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If $\mathrm{A}$ and $\mathrm{B}$ are any two events such that
$\mathrm{P}(\overline{\mathrm{A}})=0.4, \mathrm{P}(\overline{\mathrm{B}})=0.3, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.9$, then what is the
value of $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$ equal to ?
Options:
$\mathrm{P}(\overline{\mathrm{A}})=0.4, \mathrm{P}(\overline{\mathrm{B}})=0.3, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.9$, then what is the
value of $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$ equal to ?
Solution:
2847 Upvotes
Verified Answer
The correct answer is:
$0.6$
We have $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
By (De-Morgan's law)
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-(1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}))=.4+.3-(1-.9)=.6$
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
By (De-Morgan's law)
$=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})-(1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}))=.4+.3-(1-.9)=.6$
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