Question

If $A$ and $B$ are two events, then, $1+P(A\cap B)-P\left(B\right)-P\left(A\right)$ is equal to

• A. $P(\overline{A}\cup \overline{B})$
• B. $P(A\cap \overline{B})$
• C. $P(\overline{A}\cap B)$
• D. $P(A\cup B)$
• E. $P(\overline{A}\cap \overline{B})$

Answer 1

Answer: (E)

$P(\overline{A}\cap \overline{B})$

We know that for two events,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow P\left(A\right)+P\left(B\right)=P(A\cup B)+P(A\cap B)$
$1+P(A\cap B)-P\left(B\right)-P\left(A\right)$
$=1+P(A\cap B)-(P\left(B\right)+P\left(A\right))$
$=1+P(A\cap B)-(P(A\cup B)+P(A\cap B))$
$=1+P(A\cap B)-P(A\cup B)-P(A\cap B)$
$=1-P(A\cup B)=P\overline{(A\cup B)}=P(\overline{A}\cap \overline{B})$