Theory:

Example 1:
If \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{3}\) and \(P(A \cap B) = \frac{1}{6}\), then find the value of \(P(A \cup B)\).
 
Solution:
Addition Theorem of Probability:
 
If \(A\) and \(B\) are any two non mutually exclusive events then:
 
\(P(A \cup B)\) \(=\) \(P(A)\) \(+\) \(P(B)\) \(-\) \(P(A \cap B)\)
Substitute the known values in the above theorem.
 
\(P(A \cup B)\) \(=\) \(\frac{1}{2}\) \(+\) \(\frac{1}{3}\) \(-\) \(\frac{1}{6}\)
 
\(=\) \(\frac{3 + 2 - 1}{6}\)
 
\(=\) \(\frac{5 - 1}{6}\)
 
\(=\) \(\frac{4}{6}\)
 
\(=\) \(\frac{2}{3}\)
 
Therefore, the value of \(P(A \cup B)\) is \(\frac{2}{3}\).
Example  2:
Find the probability of getting a king or a black card or a jack card from a well shuffled deck of \(52\) cards.
 
Solution:
 
Let \(S\) be the sample space.
 
Here \(S\) \(=\) \(52\text{ cards}\).
 
\(\Rightarrow\) \(n(S) = 52\).
 
Let \(A\) be the event of getting a king card.
 
So, \(n(A) = 4\).
 
Thus, \(P(A) = \frac{n(A)}{n(S)} = \frac{4}{52}\).
 
Let \(B\) be the event of getting a black card.
 
So, \(n(B) = 13\).
 
Thus, \(P(B) = \frac{n(B)}{n(S)} = \frac{13}{52}\).
 
Let \(C\) be the event of getting a jack card.
 
So, \(n(C) = 4\).
 
Thus, \(P(C) = \frac{n(C)}{n(S)} = \frac{4}{52}\).
 
Here, \(P(A \cap B) =\) Probability of getting a black king.
 
So, \(P(A \cap B) = \frac{2}{52}\).
 
Here, \(P(B \cap C) =\) Probability of getting a black jack.
 
So, \(P(B \cap C) = \frac{2}{52}\).
 
Here, \(P(A \cap C) =\) Probability of getting a king and a jack.
 
So, \(P(A \cap C) = \frac{0}{52}\).
 
Thus, \(P(A \cap B \cap C)\) \(=\) \(\frac{0}{52}\).
Addition Theorem of Probability:
 
If \(A\), \(B\)and \(C\) are any three non mutually exclusive events then:
 
\(P(A \cup B \cup C)\) \(=\) \(P(A)\) \(+\) \(P(B)\) \(+\) \(P(C)\) \(-\) \(P(A \cap B)\) \(-\) \(P(B \cap C)\) \(-\) \(P(A \cap C)\) \(+\) \(P(A \cap B \cap C)\)
Substitute the known values in the above theorem.
 
\(P(A \cup B \cup C)\) \(=\) \(\frac{4}{52}\) \(+\) \(\frac{13}{52}\) \(+\) \(\frac{4}{52}\) \(-\) \(\frac{2}{52}\) \(-\) \(\frac{2}{52}\) \(-\) \(\frac{0}{52}\) \(+\) \(\frac{0}{52}\)
 
\(=\) \(\frac{21 - 4 + 0}{52}\)
 
\(=\) \(\frac{17}{52}\)
 
Hence, the value of \(P(A \cup B \cup C)\) is \(\frac{17}{52}\).
 
Therefore, the probability of getting a king or a black card or a jack card from a well shuffled deck of \(52\) cards is \(\frac{17}{52}\).
Example 3:
If \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{3}\) and \(P(A \cap B) = \frac{1}{6}\), then find the value of (i) \(P(\text{only }A)\) (ii) \(P(\text{only }B)\).
 
Solution:
By the theorem, we have:
 
(i) \(P(A \cap \overline B)\) \(=\) \(P(\text{only A})\) \(=\) \(P(A) - P(A \cap B)\)
 
(ii) \(P(\overline A \cap B)\) \(=\) \(P(\text{only B})\) \(=\) \(P(B) - P(A \cap B)\)
(i) \(P(\text{only A})\) \(=\) \(\frac{1}{2} - \frac{1}{6}\)
 
\(=\) \(\frac{3 - 1}{6}\)
 
\(=\) \(\frac{2}{6}\)
 
\(=\) \(\frac{1}{3}\)
 
Therefore, the value of \(P(\text{only A})\) is \(\frac{1}{3}\).
 
 
(ii) \(P(\text{only B})\) \(=\) \(\frac{1}{3} - \frac{1}{6}\)
 
\(=\) \(\frac{2 - 1}{6}\)
 
\(=\) \(\frac{1}{6}\)
 
Therefore, the value of \(P(\text{only B})\) is \(\frac{1}{6}\).