Question

The probability of the simultaneous occurrence of two events $A$ and $B$ is $p$. If the probability that exactly one of the events occurs is $q$, then which of the following is not correct?

• A. $P \left( A ^{\prime}\right)+ P \left( B ^{\prime}\right)=2+2 q - p$
• B. $P \left( A'\right)+ P \left( B'\right)=2-2 p - q$
• C. $P ( A \cap B| A \cup B )=\frac{ p }{ p + q }$
• D. $P \left( A' \cap B'\right)=1- p - q$

Answer 1

Answer: (A)

It is given that
$P ( A \cap B )=p$ and $P \left( A' \cap B \right)+ P \left( A \cap B'\right)= q$
since $P \left( A' \cap B \right)= P ( B )- P ( A \cap B )$, we get
$=P(B)-P(A \cap B)+P(A)-P(A \cap B)$
$q=P(A)+P(B)=q+2 p$
$P \left(A'\right)+P\left(B'\right)=1-P(A)+1-P(B)=2-q-2 p$
showing that (b) is correct. The answer (c) is also correct because
$P(A \cap B |A \cup B)=\frac{P[(A \cap B) \cap(A \cup B)]}{P(A \cup B)}=\frac{P(A \cap B)}{P(A \cup B)}$
$=\frac{P(A \cap B)}{P(A)+P(B)-P(A \cap B)}=\frac{p}{q+2 p-p}=\frac{p}{p+q}$
Finally, (d) is correct because
$P\left(A' \cap B'\right)=1-P(A \cup B)$
$=1-\left[P(A)+P(B)-P\left(A'-P(A \cap B)\right]=1-( q +2 p - p )=1- p - q \right.$