In a city of 10,000 families,40% families buy | vedclass

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(B) Given total families $N = 10,000$.$n(A) = 40\% 	ext{ of } 10,000 = 4,000$$n(B) = 20\% 	ext{ of } 10,000 = 2,000$$n(C) = 10\% 	ext{ of } 10,000

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(B) Given total families $N = 10,000$.$n(A) = 40\% 	ext{ of } 10,000 = 4,000$$n(B) = 20\% 	ext{ of } 10,000 = 2,000$$n(C) = 10\% 	ext{ of } 10,000 = 1,000$$n(A \cap B) = 5\% 	ext{ of } 10,000 = 500$$n(B \cap C) = 3\% 	ext{ of } 10,000 = 300$$n(A \cap C) = 4\% 	ext{ of } 10,000 = 400$$n(A \cap B \cap C) = 2\% 	ext{ of } 10,000 = 200$We need to find the number of families that buy only newspaper $A$,which is given by $n(A \cap B^c \cap C^c)$.Using the formula: $n(A 	ext{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$$n(A 	ext{ only}) = 4,000 - 500 - 400 + 200$$n(A 	ext{ only}) = 4,000 - 900 + 200 = 3,300$.

In a city of $10,000$ families,$40\%$ families buy newspaper $A$,$20\%$ buy newspaper $B$,$10\%$ buy newspaper $C$,$5\%$ buy newspapers $A$ and $B$,$3\%$ buy $B$ and $C$,and $4\%$ buy $A$ and $C$. If $2\%$ families buy all three newspapers,find the number of families that buy only newspaper $A$.

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