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

Answer 1

Answer: (A)

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