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

Answer 1

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

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

We have P (exactly one of $A$, $B$ occurs)
$=P(A\cap B^{\prime })\cup (A^{\prime }\cap B^{\prime })]=P(A\cap B^{\prime })+P(A^{\prime }\cap B)$
$=P\left(A\right)-P(A\cap B)+P\left(B\right)-P(A\cap B)$
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)=P(A\cup B)-P(A\cap B)$
Also P (exactly one of A, B occurs)
$=[1-P(A^{\prime }\cap B^{\prime }\left)\right]-[1-P(A^{\prime }\cup B^{\prime }\left)\right]$
$=P(A^{\prime }\cup B^{\prime })-P(A^{\prime }\cap B^{\prime })$
$=P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-2P(A^{\prime }\cap B^{\prime })$.