- A$\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
- B$1-\text{P}(\text{A})-\text{P}(\text{B})$
- C$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
- D$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
Solution:

From the diagram, we get $\text{A}\cap\text{B}$ and $\overline{\text{A}}\cap\text{B}$ are mutually exclusive events such that $(\text{A}\cap\text{B})\cup(\overline{\text{A}}\cap\text{B})=\text{B}.$ therefore by
$\text{P}(\text{A}\cap\text{B})+\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}$
$\therefore\ \text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
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