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If $\mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{4}{5}$, then $\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=$
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$P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$
$P(A \cap B)=P(A \mid B) P(B)=\frac{1}{2} \cdot \frac{3}{5}=\frac{3}{10}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\mathrm{P}(\mathrm{A})=\frac{4}{5}-\frac{3}{10}=\frac{1}{2} . \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=\frac{1}{2}$
We know, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})$
[as $A \cap B$ and $A^{\prime} \cap B$ are mutually exclusive events]
$\Rightarrow \quad \frac{3}{10}+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}$
$\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}$
Now, $\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B})-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$
$=\frac{1}{2}+\frac{3}{5}-\frac{3}{10}=\frac{5+6-3}{10}=\frac{4}{5}$
$\mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\frac{4}{5}=\frac{1}{5}$
$\therefore \mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\frac{1}{5}+\frac{4}{5}=1$
$P(A \cap B)=P(A \mid B) P(B)=\frac{1}{2} \cdot \frac{3}{5}=\frac{3}{10}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\mathrm{P}(\mathrm{A})=\frac{4}{5}-\frac{3}{10}=\frac{1}{2} . \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=\frac{1}{2}$
We know, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})$
[as $A \cap B$ and $A^{\prime} \cap B$ are mutually exclusive events]
$\Rightarrow \quad \frac{3}{10}+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}$
$\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}$
Now, $\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B})-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$
$=\frac{1}{2}+\frac{3}{5}-\frac{3}{10}=\frac{5+6-3}{10}=\frac{4}{5}$
$\mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\frac{4}{5}=\frac{1}{5}$
$\therefore \mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\frac{1}{5}+\frac{4}{5}=1$
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