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### 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:

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, $$P(A \cap B \cap C)$$ $$=$$ $$\frac{0}{52}$$.

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 $$\frac{1}{6}$$.