Question

In a group of 50 family members, three medicines are given to fight a virus.
• 40 members received medicine A 
• 30 received medicine B 
• 20 received medicine C 
• 10 received all three medicines How many members received at least two medicines?

Hint:

Be very careful with the wording in set theory problems.
"Exactly two" means ( (A cap B setminus C) cup (B cap C setminus A) cup (C cap A setminus B) ). Its size is (x+y+z).
"At least two" means ( (A cap B) cup (B cap C) cup (C cap A) ). Its size is (x+y+z+k).
A common mistake is to stop after finding (x+y+z).

Answer 1

Step 1: Understanding the Question: We are given data about a group of 50 people and the three medicines \(A, B, C\) they received. We need to find the number of people who received at least two medicines. This means we need to find the number of people in the regions \((A \cap B) \cup (B \cap C) \cup (C \cap A)\).

Step 2: Key Formula or Approach: We can use the Principle of Inclusion-Exclusion for three sets. Let's assume everyone received at least one medicine, so \(|A \cup B \cup C| = 50\). The formula is: \[ |A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |C \cap A|) + |A \cap B \cap C| \] We want to find \(|(A \cap B) \cup (B \cap C) \cup (C \cap A)|\).
The number of people who received at least two medicines is given by: \[ N_{\ge 2} = |A \cap B| + |B \cap C| + |C \cap A| - 2|A \cap B \cap C| \]

Step 3: Detailed Explanation: Let's plug the given values into the Inclusion-Exclusion formula:
• \(|A \cup B \cup C| = 50\)
• \(|A| = 40\)
• \(|B| = 30\)
• \(|C| = 20\)
• \(|A \cap B \cap C| = 10\) \[ 50 = (40 + 30 + 20) - (|A \cap B| + |B \cap C| + |C \cap A|) + 10 \] \[ 50 = 90 - (|A \cap B| + |B \cap C| + |C \cap A|) + 10 \] \[ 50 = 100 - (|A \cap B| + |B \cap C| + |C \cap A|) \] Let \(S_2 = |A \cap B| + |B \cap C| + |C \cap A|\). \[ S_2 = 100 - 50 = 50 \] Let:
\(x\) = exactly A and B,
\(y\) = exactly B and C,
\(z\) = exactly A and C,
\(k = |A \cap B \cap C| = 10\). Then:
\(|A \cap B| = x + k\)
\(|B \cap C| = y + k\)
\(|C \cap A| = z + k\) So: \[ (x+k) + (y+k) + (z+k) = 50 \] \[ x + y + z + 3k = 50 \] Substitute \(k = 10\): \[ x + y + z + 30 = 50 \] \[ x + y + z = 20 \] This is the number of people who received exactly two medicines.
Number of people who received at least two medicines: \[ N_{\ge 2} = (x + y + z) + k = 20 + 10 = 30 \]

Step 4: Final Answer:
The number of members who received at least two medicines is \(30\).