In a certain city two newspapers A and B are published. It is known that 25% of the city population reads A and 20% of the population reads B. 8% of the population reads both A and B. It is known that 30% of those who read A but not B look into advertisements and 40% of those who read B but not A look advertisements while 50% of those who read both A and B look into advertisements . What is the percentage of the population who reads an advertisement ?

Engineering

Mathematics

Conditional Probability

Question

$\frac{139}{500}$

$\frac{361}{500}$

$\frac{139}{1000}$

$\frac{861}{1000}$

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Let P(A) and P(B) denote the precentage of city population who reda newspapers A and B.
Then from given data ,we have
P(A) = 25% $= \frac{1}{4}$ ,P(B) = 20% $= \frac{1}{5} .$ P(A∩B) = 8

∴ Percentage of those who read A but not B

P (A \cap B) = P (A) - P (A \cap B)

[see theorem 2 & 3]

= \frac{1}{4} - \frac{2}{25} = \frac{17}{100} = 17

Similarly,

P (A \cap B) = P (B) - P (A \cap B) = \frac{1}{5} - \frac{2}{25} = \frac{3}{25} = 12

If P(C) denotes the percentage of those who look into advertisement , then from the given data we obtain P(C)30 of

P (A \cap B) + 50

of P(A∩B)

= \frac{3}{10} \times \frac{17}{100} + \frac{2}{5} \times \frac{3}{25} + \frac{1}{2} \times \frac{2}{25}

= \frac{51 + 48 + 40}{1000} = \frac{139}{1000} = 13.9 % .

.
Thus the percentage of population who read an advertisement is 13.9%

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