In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy news-paper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three mewspapers, find the number of families which buy (iii) none of A, B and C

Answer

To solve the problem step by step, we will use the principle of inclusion-exclusion to find the number of families that buy none of the newspapers A, B, or C.

### Step 1: Define the total number of families
We know that there are a total of 10,000 families in the town.

### Step 2: Convert percentages to actual numbers
- Families buying newspaper A:
\[ N(A) = 40\% \text{ of } 10,000 = 0.40 \times 10,000 = 4,000 \]
- Families buying newspaper B:
\[ N(B) = 20\% \text{ of } 10,000 = 0.20 \times 10,000 = 2,000 \]
- Families buying newspaper C:
\[ N(C) = 10\% \text{ of } 10,000 = 0.10 \times 10,000 = 1,000 \]

### Step 3: Find the number of families buying combinations of newspapers
- Families buying both A and B:
\[ N(A \cap B) = 5\% \text{ of } 10,000 = 0.05 \times 10,000 = 500 \]
- Families buying both B and C:
\[ N(B \cap C) = 3\% \text{ of } 10,000 = 0.03 \times 10,000 = 300 \]
- Families buying both A and C:
\[ N(A \cap C) = 4\% \text{ of } 10,000 = 0.04 \times 10,000 = 400 \]
- Families buying all three newspapers A, B, and C:
\[ N(A \cap B \cap C) = 2\% \text{ of } 10,000 = 0.02 \times 10,000 = 200 \]

### Step 4: Use the principle of inclusion-exclusion to find families buying at least one newspaper
The formula for the union of three sets is:
\[ N(A \cup B \cup C) = N(A) + N(B) + N(C) - N(A \cap B) - N(B \cap C) - N(A \cap C) + N(A \cap B \cap C) \]
Substituting the values we calculated:
\[ N(A \cup B \cup C) = 4000 + 2000 + 1000 - 500 - 300 - 400 + 200 \]
Calculating this step-by-step:
1. Sum of individual sets:
\[ 4000 + 2000 + 1000 = 7000 \]
2. Subtracting the intersections:
\[ 7000 - 500 - 300 - 400 = 7000 - 1200 = 5800 \]
3. Adding the intersection of all three:
\[ 5800 + 200 = 6000 \]

### Step 5: Calculate the number of families buying none of the newspapers
The number of families that buy none of the newspapers is:
\[ N(\text{none}) = N(U) - N(A \cup B \cup C) = 10000 - 6000 = 4000 \]

### Final Answer
The number of families that buy none of newspapers A, B, or C is **4,000**.

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