Engineering
Mathematics
Conditional Probability
Question
In a city, three daily newspapers A, B, C are published. 42% of the people in that city read A, 51% read B and 68% read C. 30% read A and B; 28% read B and C; 36% read A and C; 8% do not read any of the three newspapers. The percentage of persons who read all the three papers is
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Solution
Let the number of persons in the city be 100
Then we have
n(A) = 42, n(B) = 51, n(C) = 68
n(A∩B) = 30, n(B∩C) = 28, n(A∩C) = 36
n(A∩B∩C) = 100 − 8 = 92
Using n(A∪B∪C) = n(A) + n(B) + n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)
Substituting the above value, we have
⇒ n(A∩B∩C) = 92− 161 + 94 = 92 − 67 = 25
Hence 25% of the people real all the three papers.
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