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If $A$ and $B$ are independent events such that
$P(A)=\frac{1}{5}, P(A \cup B)=\frac{7}{10}$, then what is $P(\bar{B})$ equal to ?
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$P(A)=\frac{1}{5}, P(A \cup B)=\frac{7}{10}$, then what is $P(\bar{B})$ equal to ?
Solution:
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Verified Answer
The correct answer is:
$\frac{3}{8}$
As A and B are independent event So, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
Now, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\frac{7}{10}=\frac{1}{5}+\mathrm{P}(\mathrm{B})-\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
$\mathrm{P}(\mathrm{B})\left(1-\frac{1}{5}\right)=\frac{7}{10}-\frac{1}{5}$
$\frac{4}{5} \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{5}{8}$
Now, $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-\frac{5}{8}=\frac{3}{8}$.
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
Now, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\frac{7}{10}=\frac{1}{5}+\mathrm{P}(\mathrm{B})-\frac{1}{5} \cdot \mathrm{P}(\mathrm{B})$
$\mathrm{P}(\mathrm{B})\left(1-\frac{1}{5}\right)=\frac{7}{10}-\frac{1}{5}$
$\frac{4}{5} \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{5}{8}$
Now, $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-\frac{5}{8}=\frac{3}{8}$.
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