To solve the problem step-by-step, we will use the principle of inclusion-exclusion to find the number of families that buy only newspapers A, B, and C, as well as those that buy none of the newspapers.
### Given Data:
- Total families = 10,000
- Families buying A = 40% of 10,000 = 4,000
- Families buying B = 20% of 10,000 = 2,000
- Families buying C = 10% of 10,000 = 1,000
- Families buying both A and B = 5% of 10,000 = 500
- Families buying both B and C = 3% of 10,000 = 300
- Families buying both A and C = 4% of 10,000 = 400
- Families buying all three newspapers = 2% of 10,000 = 200
### Step 1: Define Variables
Let:
- \( n(A) \) = Number of families buying A = 4000
- \( n(B) \) = Number of families buying B = 2000
- \( n(C) \) = Number of families buying C = 1000
- \( n(A \cap B) \) = Families buying both A and B = 500
- \( n(B \cap C) \) = Families buying both B and C = 300
- \( n(A \cap C) \) = Families buying both A and C = 400
- \( n(A \cap B \cap C) \) = Families buying all three = 200
### Step 2: Calculate Families Buying Only A
To find the number of families that buy only newspaper A:
\[
n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)
\]
Substituting the values:
\[
n(A \text{ only}) = 4000 - 500 - 400 + 200 = 3300
\]
### Step 3: Calculate Families Buying Only B
To find the number of families that buy only newspaper B:
\[
n(B \text{ only}) = n(B) - n(A \cap B) - n(B \cap C) + n(A \cap B \cap C)
\]
Substituting the values:
\[
n(B \text{ only}) = 2000 - 500 - 300 + 200 = 1400
\]
### Step 4: Calculate Families Buying Only C
To find the number of families that buy only newspaper C:
\[
n(C \text{ only}) = n(C) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)
\]
Substituting the values:
\[
n(C \text{ only}) = 1000 - 400 - 300 + 200 = 500
\]
### Step 5: Calculate Families Buying None of A, B, or C
To find the number of families that do not buy any of the newspapers, we first calculate the total number of families buying at least one newspaper using the principle of inclusion-exclusion:
\[
n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)
\]
Substituting the values:
\[
n(A \cup B \cup C) = 4000 + 2000 + 1000 - 500 - 400 - 300 + 200 = 6000
\]
Thus, the number of families buying none of the newspapers:
\[
n(\text{none}) = \text{Total families} - n(A \cup B \cup C) = 10000 - 6000 = 4000
\]
### Final Answers:
(i) Families buying A only = 3300
(ii) Families buying B only = 1400
(iii) Families buying C only = 500
(iv) Families buying none of A, B, and C = 4000
