### Theory:

Theorem:
If $$A$$ and $$B$$ are two events associated with a random experiment, then prove that:

(i) $$P(A \cap \overline B)$$ $$=$$ $$P(\text{only A})$$ $$=$$ $$P(A) - P(A \cap B)$$

(ii) $$P(\overline A \cap B)$$ $$=$$ $$P(\text{only B})$$ $$=$$ $$P(B) - P(A \cap B)$$
Proof for the theorem:
Statement (i):

The Venn diagram representing $$A \cap \overline B$$ is as follows:

To prove: $$P(A \cap \overline B)$$ $$=$$ $$P(A) - P(A \cap B)$$

Proof for statement (i):

1. $$(A \cap B) \cup (A \cap \overline B)$$ $$=$$ $$A \cap (B \cup \overline B)$$

Here, $$B \cup \overline B = S$$.

Thus, $$(A \cap B) \cup (A \cap \overline B)$$ $$=$$ $$A \cap S$$

Hence, $$(A \cap B) \cup (A \cap \overline B)$$ $$=$$ $$A$$.

2. $$(A \cap B) \cap (A \cap \overline B)$$ $$=$$ $$A \cap (B \cap \overline B)$$

Here, $$B \cap \overline B = \phi$$.

Thus, $$(A \cap B) \cap (A \cap \overline B)$$ $$=$$ $$A \cap \phi$$

Hence, $$(A \cap B) \cap (A \cap \overline B)$$ $$=$$ $$\phi$$.

Therefore, from the above properties it is clear that the events $$(A \cap B)$$ and $$(A \cap \overline B)$$ are mutually exclusive and its union is set $$A$$.

So,  $$P(A)$$ $$=$$ $$P\left[(A \cap B) \cup (A \cap \overline B)\right]$$

$$\Rightarrow$$ $$P(A)$$ $$=$$ $$P(A \cap B) + P(A \cap \overline B)$$

$$\Rightarrow$$ $$P(A \cap \overline B)$$ $$=$$ $$P(A)$$ $$-$$ $$P(A \cap B)$$

Therefore, $$P(A \cap \overline B)$$ $$=$$ $$P(\text{only A})$$ $$=$$ $$P(A) - P(A \cap B)$$.

Hence, proved.

Statement (ii):

The Venn diagram representing $$\overline A \cap B$$ is as follows:

To prove: $$P(\overline A \cap B)$$ $$=$$ $$P(B) - P(A \cap B)$$

Proof for statement(ii):

1. $$(A \cap B) \cup (\overline A \cap B)$$ $$=$$ $$(A \cup \overline A) \cap B$$

Here, $$A \cup \overline A = S$$.

Thus, $$(A \cap B) \cup (\overline A \cap B)$$ $$=$$ $$S \cap B$$

Hence, $$(A \cap B) \cup (\overline A \cap B)$$ $$=$$ $$B$$.

2. $$(A \cap B) \cap (\overline A \cap B)$$ $$=$$ $$(A \cap \overline A) \cap B$$

Here, $$A \cap \overline A = \phi$$.

Thus, $$(A \cap B) \cap (\overline A \cap B)$$ $$=$$ $$\phi \cap B$$

Hence, $$(A \cap B) \cap (\overline A \cap B)$$ $$=$$ $$\phi$$.

Therefore, from the above properties it is clear that the events $$(A \cap B)$$ and $$(\overline A \cap B)$$ are mutually exclusive and its union is set $$B$$.

So,  $$P(B)$$ $$=$$ $$P\left[(A \cap B) \cup (\overline A \cap B)\right]$$

$$\Rightarrow$$ $$P(B)$$ $$=$$ $$P(A \cap B) + P(\overline A \cap B)$$

$$\Rightarrow$$ $$P(\overline A \cap B)$$ $$=$$ $$P(B)$$ $$-$$ $$P(A \cap B)$$

Therefore, $$P(\overline A \cap B)$$ $$=$$ $$P(\text{only B})$$ $$=$$ $$P(B) - P(A \cap B)$$.

Hence, proved.