The probability of the simultaneous occurrence of two events A and B…

We have,

$P(A∩ B)= p $ and $P(A) +P(B)-2P(A ∩ B)= q$

$⇒ P(A) + P(B) -2p = q $

 $⇒ P(A) + P(B) =2p + q $

 $⇒ 1-P(\overline{A}) +1 - P(\overline{B})= 2p+q$

 $⇒P(\overline{A})+P(\overline{B})= 2 - 2p - q$

So, alternative (b) is correct.

Now,

$\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P[(A ∩B)∩(A ∪B)}{P(A ∪B)}$

$⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P(A ∩B)}{P(A ∪B)}$

$⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{P(A ∩B)}{P(A)+P(B)-P(A ∩B)}$

$⇒ P\begin{Bmatrix}(A ∩B)/(A ∪B)\end{Bmatrix}=\frac{p}{2p+q-p}=\frac{p}{p+q}$

So, alternative (c) is correct.

Finally,

$P(\overline{A} ∩\overline{B}) P(\overline{A ∪B})=1- P(A ∪B)$

$⇒P(\overline{A} ∩\overline{B})= P(\overline{A ∪B})=1 -[P(A) +P(B) -P(A ∩B)]$

$⇒P(\overline{A} ∩\overline{B})= P(\overline{A ∪B})=1-[2p+q-p]=1-p-q$