Q.Three coins are tossed. Describe

The key idea is Set Operations — representing coin-toss outcomes as a sample space and defining events as subsets.

Step 1: Sample space

$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ (8 equally likely outcomes).

Step 2: Define events as subsets

1. Mutually exclusive events cannot happen together. Example: $A = \{\text{exactly one head}\} = \{HTT, THT, TTH\}$, $B = \{\text{exactly two heads}\} = \{HHT, HTH, THH\}$. $A \cap B = \emptyset$.
2. Mutually exclusive and exhaustive events cover $S$ with no overlap. Example: $C = \{\text{no head}\} = \{TTT\}$, $D = \{\text{exactly one head}\}$, $E = \{\text{at least two heads}\} = \{HHT, HTH, THH, HHH\}$. $C \cup D \cup E = S$, and all pairwise intersections are empty.
3. Not mutually exclusive means they can occur together. Example: $F = \{\text{first coin head}\} = \{HHH, HHT, HTH, HTT\}$, $G = \{\text{exactly two heads}\}$. $F \cap G = \{HHT, HTH\} \neq \emptyset$. …

When three coins are tossed, each coin can land either heads (H) or tails (T). The total number of possible outcomes is $2^3 = 8$. The sample space $S$ is:

An event is any subset of $S$. Two events are mutually exclusive if they have no outcome in common — their intersection is empty. Events are exhaustive if their union equals the entire sample space $S$. If they are not exhaustive, their union is a proper subset of $S$.

Let’s construct each required set step by step.

1. Two events which are mutually exclusive

   Pick any two events that cannot happen together. For instance, let

   $A = \{\text{all heads}\} = \{HHH\}$ and

   $B = \{\text{all tails}\} = \{TTT\}$.

   Since $A \cap B = \varnothing$, they are mutually exclusive. They are not exhaustive because many outcomes (like HHT) are in neither.

2. Three events which are mutually exclusive and exhaustive

   We need three disjoint events whose union is $S$. A natural way is to group outcomes by the number of heads:

   • $E_0 = \{\text{0 heads}\} = \{TTT\}$
   • $E_1 = \{\text{exactly 1 head}\} = \{HTT, THT, TTH\}$
   • $E_2 = \{\text{exactly 2 heads}\} = \{HHT, HTH, THH\}$
   • $E_3 = \{\text{exactly 3 heads}\} = \{HHH\}$ But that gives four events. To have exactly three, combine two of them. For example, let:
   • $X = \{\text{0 or 1 head}\} = \{TTT, HTT, THT, TTH\}$
   • $Y = \{\text{exactly 2 heads}\} = \{HHT, HTH, THH\}$
   • $Z = \{\text{exactly 3 heads}\} = \{HHH\}$ Check: $X \cap Y = \varnothing$, $X \cap Z = \varnothing$, $Y \cap Z = \varnothing$, and $X \cup Y \cup Z = S$. So these three are mutually exclusive and exhaustive.

3. Two events which are not mutually exclusive

   They must share at least one outcome. Let

   $P = \{\text{first coin is H}\} = \{HHH, HHT, HTH, HTT\}$ and

   $Q = \{\text{second coin is H}\} = \{HHH, HHT, THH, THT\}$.

   Their intersection $P \cap Q = \{HHH, HHT\}$ is non-empty, so they are not mutually exclusive.

4. Two events which are mutually exclusive but not exhaustive

   They must be disjoint, but their union should miss at least one outcome. Take

   $R = \{\text{exactly 2 heads}\} = \{HHT, HTH, THH\}$ and

   $S = \{\text{exactly 0 heads}\} = \{TTT\}$.

   $R \cap S = \varnothing$, so mutually exclusive. Their union $\{HHT, HTH, THH, TTT\}$ does not include outcomes like HHH or HTT, so not exhaustive.

5. Three events which are mutually exclusive but not exhaustive

   We need three pairwise disjoint events whose union is a proper subset of $S$. For instance:

   • $U = \{\text{first coin H, second H}\} = \{HHH, HHT\}$
   • $V = \{\text{first coin T, second T}\} = \{TTH, TTT\}$ …