Question

If $A$ and $B$ are two events, the probability that exactly one of them occurs is given by

• A. $P\left(A\right)+P\left(B\right)-2P(A\cap B)$
• B. $P(A\cap B)+2P(\overline{A}\cap B)$
• C. $P(A\cup B)-P(A\cap B)$
• D. $P\left(\overline{A}\right)+P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $P\left(A\right)+P\left(B\right)-2P(A\cap B)$
B $P(A\cap B)+2P(\overline{A}\cap B)$
C $P(A\cup B)-P(A\cap B)$
D $P\left(\overline{A}\right)+P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})$

Exactly one of the events of $E$
$A$ and $B$ is represented by $A\cap \overline{B}+\overline{A}\cap B$
hence $P(A\cap \overline{B})+P(\overline{A}\cap B)=\{P\left(A\right)-P(A\cap B)\}+\{P\left(B\right)-P(A\cap B)\}=P\left(A\right)+P\left(B\right)-2P(A\cap B)$

which is option (A).
$\Rightarrow P(A\cap \overline{B})+P(\overline{A}\cap B)=[P\left(A\right)+P\left(B\right)-P(A\cap B)]-P(A\cap B)=P(A\cup B)-P(A\cap B)$

which is option (C).
$P(A\cap \overline{B})+P(\overline{A}\cap B)=1-P\left(\overline{A}\right)+1-P\left(\overline{B}\right)-2\{1-P(\overline{A}\cup \overline{B})\}=2P(\overline{A}\cup \overline{B})-P\left(\overline{A}\right)-P\left(\overline{B}\right)=\{2P\left(\overline{A}\right)+2P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})\}-P\left(\overline{A}\right)-P\left(\overline{B}\right)=P\left(\overline{A}\right)+P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})$

which is option (D).
and $P(A\cap \overline{B})+P(\overline{A}\cap B)=1-P(\overline{A}\cup B)+P(\overline{A}\cap B)\because \overline{A\cap \overline{B}}=\overline{A}\cup B$

by demorgan's law $P(A\cap \overline{B})=1-P(\overline{A}\cup B)\Rightarrow P(A\cap \overline{B})+P(\overline{A}\cap B)=\{1-P\left(\overline{A}\right)-P\left(B\right)+P(\overline{A}\cap B)\}+P(\overline{A}\cap B)=P\left(A\right)-P\left(B\right)+2P(\overline{A}\cap B)$