Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 14 – Probability (Exercise 14.1) as per the latest NCERT and CBSE syllabus (2025–2026). This exercise introduces students to the basic concepts of probability based on set theory, sample space, events, and outcomes. It helps in understanding empirical and theoretical probability, with clear, step-by-step explanations of each question to strengthen conceptual clarity. These detailed NCERT solutions are ideal for CBSE Class 11 exams and competitive exams such as JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes.
NCERT Question 1 : A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
Solution:
Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
Hence,
$$\text{Sample Space} = \lbrace 1,2,3,4,5,6 \rbrace$$
As per the conditions given in the question,
E be the event “die shows 4”
$$E = \lbrace 4 \rbrace$$
F be the event “die shows even number”
$$F = \lbrace 2,4,6 \rbrace$$
Now,
$$E \cap F = \lbrace 4 \rbrace \cap \lbrace 2,4,6 \rbrace = \lbrace 4 \rbrace$$
Since
$$\lbrace 4 \rbrace \neq \varnothing$$
there is a common element in E and F.
Therefore, E and F are not mutually exclusive events.
Understanding mutually exclusive events strengthens your foundation in Class 11 probability, useful for JEE, NDA, and CUET exams.
NCERT Question 2 : A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3
Also find $$(A \cup B), (A \cap B), (B \cup C), $$ $$(E \cap F), (D \cap E), (A – C), (D – E), (E \cap F’), (F’)$$
Solution:
Possible outcomes:
$\text{Sample Space} = \lbrace 1,2,3,4,5,6 \rbrace$
(i) $A = \lbrace 1,2,3,4,5,6 \rbrace$
(ii) $B = \varnothing$
(iii) $C = \lbrace 3,6 \rbrace$
(iv) $D = \lbrace 1,2,3 \rbrace$
(v) $E = \lbrace 6 \rbrace$
(vi) $F = \lbrace 3,4,5,6 \rbrace$
Now,
$\textstyle A \cap B = \varnothing$, $\textstyle B \cup C = \lbrace 3,6 \rbrace$, $\textstyle E \cap F = \lbrace 6 \rbrace$, $\textstyle D \cap E = \varnothing$, $\textstyle D – E = \lbrace 1,2,3 \rbrace$, $\textstyle A – C = \lbrace 1,2,4,5 \rbrace$, $\textstyle F’ = \lbrace 1,2 \rbrace$, $\textstyle E \cap F’ = \varnothing$
Hence, all relations are correctly derived.
Set operations such as intersection, union, and complement are the foundation of probability and sample space concepts in Class 11 Maths, useful for JEE, CUET, and NDA preparation.
NCERT Question 3 : An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
A: the sum is greater than 8
B: 2 occurs on either die
C: the sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Solution:
Sample space when two dice are thrown:
$$S = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$$ $$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$$ $$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$$ $$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$$ $$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$$ $$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
A: sum greater than 8
$$A = \lbrace (3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6) \rbrace$$
B: 2 occurs on either die
$$B = \lbrace (2,1),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2),(2,2) \rbrace$$
C: sum is at least 7 and a multiple of 3
$$C = \lbrace (3,6),(4,5),(5,4),(6,3),(6,6) \rbrace$$
Now,
$$\textstyle A \cap B = \varnothing \Rightarrow \text{A and B are mutually exclusive}$$
$$\textstyle B \cap C = \varnothing \Rightarrow \text{B and C are mutually exclusive}$$
$$\textstyle A \cap C = \lbrace (3,6),(4,5),(5,4),(6,3),(6,6) \rbrace \neq \varnothing $$ $$\Rightarrow \text{A and C are not mutually exclusive}$$
This example demonstrates mutually exclusive events and outcomes when rolling dice, an essential concept for probability and random experiments in Class 11 Maths, JEE, and NDA exams.
NCERT Question 4 : Three coins are tossed once. Let A denote the event “three heads show”, B denote the event “two heads and one tail show”, C denote the event “three tails show”, and D denote the event “a head shows on the first coin”. Which events are
(i) Mutually exclusive?
(ii) Simple?
(iii) Compound?
Solution:
Possible outcomes when three coins are tossed:
$$S = \lbrace HHH,HHT,HTH,THH,HTT,THT,TTH,TTT \rbrace$$
Now,
$A = \lbrace HHH \rbrace$
$B = \lbrace HHT,HTH,THH \rbrace$
$C = \lbrace TTT \rbrace$
$D = \lbrace HHH,HHT,HTH,HTT \rbrace$
(i) Mutually exclusive:
Mutually exclusive events are events that cannot happen at the same time, meaning the occurrence of one event prevents the other from occurring. These are also known as disjoint events.
$$\textstyle A \cap B = \varnothing \Rightarrow \text{A and B mutually exclusive}$$
$$\textstyle A \cap C = \varnothing \Rightarrow \text{A and C mutually exclusive}$$
$$\textstyle A \cap D = \lbrace HHH \rbrace \neq \varnothing \Rightarrow \text{A and D not mutually exclusive}$$
$$\textstyle B \cap C = \varnothing \Rightarrow \text{B and C mutually exclusive}$$
$$\textstyle B \cap D = \lbrace HHT,HTH \rbrace \neq \varnothing \Rightarrow \text{B and D not mutually exclusive}$$
$$\textstyle C \cap D = \varnothing \Rightarrow \text{C and D mutually exclusive}$$
(ii) Simple events:
it is an event that consists of only one sample point from the set of all possible outcomes
$$A = \lbrace HHH \rbrace, \quad C = \lbrace TTT \rbrace$$
(iii) Compound events:
It is an event is an event that has more than one outcome, or a combination of two or more simple events that occur together
$$B = \lbrace HHT,HTH,THH \rbrace, \quad D = \lbrace HHH,HHT,HTH,HTT \rbrace$$
This question helps identify simple, compound, and mutually exclusive events in probability — a key topic for Class 11, NDA, JEE, and CUET preparation.
NCERT Question.5 : Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
Solution:
When a coin is thrown, the possible outcomes are either Head (H) or Tail (T).
Now, according to the question, three coins are tossed once. So, the possible sample space is:
$$\text{Space} = \lbrace HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \rbrace$$
(i) Two events which are mutually exclusive
Let us consider A be the event of getting only heads:
$$A = \lbrace HHH \rbrace$$
And B be the event of getting only tails:
$$B = \lbrace TTT \rbrace$$
Now,
$$A \cap B = \varnothing$$
There is no common element in A and B, hence these two events are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive
Let us consider
$$P = \lbrace HTT, TTH, THT \rbrace$$
$$Q = \lbrace HHT, HTH, THH, HHH \rbrace$$
$$R = \lbrace TTT \rbrace$$
Now,
$$P \cap Q = \varnothing$$
$$Q \cap R = \varnothing$$
$$P \cap R = \varnothing$$
Hence, P, Q, and R are mutually exclusive events.
Also,
$$P \cup Q \cup R = \lbrace HTT, TTH, THT, HHT, HTH, THH, HHH, TTT \rbrace = \text{Space}$$
Therefore, P, Q, and R are mutually exclusive and exhaustive events.
(iii) Two events which are not mutually exclusive
Let A be the event of getting at least two heads:
$$A = \lbrace HHH, HHT, THH, HTH \rbrace$$
Let B be the event of getting only heads:
$$B = \lbrace HHH \rbrace$$
Now,
$$A \cap B = \lbrace HHH \rbrace \neq \varnothing$$
There is a common element in A and B, so these two are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive
Let P be the event of getting only heads:
$$P = \lbrace HHH \rbrace$$
Let Q be the event of getting only tails:
$$Q = \lbrace TTT \rbrace$$
Then,
$$P \cap Q = \varnothing$$
So, P and Q are mutually exclusive.
But,
$$P \cup Q = \lbrace HHH, TTT \rbrace \neq \text{Space}$$
Hence, these two events are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive
Let X be the event of getting only heads:
$$X = \lbrace HHH \rbrace$$
Let Y be the event of getting only tails:
$$Y = \lbrace TTT \rbrace$$
Let Z be the event of getting exactly two heads:
$$Z = \lbrace HHT, THH, HTH \rbrace$$
Now,
$$X \cap Y = \varnothing$$
$$X \cap Z = \varnothing$$
$$Y \cap Z = \varnothing$$
Therefore, X, Y, and Z are mutually exclusive.
But,
$$X \cup Y \cup Z = \lbrace HHH, TTT, HHT, THH, HTH \rbrace \neq \text{Space}$$
Hence, X, Y, and Z are mutually exclusive but not exhaustive.
This problem highlights how to classify mutually exclusive and exhaustive events in probability. Mastering these concepts strengthens your foundation for Class 11 Maths, JEE Main, NDA, and CUET preparation.
NCERT Question.6 : Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le 5$.
Describe the events
(i) $A’$ (ii) not $B$ (iii) $A$ or $B$ (iv) $A$ and $B$
(v) $A$ but not $C$ (vi) $B$ or $C$ (vii) $B$ and $C$
(viii) $A \cap B’ \cap C’$
Solution:
Sample space for two dice (first die, second die):
$$S = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$$ $$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$$ $$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$$ $$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$$ $$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$$ $$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
Events:
$$\textstyle A = \lbrace (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),$$ $$(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
$$\textstyle B = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),$$ $$(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \rbrace$$
Sum $\le 5$ gives (sums 2,3,4,5):
$$\textstyle C = \lbrace (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1) \rbrace$$
Now find the requested events:
(i) Complement of $A$ (i.e. $A’$): first die not even $\Rightarrow$ first die odd, so
$$\textstyle A’ = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),$$ $$(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \rbrace = B$$
(ii) Not $B$ (i.e. $B’$): complement of $B$ is first die even, so
$$\textstyle B’ = \lbrace (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),$$ $$(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace = A$$
(iii) $A$ or $B$ ($A \cup B$): either first die even or odd — all ordered pairs, so
$$\textstyle A \cup B = S$$
(iv) $A$ and $B$ ($A \cap B$): even and odd on first die cannot occur together, so
$$\textstyle A \cap B = \varnothing$$
(v) $A$ but not $C$ ($A – C = A \cap C’$): remove from $A$ those pairs whose sum $\le 5$. Elements of $A$ that are in $C$ are $(2,1),(2,2),(2,3),(4,1)$. Thus
$$\textstyle A – C = \lbrace (2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),$$ $$(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
(vi) $B$ or $C$ ($B \cup C$): all pairs with first die odd plus the pairs in $C$ (no duplicates):
$$\textstyle B \cup C = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),$$ $$(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(4,1) \rbrace$$
(Notice $(4,1)$ from $C$ is added; all other $C$ pairs with first die 1 or 3 are already in $B$.)
(vii) $B$ and $C$ ($B \cap C$): pairs common to $B$ and $C$ (i.e. in $C$ with first die odd):
$$\textstyle B \cap C = \lbrace (1,1),(1,2),(1,3),(1,4),(3,1),(3,2) \rbrace$$
(viii) $A \cap B’ \cap C’$: since $B’ = A$, this equals $A \cap A \cap C’ = A \cap C’$, i.e. same as (v):
$$\textstyle A \cap B’ \cap C’ = \lbrace (2,4),(2,5),(2,6),(4,2),(4,3),(4,4),$$ $$(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
These set operations on two-dice outcomes demonstrate complements, unions, intersections, and differences — essential practice for mastering probability, sample space reasoning, and event classification for CBSE Class 11 and competitive exams like JEE Main and NDA.
NCERT Question.7 : Refer to Question 6 above. State True or False (give reason):
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) (A = B’)
(iv) A and C are mutually exclusive
(v) A and (B’) are mutually exclusive
(vi) (A’, B’, C) are mutually exclusive and exhaustive
Solution (with reasons)
Recall (from Q6):
$$S = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$$ $$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$$ $$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$$ $$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$$ $$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$$ $$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \rbrace$$
$$\textstyle A = \lbrace (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),$$ $$(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\rbrace$$
$$\textstyle B = \lbrace (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),$$ $$(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\rbrace$$
$$\textstyle C = \lbrace (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\rbrace$$
(i) A and B are mutually exclusive — True.
Reason: (A) contains ordered pairs whose first die is even, (B) contains ordered pairs whose first die is odd. No ordered pair can have the first die both even and odd, so
$$\textstyle A \cap B = \varnothing.$$
Therefore they are mutually exclusive.
(ii) A and B are mutually exclusive and exhaustive — True.
Reason: From (i) they are mutually exclusive. Also every ordered pair has either an even or an odd first die, so
$$\textstyle A \cup B = S.$$
Hence they are both mutually exclusive and exhaustive.
(iii) ($A = B’$) — True.
Reason: ($B’$) is the complement of (B) relative to (S); since (B) = “first die odd”, (B’) = “first die not odd” = “first die even”, which is exactly (A). So (A = B’).
(iv) $A$ and $C$ are mutually exclusive — False.
Reason: Some outcomes belong to both (A) and (C). For example ((2,1),(2,2),(2,3),(4,1)) are in (C) and have even first die, so they are in (A). Thus
$$\textstyle A \cap C = \lbrace (2,1),(2,2),(2,3),(4,1)\rbrace \neq \varnothing,$$
so (A) and (C) are not mutually exclusive.
(v) $A$ and $(B’)$ are mutually exclusive — False.
Reason: From (iii) ($B’ = A$). So ($A \cap B’ = A \cap A = A \neq \varnothing$). Two identical nonempty sets are not mutually exclusive.
(vi) ($A’, B’, C$) are mutually exclusive and exhaustive — False.
Reason (mutual exclusivity): ($A’ = B$) (since ($A’ =$) first die odd) and ($B’ = A$). We check intersections: ($A’ \cap B’ = B \cap A = \varnothing$) (they are disjoint), but ($B’ \cap C = A \cap C \neq \varnothing$) (see (iv)). So the three are not mutually exclusive.
Reason (exhaustive): ($A’ \cup B’ = B \cup A = S$), so ($A’ \cup B’ \cup C = S$). They are exhaustive as a triple only because ($A’ \cup B’ = S$) already covers everything; however mutual exclusivity fails. Therefore the combined claim (mutually exclusive and exhaustive) is false.
Understanding complements and intersections on ordered-pair sample spaces is essential for probability, sample-space reasoning, and event classification for Class 11 and exam prep like JEE and NDA.
