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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 NEWSPAPER C, 5% FAMILIES BUY A AND B , 3% BUY B AND C AND 4% BUY A AND C .IF 2% FAMILIES BUY ALL THE NEWSPAPERS FIND THE NUMBER OF FAMILIES WHICH BUY

1. A ONLY 2. B ONLY 3. NONE OF A ,B , AND C.

$Given N=10000 n(A)=40% of 10000=4000 n(B)=20% of 10000 = 2000 n (C) = 10 % of 10000=1000 n(A \cap B)=500 n(B \cap C)=300 n(C \cap A)=400 and n(A \cap B \cap C)=200 (i) n (A \cap \bar{B} \cap \bar{C}) = n (A \cap \bar{(B \cup C)}) = n (A) - n (A \cap (B \cup C)) = n (A) - [n (A \cap B) \cup n (A \cap C)] = n (A) - [n (A \cap B) + n (A \cap C) - n (A \cap B) \cap (A \cap C)] = n (A) - [n (A \cap B) + n (A \cap C) - n (A \cap B \cap C)] = 4000 - 500 - 400 + 200 = 3300 (i i) n (\bar{A} \cap B \cap \bar{C}) = n (B \cap \bar{(A \cup C)}) = n (B) - n (B \cap (A \cup C)) = n (B) - [n (B \cap A) \cup n (B \cap C)] = n (B) - [n (A \cap B) + n (B \cap C) - n (A \cap B) \cap (B \cap C)] = n (B) - [n (A \cap B) + n (B \cap C) - n (A \cap B \cap C)] = 2000 - 500 - 300 + 200 = 1400 (i i i) n (\bar{A} \cap \bar{B} \cap \bar{C}) = n (\bar{A \cup B \cup C)} = N - n (A \cup B \cup C) = n - [n (A) + n (B) + n (C) - n (A \cup B) - n (B \cup C) - n (C \cup A) + n (A \cap B \cap C)] = 10000 - [4000 + 2000 + 1000 - 500 - 400 - 300 + 200] = 10000 - 6000 = 4000$

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