Probability Ques 34
In a certain city only two newspapers $A$ and $B$ are published, it is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$, while $8\%$ reads both $A$ and $B$. It is also 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 into advertisements while $50\%$ of those who read both $A$ and $B$ look into advertisements. What is the percentage of the population reads an advertisement?
(1984, 4M)
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Answer:
Correct Answer: 34.$(13.9\%)$
Solution:
- Let $P(A)$ and $P(B)$ denote respectively the percentage of city population that reads newspapers $A$ and $B$.
Then,
$ \begin{aligned} P(A) & =\frac{25}{100}=\frac{1}{4}, P(B)=\frac{20}{100}=\frac{1}{5}, \\ P(A \cap B) & =\frac{8}{100}=\frac{2}{25}, \end{aligned} $
$P(A \cap B)=P(A)-P(A \cap B)=\frac{1}{4}-\frac{2}{25}=\frac{17}{100}$,
$P(\bar{A} \cap B)=P(B)-P(A \cap B)=\frac{1}{5}-\frac{2}{25}=\frac{3}{25}$
Let $P(C)$ be the probability that the population who reads advertisements.
$\therefore \quad P(C)=30 \%$ of $P(A \cap \bar{B})+40 \%$ of $P(\bar{A} \cap B)$ $+50 \%$ of $P(A \cap B)$
[since, $A \cap \bar{B}, \bar{A} \cap B$ and $A \cap B$ are all mutually exclusive]
$\Rightarrow \quad P(C)=\frac{3}{10} \times \frac{17}{100}+\frac{2}{5} \times \frac{3}{25}+\frac{1}{2} \times \frac{2}{25}=\frac{139}{1000}=13.9 \%$
BETA
