How can I express the probabilities of $P(A' \cap B')$ and $P(A' \cap (A \cup B))$?

Question

If there are two independent situations I can use to express those probabilities, how exactly should I approach this?

a. Assuming $P(A)$, $P(B)$, and $P(A \cap B)$ are known

b. Assuming $P(A)$, $P(B)$, and $P(A \cup B)$ are known

Very confused and any help would be greatly appreciated

Answer 1

$$\begin{align}P(A'\cap B')&=P(A')-P(A'\cap B)\\ &=1-P(A)-(P(B)-P(B\cap A))\\ &=1-P(A)-P(B)+P(A\cap B) \end{align}$$

$$\begin{align}P(A'\cap (A\cup B))&=P((A'\cap A)\cup (A'\cap B))\\ &=P(\emptyset )+P(A'\cap B)\\ &=P(B)-P(A\cap B) \end{align}$$

Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ to convert $P(A\cap B)$ into $P(A\cup B)$ and vice versa.

Also if you assueme independence, the complimentary events are also independent. In such case,

$$P(A′\cap B′)=P(A′)P(B′)=(1−P(A))(1−P(B))$$

$$P(A′∩(A∪B))=P((A′∩A)∪(A′∩B))=P(∅)+P(A′∩B)=(1−P(A))P(B)$$

Answer 2

Remark: If you are able to express something in terms of $P(A), P(B), P(A \cap B)$, you are able to express the same thing in terms of $P(A), P(B), P(A \cup B)$ as $$P(A \cup B)= P(A)+P(B)-P(A \cap B)$$

For part a,

$$P(A' \cap B')=P((A \cup B)')=1-P(A \cup B) $$

For part b,

$$P(A' \cap (A \cup B))=P(A' \cap B)=P(B)-P(A \cap B)$$