Question

For two independent events $A$ and $B$, which of the following pair of events need not be independent?

• A. $A^{\prime },B^{\prime }$
• B. $A,B^{\prime }$
• C. $A^{\prime },B$
• D. $A-B,B-A$

Answer 1

The correct option is D $A-B,B-A$

$A$ and $B$ are independent event.

$\Rightarrow P(A\cap B)=P\left(A\right).P\left(B\right)$

Option A:

$P(A^{\prime }\cap B^{\prime })$ $=P\left(\right(A\cup B{)}^{\prime })=1-P(A\cup B)=1-[P\left(A\right)+P\left(B\right)-P(A\cap B)]=1-P(A)-P(B)+P(A\left)P\right(B)=(1-P\left(A\right)\left)\right(1-P\left(B\right))=P(A^{\prime }\left)P\right(B^{\prime })$

Hence, $A^{\prime },B^{\prime }$ are also independent events.

Option B:

$P(A\cap B^{\prime })$ $=P\left(\right(A^{\prime }\cup B{)}^{\prime })=1-P(A^{\prime }\cup B)=1-[P\left(A^{\prime }\right)+P\left(B\right)-P(A^{\prime }\cap B)]=1-[1-P\left(A\right)+P\left(B\right)-P\left(B\right)+P(A\cap B)]=P(A)-P(A\cap B)=P(A\left)\right(1-P\left(B\right))=P(A\left)P\right(B^{\prime })$

Hence, $A,B^{\prime }$ are also independent.

Option C:

Similar as above, $A^{\prime },B$ can also be proven to be independent.

Option D:

Event $A-B$ implies the happenings in event $A$ which do not occur in event $B$.

Event $B-A$ implies the happenings in event $B$ which do not occur in event $A$.

From the above, it is clear that $A-B$ and $B-A$ are mutually exclusive events. Hence $P\left(\right(A-B)\cap (B-A\left)\right)=0$

Hence, $A-B,B-A$ are not independent events.

Option $D$ is the correct answer.