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Difficulty: MediumSet Theory Concepts and Venn Diagrams

A cybersecurity firm audited a sample of 300300 corporate networks for compliance across three security standards: Network Encryption (EE), Multi-Factor Authentication (MM), and Access Logging (LL). The audit revealed that 160160 networks met standard EE, 140140 met standard MM, and 120120 met standard LL. Exactly 2525 networks met all three standards, while 4545 networks met none of the three standards. How many of the audited networks met exactly two of the three security standards?

  1. A
    90
  2. 115Answer
  3. C
    140
  4. D
    165
  5. E
    190

Answer

115 networks met exactly two of the three security standards.
The total number of networks meeting at least one standard is 300−45=255300 - 45 = 255. By the three-set inclusion-exclusion principle, 255=160+140+120−S2+25255 = 160 + 140 + 120 - S_2 + 25, where S2S_2 is the sum of the pairwise intersections ∣E∩M∣+∣M∩L∣+∣E∩L∣|E \cap M| + |M \cap L| + |E \cap L|. Solving for S2S_2 yields S2=190S_2 = 190. Since each pairwise intersection includes the 25 networks that met all three standards, the number of networks meeting exactly two standards is 190−3(25)=115190 - 3(25) = 115.

Step-by-Step Solution

1
Calculate the total number of networks that met at least one security standard.
∣E∪M∪L∣=300−45=255|E \cup M \cup L| = 300 - 45 = 255
Subtracting networks that met none of the standards from the total sample size gives the union of the three sets.
2
Apply the Principle of Inclusion-Exclusion for three sets to find the sum of pairwise overlapping regions.
255=160+140+120−(∣E∩M∣+∣M∩L∣+∣E∩L∣)+25255 = 160 + 140 + 120 - (|E \cap M| + |M \cap L| + |E \cap L|) + 25, so ∣E∩M∣+∣M∩L∣+∣E∩L∣=445−255=190|E \cap M| + |M \cap L| + |E \cap L| = 445 - 255 = 190
The formula ∣E∪M∪L∣=∣E∣+∣M∣+∣L∣−(∣E∩M∣+∣M∩L∣+∣E∩L∣)+∣E∩M∩L∣|E \cup M \cup L| = |E| + |M| + |L| - (|E \cap M| + |M \cap L| + |E \cap L|) + |E \cap M \cap L| relates all known quantities.
3
Calculate the number of networks that met exactly two standards.
Exactly 2=(∣E∩M∣+∣M∩L∣+∣E∩L∣)−3∣E∩M∩L∣=190−3(25)=190−75=115\text{Exactly 2} = (|E \cap M| + |M \cap L| + |E \cap L|) - 3|E \cap M \cap L| = 190 - 3(25) = 190 - 75 = 115
Each of the three pairwise intersections includes the triple intersection. Subtracting three times the triple intersection isolates the regions representing membership in exactly two sets.

Key Concept

Three-Set Principle of Inclusion-Exclusion
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Topics

Set Theory Concepts and Venn Diagrams

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