CBSE Class 11 Maths Chapter 14 Probability – Solutions 2025–26

Exercise 14.1

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?

Ans: 1,2,3,4,5 and 6 are the possible outcomes when the die is thrown.

Therefore, S = {1,2,3,4,5,6} 

According to the question,

E is event die shows 4

\[E = \left( 4 \right)\]

F is die shows even number

\[F = \left( {2,4,6} \right)\]

\[E \cap F = \left( 4 \right) \cap \left( {2,4,6} \right)\]

\[E \cap F = 4\]

Therefore, we can conclude that E and F are not mutually exclusive event.

2. A die is thrown. Describe the following events:

(i) A: a number less than 7.

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

A: a number less than 7

All the numbers in the die are less than 7,

\[A = \left( {1,2,3,4,5,6} \right)\]

(ii) B: a number greater than 7.

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

B: a number greater than 7

There is no number on dice greater than 7.

\[B = \phi \]

(iii) C: a multiple of 3 

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

C : a number multiple of 3

Multiple of 3 between 1 to 6 is 3 and 6

Therefore,

\[C = \left( {3,6} \right)\]

(iv) D: a number less than 4.

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

D: a number less than 4

\[D = \left( {1,2,3} \right)\]

(v) E: an even number greater than 4.

Ans: \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

E: an even number greater than 4

There are only one number which is even greater than 4.

Therefore,

\[E = \left( 6 \right)\]

(vi) F: a number not less than 3.

Ans: Let us 1,2,3,4,5 and 6 are the possible outcomes when the die is thrown.

Therefore, \[S = \left( {1,2,3,4,5,6} \right)\]

According to the question,

F: a number not less than 3

\[F = \left( {3,4,5,6} \right)\]

Also find \[A \cup B,{\text{ }}A\; \cap B,{\text{ }}B \cup C,{\text{ }}E\; \cap {\text{ }}F,{\text{ }}D\; \cap {\text{ }}E,{\text{ }}D--E,{\text{ }}A--C,{\text{ }}E\; \cap {\text{ }}F',{\text{ }}F'\]

Ans: As we have to find, \[{\text{ }}A \cup B,{\text{ }}A\; \cap B,{\text{ }}B \cup C,{\text{ }}E\; \cap {\text{ }}F,{\text{ }}D\; \cap {\text{ }}E,{\text{ }}D--E,{\text{ }}A--C,{\text{ }}E\; \cap {\text{ }}F',{\text{ }}F'\]

Therefore,

\[A\; \cap B{\text{ }} = \;\left( {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right)\; \cap \;\left( \phi  \right)\]

\[A\; \cap B = {\text{ }}\left( \phi  \right)\]

\[B \cup C{\text{ }} = \;\left( \phi  \right) \cup \;\left( {3,{\text{ }}6} \right)\]

\[B \cup C = \left( {3,6} \right)\]

\[E\; \cap F = \;\left( 6 \right)\; \cap \;\left( {3,4,5,6} \right)\]

\[E\; \cap F = \left( 6 \right)\]

\[D\; \cap {\text{ }}E{\text{ }} = \;\left( {1,2,3} \right)\; \cap \;\left( 6 \right)\]

\[D\; \cap {\text{ }}E = \left( \phi  \right)\]

\[D--E{\text{ }} = \;\left( {1,2,3} \right)--\left( 6 \right)\]

\[D--E = \;\left( {1,2,3} \right)\]

\[A--C = \left( {1,2,3,4,5,6} \right)--\;\;\left( {3,6} \right)\]

\[A--C = \left( {1,2,4,5} \right)\]

\[F' = S--F\]

\[F' = \left( {1,2,3,4,5,6} \right)--\left( {3,4,5,6} \right)\]

\[F' = \left( {1,2} \right)\]

\[E\; \cap F' = \;\left( 6 \right)\; \cap \;\left( {1,2} \right)\]

\[E\; \cap F' = \;\left( \phi  \right)\]

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?

Ans: Let \[1,2,3,4,5\] and 6 are the possible outcomes when the die is thrown.

As in this question pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: The sum is greater than 8

$A=   (3,6) ,( 4,5),(4,6),( 5,4),(5,5) \\$ 

       $(5,6) ,( 6,3),(6,4),(6,5),(6,6) \\$ 

B: 2 occur on the die

$B=   (2,1) ,( 2,2),(2,3),( 2,4),(2,5) \\$ 

       $(1,2) ,( 3,2),(4,2),(5,2),(6,2),(2,6) \\$ 

C: the sum is at least 7 and multiple of 3

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

Now, we can find pairs of the events which are mutually exclusive or not.

(i) \[A \cap \;B = \phi \]

As, there is no common element in A and B

Therefore, A and B are mutually exclusive

(ii) \[B\; \cap \;C = \;\phi \]

Since there is no common element between

Therefore, B and C are mutually exclusive.

(iii) \[A\; \cap \;C =  \left\{ {\left( {3,6} \right),{\text{ }}\left( {4,5} \right),{\text{ }}\left( {5,4} \right),{\text{ }}\left( {6,3} \right),{\text{ }}\left( {6,6} \right)}   \right\}\]

\[A\; \cap \;C\; \ne \phi \]

As, A and C have common elements.

Therefore A and C are not mutually exclusive.

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?

Ans: As three coins are tossed once so the possible sample space will be,

\[S{\text{ }} =  \left\{ HHH,HHT,HTH,THH, HTT,THT,TTH,TTH   \right\}\]

According to question,

A: ‘three heads’

\[A = \left( {HHH} \right)\]

B: two heads and one tail

\[B = \left( {HHT,THH,HTH} \right)\]

C :three tails

\[C = \left( {TTT} \right)\]

D: a head shows on the first coin

\[D = \left( {HHH,HHT,HTH,HTT} \right)\]

(i) Mutually Exclusive

\[A\; \cap \;B = \left( {HHH} \right)\; \cap \;\left( {HHT,THH,HTH} \right)\]

\[A\; \cap \;B = \phi \]

Therefore, A and B are mutually exclusive.

\[A \cap \;C = \left( {HHH} \right)\; \cap \;\left( {TTT} \right)\]

\[A \cap \;C = \phi \]

Therefore, A and C are mutually exclusive.

\[A \cap D{\text{ }} = \left( {HHH} \right)\; \cap \;\left( {HHH,HHT,HTH,HTT} \right)\]

\[A \cap D = \left( {HHH} \right)\]

\[A \cap D{\text{ }} \ne \;\phi \]

Therefore, A and D are not mutually exclusive

\[B \cap C = \left( {HHT,HTH,THH} \right) \cap \left( {TTT} \right)\]

\[B \cap C = \phi \]

Therefore, B and C are mutually exclusive.

\[B \cap D = \left( {HHT,THH,HTH} \right) \cap \left( {HHH,HHT,HTH,\;HTT} \right)\]

\[B \cap D = \left( {HHT,HTH} \right)\]

\[B \cap D{\text{ }} \ne \;\phi \]

Therefore, B and D are not mutually exclusive.

\[C \cap D = \left( {TTT} \right)\; \cap \;\left( {HHH,HHT,HTH,\;HTT} \right)\]

\[C \cap D = \;\phi \]

Therefore, C and D are mutually exclusive.

(ii) Simple Event

If an event has only one outcome then it is called a simple event.

\[A = \left( {HHH} \right)\]

\[C{\text{ }} = {\text{ }}\left( {TTT} \right)\]

Both A and C have only one element,

Therefore, they are simple events.

(iii) Compound Events

If an event has more than one outcome then it is called a Compound event

\[B = \left( {HHT,HTH,THH} \right)\]

\[D = \left( {HHH,HHT,HTH,\;HTT} \right)\]

Both B and D have more than one element,

Therefore, they are compound events.

5. Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

As three coins are tossed once so the possible sample space contains,

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

Two events which are mutually exclusive.

Let A be the event of getting only head

\[A =  \left\{ HHH  \right\}\]

Let B be the event of getting only tail

\[B =  \left\{ TTT  \right\}\]

Therefor,

\[A \cap B = \;\phi \]

So, A and B are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive.

As three coins are tossed once so the possible sample space contains,

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

Three events which are mutually exclusive and exhaustive

Now,

Let P be the event of getting exactly two tails

\[P =  \left\{ HTT,TTH,THT  \right\}\]

Let Q be the event of getting at least two heads

\[Q =  \left\{ HHT,HTH,THH,HHH  \right\}\]

Let R be the event of getting three tail

\[C =  \left\{ TTT  \right\}\]

\[P \cap Q \cap R =  \left\{ HTT,TTH,THT  \right\} \cap  \left\{ HHT,HTH,THH,HHH  \right\} \cap  \left\{ TTT  \right\}\]

\[P \cap Q \cap R = \;\phi \]

Therefore, P, Q and R are mutually exclusive.

And also,

\[P \cup Q \cup R =  \left\{ HTT,TTH,THT  \right\} \cup  \left\{ HHT,HTH,THH,HHH  \right\} \cup  \left\{ TTT  \right\}\]

\[P \cup Q \cup R =  \left\{ HHH,HHT,HTH,\;HTT,THH,THT,TTH,TTT  \right\}\]

\[P \cup Q \cup R = S\]

Hence P, Q and R are exhaustive events.

(iii) Two events, which are not mutually exclusive.

As three coins are tossed once so the possible sample space contains,

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

Two events, which are not mutually exclusive

Let A be the event of getting at least two heads

\[A =  \left\{ HHH,HHT,THH,HTH  \right\}\]

LetB be the event of getting only head

\[B =  \left\{ HHH  \right\}\]

\[A \cap B =  \left\{ HHH,HHT,THH,HTH  \right\} \cap  \left\{ HHH  \right\}\]

\[A \cap B =  \left\{ HHH  \right\}\]

\[A\; \cap \;B{\text{ }} \ne \;\phi \]

Therefore, A and B are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

As three coins are tossed once so the possible sample space contains,

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

Two events which are mutually exclusive but not exhaustive

Let P be the event of getting only Head

\[P =  \left\{ HHH  \right\}\]

LetQ be the event of getting only tail

\[Q =  \left\{ TTT  \right\}\]

\[P \cap Q =  \left\{ HHH  \right\} \cap  \left\{ TTT  \right\}\]

\[P \cap Q = \phi \]

Therefore, P and Q are mutually exclusive events.

But,

\[P \cup Q =  \left\{ HHH  \right\} \cup  \left\{ TTT  \right\}\]

\[P \cup Q = \; \left\{ {HHH,TTT}   \right\}\]

\[P \cup Q \ne S\]

Therefore,P and Q are not exhaustive events.

(v) Three events which are mutually exclusive but not exhaustive.

As three coins are tossed once so the possible sample space contains,

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

Three events which are mutually exclusive but not exhaustive

Let X be the event of getting only head

\[X =  \left\{ HHH  \right\}\]

Let Y be the event of getting only tail

\[Y =  \left\{ TTT  \right\}\]

Let Z be the event of getting exactly two heads

\[Z =  \left\{ HHT,THH,HTH  \right\}\]

Now,

\[X \cap Y \cap Z =  \left\{ HHH  \right\} \cap  \left\{ TTT  \right\} \cap  \left\{ HHT,THH,HTH  \right\}\]

\[X \cap Y \cap Z = \phi \]

Therefore, they are mutually exclusive

Also

\[X \cup Y \cup Z =  \left\{ HHH  \right\} \cup  \left\{ TTT  \right\} \cup  \left\{ HHT,THH,HTH  \right\}\]

\[X \cap Y \cap Z =  \left\{ HHH,TTT,HHT,THH,HTH\,  \right\}\]

\[X \cap Y \cap Z \ne S\]

So, X, Y and Z are not exhaustive.

Therefore, X, Y and X are mutually exclusive but not exhaustive.

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 ≤ 5.

Describe the events: 

Ans: Let 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question is given that pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: Getting an even number on first die

$A =   (2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

       $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

B: Getting odd number on first die

$B =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$

C: Getting the sum of the number on dice \[ \leqslant 5\]

$C =   (1,1) ,( 1,2),(1,3),( 1,4),(2,1)\\$ 

      $ (2,2),(2,3),(3,1),(3,2),(4,1) \\$

(i) \[A'\]

Ans: $A = (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

(ii) not B

Ans: $B=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

       $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

(iii) A or B

Ans: \[A{\text{ or }}B\left( {A \cup B} \right)\] =$ (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) \\$

(iv) A and B

Ans: \[A{\text{ and }}B\left( {A \cap B} \right) = \phi \]

(v) A but not C 

Ans: \[A{\text{ but not }}C\left( {A - C} \right)\] =$    (2,4),(2,5),(2,6),(4,2),(4,3),(4,2),(4,5) \\$

 $ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

(vi) B or C

Ans: \[B{\text{ or }}C\left( {B \cup C} \right)\] =$ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\$

 $(2,1),(2,2),(2,3),(3,1),(3,2),(3,3) \\$

 $(3,4),(3,5),(3,6),(4,1),(5,1),(5,2) \\$

 $(5,3),(5,4),(5,5),(5,6)\\$

(vii) B and C 

Ans: \[B{\text{ and }}C\left( {B \cap C} \right) =  \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {3,1} \right),\left( {3,2} \right)}   \right\}\]

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

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

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

\[A \cap C' =  \left\{  \left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right)\\   \left( {4,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)     \right\}\]

7. Refer to question 6 above, state true or false: (give reason for your answer)

Ans: Let 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question is given that pair of die is thrown, so sample space will be,

$S =   (1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (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) \\$

A: Getting an even number on first die

$A=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

B: Getting odd number on first die

$B=(1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

C: Getting the sum of the number on dice \[ \leqslant 5\]

$C =   (1,1) ,( 1,2),(1,3),( 1,4),(2,1)\\$ 

      $ (2,2),(2,3),(3,1),(3,2),(4,1) \\$

(i) A and B are mutually exclusive

Ans: \[\left( {A \cap B} \right) = \phi \]

Therefore, A & B are mutually exclusive.

(ii) A and B are mutually exclusive and exhaustive

Ans: \[\left( {A \cap B} \right) = \phi \]

Therefore, A & B are mutually exclusive.

\[A \cup B \]= $(1,1)(1,2)(1,3),(1,4),(1,5),(1,6)\\$

    $ (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) \\$

\[A \cup B = S\]

Therefore, A and B are exhaustive also.

(iii) \[A = B'\]

Ans: $A = (2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

$B'=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

Therefore, \[A = B'\]

(iv) A and C are mutually exclusive

Ans: As we have,

\[A \cap C =  \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {4,1} \right)}   \right\}\]

\[A \cap C \ne \phi \]

Therefore, A and C are not mutually exclusive

(v) A and \[B'\] are mutually exclusive.

Ans: As we have,

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

\[\therefore A \cap B' \ne \phi \]

Therefore, A and \[B'\] are not mutually exclusive.

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

Ans: $A'=(1,1) ,( 1,2),(1,3),( 1,4),(1,5),(1,6) \\$ 

      $ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\$

       $ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\$ 

$B'=(2,1) ,( 2,2),(2,3),( 2,4),(2,5),(2,6) \\$ 

      $ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\$

      $ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\$

$C=(1,1),(1,2),(1,3),(1,4),(2,1)\\$

$(2,2),(2,3),(3,1),(3,2),(4,1)\\$

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

But, \[B' \cap C =  \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right)}   \right\}\]

\[B' \cap C \ne \phi \]

\[A'{\text{ and }}B'\] are mutually exclusive.

\[B'{\text{ and }}C\] are not mutually exclusive.

Therefore, \[A',B',C\] are not mutually exclusive.

\[A' \cup B' \cup C\] =  $(1,1)(1,2)(1,3),(1,4),(1,5),(1,6)\\$

    $ (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) \\$

\[A' \cup B' \cup C = S\]

Therefore, \[A',B',C\] are exhaustive.

NCERT Solutions for Class 11 Maths Chapter 14 Probability Exercise 14.1

Opting for the NCERT solutions for Ex 14.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 14 Exercise 14.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

NCERT Solution Class 11 Maths of Chapter 14 All Exercises

Important Study Material Links for Chapter 14: Probability

• Probability Important Questions

• Probability Important Formulas

• Probability Revision Notes

• Probability NCERT Exemplar Solutions

• Probability RD Sharma Solutions

• Probability RS Aggarwal Solutions

NCERT Class 11 Maths Solutions Chapter-wise Links - Download the FREE PDF

Important Related Links for CBSE Class 11 Maths