If A and B are two events, then P( A )=

MCQ

If A and B are two events, then $\text{P}(\overline{\text{A}}\cap\text{B})=$

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
$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

✓

Answer

$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

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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