Question

The probability that exactly one of the independent events $A$ and $B$ occurs is equal to

• A. $P\left(A\right)+P\left(B\right)-2P(A\cap B)$
• B. $P\left(A\right)+P\left(B\right)-P(A\cap B)$
• C. $P\left(\overline{A}\right)+P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})$
• D. None of the above

Answer 1

Answer: (A), (C)

$P\left(\overline{A}\right)+P\left(\overline{B}\right)-2P(\overline{A}\cap \overline{B})$
$P\left(A\right)+P\left(B\right)-2P(A\cap B)$

$P(A\cap \overline{B})+P(B\cap \overline{A})=P\left(A\right)-P(A\cap B)+P\left(B\right)-P(A\cap B)=P\left(A\right)+P\left(B\right)-2P(A\cap B)$

Also

$P(A\cap \overline{B})+P(B\cap \overline{A})=1-P\left(\overline{A}\right)+1-P\left(\overline{B}\right)-2\{1-P(\overline{A}\cup \overline{B})\}=1-P\left(\overline{A}\right)+1-P\left(\overline{B}\right)-2\{1-P\left(\overline{A}\right)-P\left(\overline{B}\right)+P(\overline{A}\cap \overline{B})\}=P\left(\overline{A}\right)+P\left(\overline{B}\right)-P(\overline{A}\cap \overline{B})$