people who like green pens, so \(n(G) = 140\). 4. The intersection \(B \cap G\) represents those who like both, so \(n(B \cap G) = 70\). 5. Calculate the number of people who like at least one of the pens (either blue or green or both): \[ n(B \cup G) = 120 + 140 - 70 \] \[ n(B \cup G) = 260 - 70 = 190 \] This means 190 people like either blue, green, or both pens. 6. The number of people who like neither is the total number of people minus those who like at least one: \[ \text{Neither} = n(U) - n(B \cup G) \] \[ \text{Neither} = 200 - 190 = 10 \]
Step 4: Final Answer:
The number of people who like neither green nor blue pens is 10.
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