Chapter 14 — STATISTICS

Two events are said to be mutually exclusive if they cannot occur simultaneously. Mathematically,

\[ E \cap F = \varnothing \]

This means there is no common outcome between the events. If even a single outcome is common, the events are not mutually exclusive.

• Step 1: Write the sample space of experiment
• Step 2: Define both events explicitly
• Step 3: Find intersection \(E \cap F\)
• Step 4: Check if intersection is empty or not
• Step 5: Conclude using definition

The sample space for a single throw of a fair die is:

\[ S = \{1,2,3,4,5,6\} \]

Event \(E\): Die shows 4

\[ E = \{4\} \]

Event \(F\): Die shows an even number

\[ F = \{2,4,6\} \]

Now, we find the intersection of events \(E\) and \(F\).

\[ E \cap F = \{4\} \cap \{2,4,6\} \]

The common element in both sets is:

\[ E \cap F = \{4\} \]

Since the intersection contains an element, i.e., it is not empty, we conclude:

\[ E \cap F \neq \varnothing \]

Therefore, events \(E\) and \(F\) are not mutually exclusive.

Also, if event \(E\) occurs (i.e., outcome is 4), then event \(F\) automatically occurs, because 4 is an even number.

• This concept is directly used in probability addition rule: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
• In Board exams, 1–2 mark conceptual questions often test mutual exclusivity
• In JEE Main, this idea appears in event classification and conditional probability
• In advanced problems, confusion between independent and mutually exclusive events is common

In probability, events are subsets of the sample space. Operations on events follow set theory:

• \(A \cup B\): outcomes in either \(A\) or \(B\)
• \(A \cap B\): outcomes common to both
• \(A - B\): outcomes in \(A\) but not in \(B\)
• \(A'\): complement of \(A\), i.e., outcomes not in \(A\)

• Step 1: Write sample space
• Step 2: Define each event explicitly
• Step 3: Apply set operations carefully
• Step 4: Compute complement using sample space
• Step 5: Verify results logically

The sample space for a single throw of a die is:

\[ S = \{1,2,3,4,5,6\} \]

Now define each event:

(i) Number less than 7: \[ A = \{1,2,3,4,5,6\} \]

(ii) Number greater than 7: No such outcome exists in \(S\), therefore \[ B = \varnothing \]

(iii) Multiple of 3: \[ C = \{3,6\} \]

(iv) Number less than 4: \[ D = \{1,2,3\} \]

(v) Even number greater than 4: Even numbers are \(2,4,6\), among these only 6 satisfies condition: \[ E = \{6\} \]

(vi) Number not less than 3: \[ F = \{3,4,5,6\} \]

Now compute required operations step by step:

\[ \begin{aligned} A \cup B &= \{1,2,3,4,5,6\} \cup \varnothing \\&= \{1,2,3,4,5,6\} \end{aligned} \]

\[ \begin{aligned} A \cap B &= \{1,2,3,4,5,6\} \cap \varnothing \\&= \varnothing \end{aligned} \]

\[ \begin{aligned} B \cup C &= \varnothing \cup \{3,6\} \\&= \{3,6\} \end{aligned} \]

\[ \begin{aligned} E \cap F &= \{6\} \cap \{3,4,5,6\} \\&= \{6\} \end{aligned} \]

\[ \begin{aligned} D \cap E &= \{1,2,3\} \cap \{6\} \\&= \varnothing \end{aligned} \]

\[ \begin{aligned} A - C &= \{1,2,3,4,5,6\} - \{3,6\} \\&= \{1,2,4,5\} \end{aligned} \]

\[ \begin{aligned} D - E &= \{1,2,3\} - \{6\} \\&= \{1,2,3\} \end{aligned} \]

First find complement of \(F\):

\[ \begin{aligned} F' &= S - F \\&= \{1,2,3,4,5,6\} - \{3,4,5,6\} \\&= \{1,2\} \end{aligned} \]

Now, \[ \begin{aligned} E \cap F' &= \{6\} \cap \{1,2\} \\&= \varnothing \end{aligned} \]

• Fundamental for understanding event algebra used in probability formulas
• Board exams frequently test complement and set difference
• JEE Main questions often combine these with probability calculations
• Helps avoid common mistakes like misinterpreting "not less than" and complements
• Base for conditional probability and Bayes’ theorem in higher problems

When two dice are thrown, each outcome is represented as an ordered pair: \[ (x,y), \quad x,y \in \{1,2,3,4,5,6\} \]

Total number of outcomes: \[ n(S) = 6 \times 6 = 36 \]

Events are subsets of this sample space and are constructed using conditions on:

• Sum of outcomes
• Specific number appearance
• Combination of conditions

• Step 1: Represent outcomes as ordered pairs
• Step 2: Identify valid sums or conditions
• Step 3: List all satisfying ordered pairs systematically
• Step 4: Verify no pair is missed or repeated
• Step 5: Check intersections for mutual exclusivity

Let each outcome be represented as an ordered pair \((x,y)\), where \(x\) is the result on first die and \(y\) on second die.

The sample space consists of all such ordered pairs: \[ S = \{(x,y) \mid x,y = 1,2,3,4,5,6\} \]

(i) Event \(A\): Sum greater than 8

Condition: \[ x + y > 8 \Rightarrow x+y = 9,10,11,12 \]

Listing systematically:

Sum = 9: \((3,6),(4,5),(5,4),(6,3)\)
Sum = 10: \((4,6),(5,5),(6,4)\)
Sum = 11: \((5,6),(6,5)\)
Sum = 12: \((6,6)\)

Therefore, \[ A = \{(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\} \]

(ii) Event \(B\): 2 occurs on either die

This means either:
First die = 2 OR Second die = 2

First die = 2: \((2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\)

Second die = 2: \((1,2),(2,2),(3,2),(4,2),(5,2),(6,2)\)

Combining and removing repetition:

\[ B = \{(1,2),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2)\} \]

(iii) Event \(C\): Sum ≥ 7 and multiple of 3

Multiples of 3 ≥ 7 are: \[ 9, 12 \]

Sum = 9: \((3,6),(4,5),(5,4),(6,3)\)

Sum = 12: \((6,6)\)

Therefore, \[ C = \{(3,6),(4,5),(5,4),(6,3),(6,6)\} \]

Mutual Exclusivity Check

\(A \cap B\):

Event \(A\) contains only sums ≥ 9 (no 2 involved), while \(B\) contains outcomes with 2. No ordered pair is common.

\[ A \cap B = \varnothing \]

\(B \cap C\):

Event \(C\) has outcomes with sums 9 and 12. None of these contain 2.

\[ B \cap C = \varnothing \]

Therefore, \(A\) and \(B\), and also \(B\) and \(C\), are mutually exclusive events.

• Very important for JEE: systematic listing of ordered pairs avoids missing outcomes
• Board exams often test sum-based event construction
• Understanding “either” vs “both” is critical in probability
• Foundation for probability calculation using favorable outcomes
• Helps in solving conditional probability and independence questions

For multiple coin tosses, outcomes are represented as ordered sequences of \(H\) and \(T\).

• Simple Event: contains exactly one outcome
• Compound Event: contains more than one outcome
• Mutually Exclusive Events: events with no common outcomes

• Step 1: Write full sample space (all 8 outcomes)
• Step 2: Define each event explicitly
• Step 3: Compare intersections pairwise
• Step 4: Count elements to classify simple/compound

The sample space for three coin tosses is:

\[ \begin{aligned} S = \{&(H,H,H),(H,H,T),(H,T,H),\\&(H,T,T),(T,H,H),(T,H,T),\\&(T,T,H),(T,T,T)\} \end{aligned} \]

Define events:

\[ A = \{(H,H,H)\} \]

\[ B = \{(H,H,T),(H,T,H),(T,H,H)\} \]

\[ C = \{(T,T,T)\} \]

First coin must be Head:

\[ D = \{(H,H,H),(H,H,T),(H,T,H),(H,T,T)\} \]

(i) Mutually Exclusive Events

Two events are mutually exclusive if their intersection is empty.

\[ A \cap B = \varnothing,\quad A \cap C = \varnothing,\quad B \cap C = \varnothing \]

Also,

\[ C \cap D = \varnothing \]

However,

\[ A \cap D = \{(H,H,H)\} \neq \varnothing \]

\[ B \cap D = \{(H,H,T),(H,T,H)\} \neq \varnothing \]

Therefore, mutually exclusive pairs are: \(A\) and \(B\), \(A\) and \(C\), \(B\) and \(C\), and \(C\) and \(D\).

(ii) Simple Events

Events containing only one outcome:

\[ A,\; C \]

(iii) Compound Events

Events containing more than one outcome:

\[ B,\; D \]

• Very important for Board exams: classification of events is frequently asked
• Tree diagram approach is highly useful in JEE probability questions
• Helps in understanding independence vs mutual exclusivity (common confusion)
• Foundation for binomial probability and conditional probability
• Reduces errors in counting outcomes in multi-step experiments

• Mutually Exclusive: No common outcomes, \(A \cap B = \varnothing\)
• Exhaustive Events: Union equals sample space, \(A_1 \cup A_2 \cup \cdots = S\)
• Not Mutually Exclusive: At least one common outcome exists

• Step 1: Write complete sample space
• Step 2: Construct events using head-count logic
• Step 3: Verify intersection (mutual exclusivity)
• Step 4: Verify union (exhaustiveness)

Sample space:

\[ S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\} \]

(i) Two mutually exclusive events

Let \[ A = \{HHH,HHT,HTH,THH\} \quad (\text{at least two heads}) \] \[ B = \{HTT,THT,TTH,TTT\} \quad (\text{at least two tails}) \]

These events have no common outcome: \[ A \cap B = \varnothing \]

(ii) Three mutually exclusive and exhaustive events

Let

\[ A_1 = \{TTT\} \quad (\text{no head}) \] \[ A_2 = \{HTT,THT,TTH\} \quad (\text{exactly one head}) \] \[ A_3 = \{HHH,HHT,HTH,THH\} \quad (\text{at least two heads}) \]

Clearly, \[ A_1 \cap A_2 = A_2 \cap A_3 = A_1 \cap A_3 = \varnothing \]

Also, \[ A_1 \cup A_2 \cup A_3 = S \]

Hence, they are mutually exclusive and exhaustive.

(iii) Two events which are not mutually exclusive

Let

\[ C = \{HHH,HHT,HTH,THH,HTT,THT,TTH\} \quad (\text{at least one head}) \] \[ D = \{HHT,HTH,THH\} \quad (\text{exactly two heads}) \]

Common outcomes exist: \[ C \cap D = \{HHT,HTH,THH\} \neq \varnothing \]

(iv) Two mutually exclusive but not exhaustive events

Let

\[ E = \{HHH\}, \quad F = \{TTT\} \]

Then, \[ E \cap F = \varnothing \]

But, \[ E \cup F = \{HHH,TTT\} \neq S \]

Hence, not exhaustive.

(v) Three mutually exclusive but not exhaustive events

Let

\[ G = \{HHH\}, \quad H = \{HHT,HTH,THH\}, \quad I = \{TTT\} \]

Clearly, \[ G \cap H = H \cap I = G \cap I = \varnothing \]

But, \[ G \cup H \cup I = \{HHH,HHT,HTH,THH,TTT\} \neq S \]

Outcomes like \(HTT,THT,TTH\) are missing.

• Very common Board question: classification of events
• JEE Main often tests exhaustive vs non-exhaustive confusion
• Essential for applying total probability theorem
• Helps avoid logical mistakes in event construction
• Builds foundation for conditional probability and partitions

• Complement: \(A' = S - A\)
• Union: \(A \cup B\) → either event occurs
• Intersection: \(A \cap B\) → both occur
• Difference: \(A - C = A \cap C'\)
• Important: Even and odd outcomes partition the sample space

• Step 1: Represent outcomes as ordered pairs
• Step 2: Define A, B, C explicitly
• Step 3: Use complement relations \(A'=B,\; B'=A\)
• Step 4: Apply set operations carefully
• Step 5: Simplify expressions before listing

Sample space: \[ S = \{(x,y) \mid x,y = 1,2,3,4,5,6\} \]

Define events:

Even numbers on first die: \[ A = \{(2,y),(4,y),(6,y)\} \]

Odd numbers on first die: \[ B = \{(1,y),(3,y),(5,y)\} \]

Sum ≤ 5:

\[ \begin{aligned} C = \{&(1,1),(1,2),(1,3),(1,4),\\&(2,1),(2,2),(2,3),\\&(3,1),(3,2),\\&(4,1)\} \end{aligned} \]

(i) \(A'\)

Complement of A: \[ A' = B \]

(ii) not \(B\)

\[ B' = A \]

(iii) \(A \cup B\)

All outcomes are either even or odd: \[ A \cup B = S \]

(iv) \(A \cap B\)

No number can be both even and odd: \[ A \cap B = \varnothing \]

(v) \(A - C = A \cap C'\)

Remove all outcomes of C from A:

\[ \begin{aligned} A - C = \{&(2,4),(2,5),(2,6),\\ &(4,2),(4,3),(4,4),(4,5),(4,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \]

(vi) \(B \cup C\)

Combine all outcomes of B and C:

\[ B \cup C = \{(1,y),(3,y),(5,y)\} \cup C \]

(Explicit listing same as given, covering all odd-first outcomes plus small sums)

(vii) \(B \cap C\)

Outcomes common to both:

\[ \begin{aligned} B \cap C = \{&(1,1),(1,2),(1,3),(1,4),\\&(3,1),(3,2)\} \end{aligned} \]

(viii) \(A \cap B' \cap C'\)

Since \(B' = A\),

\[ A \cap B' \cap C' = A \cap A \cap C' = A \cap C' \]

Hence,

\[ A \cap B' \cap C' = A - C \]

Which is:

\[ \begin{aligned} \{&(2,4),(2,5),(2,6),\\ &(4,2),(4,3),(4,4),(4,5),(4,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \]

• Highly important for JEE: simplifying expressions like \(A \cap B' \cap C'\)
• Board exams test complement and difference frequently
• Recognizing partitions (even/odd) saves time
• Reduces heavy listing using logical simplification
• Foundation for probability calculation using set algebra

• Mutually Exclusive: \(A \cap B = \varnothing\)
• Exhaustive: \(A \cup B = S\)
• Complement: \(A' = S - A\)
• Partition: Events that are mutually exclusive and exhaustive

• Step 1: Recall from Q6 that \(A\) = even first die, \(B\) = odd first die
• Step 2: Use logic: even vs odd partitions the sample space
• Step 3: Check intersection (for exclusivity)
• Step 4: Check union (for exhaustiveness)
• Step 5: Use complement relations \(A'=B,\; B'=A\)

(i) \(A\) and \(B\) are mutually exclusive

A: first die even, B: first die odd. No number can be both even and odd.

\[ A \cap B = \varnothing \]

Answer: True

(ii) \(A\) and \(B\) are mutually exclusive and exhaustive

Every outcome has first die either even or odd.

\[ A \cup B = S \]

Also from (i), intersection is empty.

Answer: True

(iii) \(A = B'\)

Complement of B (odd first die) gives even first die.

\[ B' = A \]

Answer: True

(iv) \(A\) and \(C\) are mutually exclusive

Check common outcomes:

\(C =\) sum ≤ 5 includes \((2,1),(2,2),(2,3)\), etc.

These belong to A (first die even).

\[ A \cap C \neq \varnothing \]

Answer: False

(v) \(A\) and \(B'\) are mutually exclusive

Since \(B' = A\),

\[ A \cap B' = A \cap A = A \neq \varnothing \]

Identical events are never mutually exclusive.

Answer: False

(vi) \(A', B', C\) are mutually exclusive and exhaustive

First simplify:

\[ A' = B,\quad B' = A \]

So the statement becomes: \(B, A, C\)

Now check intersection:

\[ A \cap C \neq \varnothing \]

Hence not mutually exclusive.

Also union does not cover full sample space.

Answer: False

• Very high probability Board question: True/False with reasoning
• JEE Main frequently tests complement-based simplification
• Identifying partitions (even/odd) is a powerful shortcut
• Helps avoid confusion between identical and mutually exclusive events
• Strong foundation for conditional probability and Bayes theorem