Let's consider the experiment of tossing a coin three times. We first need to determine the sample space, which is the set of all possible outcomes.

The possible outcomes when tossing a coin three times are:

• HHH (Head, Head, Head)
• HHT (Head, Head, Tail)
• HTH (Head, Tail, Head)
• THH (Tail, Head, Head)
• HTT (Head, Tail, Tail)
• THT (Tail, Head, Tail)
• TTH (Tail, Tail, Head)
• TTT (Tail, Tail, Tail)

The sample space, denoted by $S$, is:

$\qquad S = \{ \text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} \}$

The total number of outcomes is $2^3 = 8$.

Understanding the Defined Coin Toss Events

Now let's define the given events based on the sample space:

Event A: No head appears

This means all three tosses are tails.

$\qquad A = \{ \text{TTT} \}$

Event B: Exactly one head appears

This means one head and two tails in any order.

$\qquad B = \{ \text{HTT, THT, TTH} \}$

Event C: At least two heads appear

This means two heads and one tail, or three heads.

Outcomes with two heads: HHT, HTH, THH

Outcomes with three heads: HHH

$\qquad C = \{ \text{HHT, HTH, THH, HHH} \}$

Let's check the relationship between these events. Events A, B, and C are mutually exclusive (no common outcomes) and their union covers the entire sample space ($A \cup B \cup C = S$).

Calculating Complements of Events

We need the complements of events B and C, denoted by $B'$ and $C'$. The complement of an event contains all outcomes in the sample space that are not in the event.

Event B': Not exactly one head

This includes outcomes with 0 heads, 2 heads, or 3 heads.

$\qquad B' = S \setminus B = \{ \text{HHH, HHT, HTH, THH, TTT} \}$

Alternatively, $B'$ consists of outcomes in A or C:

$\qquad B' = A \cup C = \{ \text{TTT} \} \cup \{ \text{HHT, HTH, THH, HHH} \} = \{ \text{TTT, HHT, HTH, THH, HHH} \}$

Event C': Not at least two heads

This means less than two heads, i.e., 0 heads or 1 head.

$\qquad C' = S \setminus C = \{ \text{TTT, HTT, THT, TTH} \}$

Alternatively, $C'$ consists of outcomes in A or B:

$\qquad C' = A \cup B = \{ \text{TTT} \} \cup \{ \text{HTT, THT, TTH} \} = \{ \text{TTT, HTT, THT, TTH} \}$

Summary of Events and Complements:

Evaluating the Options Using Set Operations

Now we evaluate each option using the sets defined above.

Option 1: $(A \cup B) \cap (A \cup C) = B \cup C$

• Left Hand Side (LHS): $(A \cup B) \cap (A \cup C)$
• $A \cup B = \{ \text{TTT} \} \cup \{ \text{HTT, THT, TTH} \} = \{ \text{TTT, HTT, THT, TTH} \} = C'$
• $A \cup C = \{ \text{TTT} \} \cup \{ \text{HHT, HTH, THH, HHH} \} = \{ \text{TTT, HHT, HTH, THH, HHH} \} = B'$
• LHS = $C' \cap B' = \{ \text{TTT, HTT, THT, TTH} \} \cap \{ \text{TTT, HHT, HTH, THH, HHH} \} = \{ \text{TTT} \} = A$
• Right Hand Side (RHS): $B \cup C$
• $B \cup C = \{ \text{HTT, THT, TTH} \} \cup \{ \text{HHT, HTH, THH, HHH} \} = \{ \text{HTT, THT, TTH, HHT, HTH, THH, HHH} \}$
• LHS ($A$) $\neq$ RHS ($B \cup C$). Option 1 is incorrect.

Option 2: $(A \cap B') \cup (A \cup C') = B' \cup C'$

• Left Hand Side (LHS): $(A \cap B') \cup (A \cup C')$
• $A \cap B' = \{ \text{TTT} \} \cap \{ \text{TTT, HHT, HTH, THH, HHH} \} = \{ \text{TTT} \} = A$
• $A \cup C' = \{ \text{TTT} \} \cup \{ \text{TTT, HTT, THT, TTH} \} = \{ \text{TTT, HTT, THT, TTH} \} = C'$
• LHS = $A \cup C' = \{ \text{TTT} \} \cup \{ \text{TTT, HTT, THT, TTH} \} = \{ \text{TTT, HTT, THT, TTH} \} = C'$
• Right Hand Side (RHS): $B' \cup C'$
• $B' \cup C' = \{ \text{TTT, HHT, HTH, THH, HHH} \} \cup \{ \text{TTT, HTT, THT, TTH} \} = \{ \text{TTT, HHT, HTH, THH, HHH, HTT, THT, TTH} \} = S$
• LHS ($C'$) $\neq$ RHS ($S$). Option 2 is incorrect.

Option 3: $A \cap (B' \cup C') = A \cup B \cup C$

• Left Hand Side (LHS): $A \cap (B' \cup C')$
• $B' \cup C' = \{ \text{TTT, HHT, HTH, THH, HHH} \} \cup \{ \text{TTT, HTT, THT, TTH} \} = \{ \text{TTT, HHT, HTH, THH, HHH, HTT, THT, TTH} \} = S$
• LHS = $A \cap S = \{ \text{TTT} \} \cap S = \{ \text{TTT} \} = A$
• Right Hand Side (RHS): $A \cup B \cup C$
• Since A, B, C partition S, $A \cup B \cup C = S$.
• LHS ($A$) $\neq$ RHS ($S$). Option 3 is incorrect.

Option 4: $A \cap (B' \cup C') = B' \cap C'$

• Left Hand Side (LHS): $A \cap (B' \cup C')$
• As calculated in Option 3, $B' \cup C' = S$.
• LHS = $A \cap S = \{ \text{TTT} \} \cap S = \{ \text{TTT} \} = A$
• Right Hand Side (RHS): $B' \cap C'$
• As calculated in Option 1, $B' \cap C' = \{ \text{TTT} \} = A$
• LHS ($A$) = RHS ($A$). Option 4 is correct.

Let's verify the result using set properties. Recall $B' = A \cup C$ and $C' = A \cup B$ because A, B, C partition S.

• LHS: $A \cap (B' \cup C') = A \cap ((A \cup C) \cup (A \cup B))$
• Using associative and commutative laws for union: $A \cap (A \cup B \cup C)$
• Since $A \cup B \cup C = S$: $A \cap S = A$
• RHS: $B' \cap C' = (A \cup C) \cap (A \cup B)$
• Using the distributive law $(X \cup Y) \cap (X \cup Z) = X \cup (Y \cap Z)$ with $X=A, Y=C, Z=B$: $A \cup (C \cap B)$
• Since events B and C are mutually exclusive, $C \cap B = \emptyset$.
• RHS: $A \cup \emptyset = A$
• LHS = A and RHS = A. The equality holds.

Conclusion

By defining the sample space and the events A, B, and C, calculating their complements, and applying set operations, we found that the equation in Option 4 is correct.

$\qquad A \cap (B' \cup C') = B' \cap C'$

Revision Table: Key Concepts

Additional Information: Set Identities in Probability

Set operations like union, intersection, and complement are fundamental in probability theory. They allow us to define relationships between events and calculate probabilities of combined events using rules like the Addition Rule ($P(A \cup B) = P(A) + P(B) - P(A \cap B)$) or De Morgan's Laws $((A \cup B)' = A' \cap B')$ and $((A \cap B)' = A' \cup B')$. Understanding how to manipulate sets using these laws is crucial for solving problems involving multiple events.

In this problem, we implicitly used the distributive law for sets: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$ and $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. The identity used to simplify $B' \cap C'$ in Option 4 was $(A \cup C) \cap (A \cup B) = A \cup (C \cap B)$, which is a standard distributive property.