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If $P(A)=\frac{2}{3}, P(B)=\frac{2}{5}$ and $P(A \cup B)-P(A \cap B)=\frac{2}{5}$, then
what is $P(A \cap B)$ equal to ?
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what is $P(A \cap B)$ equal to ?
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Verified Answer
The correct answer is:
$\frac{1}{3}$
$\quad P(A \cup B)-P(A \cap B)=\frac{2}{5}$
We know that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$\frac{2}{3}+\frac{2}{5}-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}-\frac{2}{3}-\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6-16}{15} \Rightarrow \frac{-10}{15}=-\frac{2}{3} \times \frac{-1}{2}=\frac{1}{3}$
We know that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$\frac{2}{3}+\frac{2}{5}-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}-\frac{2}{3}-\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6-16}{15} \Rightarrow \frac{-10}{15}=-\frac{2}{3} \times \frac{-1}{2}=\frac{1}{3}$
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