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If $A, B$ are two events with $P(A \cup B)=0.65, P(A \cap B)=0.15$, then find the value of $\mathrm{P}\left(\mathrm{A}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{B}^{\mathrm{C}}\right)$
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Verified Answer
The correct answer is:
1.2
Fact: $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
So $P(A \cup B)+P(A \cap B)=P(A)+P(B)$
$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.65+0.15=0.8$, substitute the given values
Therefore
$$
\mathrm{P}\left(\mathrm{A}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{B}^{\mathrm{C}}\right)=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\mathrm{B})=2-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})]=2-0.8=1.2
$$
So $P(A \cup B)+P(A \cap B)=P(A)+P(B)$
$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.65+0.15=0.8$, substitute the given values
Therefore
$$
\mathrm{P}\left(\mathrm{A}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{B}^{\mathrm{C}}\right)=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\mathrm{B})=2-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})]=2-0.8=1.2
$$
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