We can treat the union of three events, A B C, as the union of two events: event A and event (B C). This allows us to apply the addition theorem for two events, which states that P(X Y) = P(X) + P(Y) - P(X Y).

We can treat the union of three events, A B C, as the union of two events: event A and event (B C). This allows us to apply the addition theorem for two events, which states that P(X Y) = P(X) + P(Y) - P(X Y).

Applying the addition theorem for two events, P(X Y) = P(X) + P(Y) - P(X Y), where X = A and Y = (B C).

We apply the addition theorem again to expand the term P(B C). This is a standard result for the union of two events.

We apply the addition theorem one more time to the expression P((A B) (A C)). Here, X = (A B) and Y = (A C).

The intersection of (A B) and (A C) simplifies to A B C, as A is common to both intersections.

If A, B, C are three events associated with a random experiment, prove that $\begin{aligned} \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{C}) \\ & -\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \end{aligned}$

Answer: $P(A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$

Step-by-step solution

Step 1: Apply the Addition Theorem for Two Events

We can treat the union of three events, $A \cup B \cup C$ , as the union of two events: event $A$ and event $(B \cup C)$ . This allows us to apply the addition theorem for two events, which states that $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ .

Step 2: Expand using the Addition Theorem

Applying the addition theorem for two events, $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ , where $X = A$ and $Y = (B \cup C)$ .

Step 3: Expand $P(B \cup C)$

We apply the addition theorem again to expand the term $P(B \cup C)$ . This is a standard result for the union of two events.

Step 4: Expand $P(A \cap (B \cup C))$ using Distributive Law

Using the distributive law of set theory, the intersection of event $A$ with the union of events $B$ and $C$ is equivalent to the union of the intersection of $A$ with $B$ and the intersection of $A$ with $C$ .

Step 5: Apply Addition Theorem to $P((A \cap B) \cup (A \cap C))$

We apply the addition theorem one more time to the expression $P((A \cap B) \cup (A \cap C))$ . Here, $X = (A \cap B)$ and $Y = (A \cap C)$ .

Step 6: Simplify the last term

The intersection of $(A \cap B)$ and $(A \cap C)$ simplifies to $A \cap B \cap C$ , as $A$ is common to both intersections.

Step 7: Substitute and Combine Terms

Now, we substitute the expanded forms of $P(B \cup C)$ and $P(A \cap (B \cup C))$ back into the main equation from Step 2. Then, we distribute the negative sign and combine like terms to arrive at the final identity.

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