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:
 
T-1_1.png 
 
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:
 
T-1_2.png
 
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.