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# NCERT Solutions for Class 11 Maths Chapter 14: Probability - Exercise 14.1

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 11 Maths Chapter 14 Probability

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.1 (Ex 14.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 14 Probability Exercise 14.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 14 - Probability Exercise: Exercise - 14.1 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
Competitive Exams after 12th Science

## NCERT Solutions for Class 11 Maths Chapter 14 Probability 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,

$A = \left( {HHH} \right)$

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.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 14: Probability - Exercise 14.1

1. What is a sample space?

The set of all potential outcomes or results of an experiment or random trial is referred to as the sample space in probability theory. The potential ordered outcomes, or sample points, are listed as elements in a set that is used to represent a sample space.

2. What do you mean by a mutually exclusive event?

Events that do not take place at the same time are said to be mutually exclusive. For instance, if a coin is tossed, the outcome will either be head or tail; we cannot get both outcomes. Since they do not occur at the same time, such events are also known as disjoint events.

3. What do you mean by "exhaustive events"?

If at least one of two events, A and B, will undoubtedly happen, then the pair is said to be exhaustive events.

If you roll a die, for example, at least one of the following things will happen:

A=(" a number less than '5' appears")

B=("a number greater than "2" appears")

4. What are not mutually exclusive or non-exclusive events?

If two events A and B have at least one common outcome, then they are said to be mutually non-exclusive events. The fact that events A and B cannot stop one another from happening leads us to conclude that they share some characteristics.

5. Why should I refer to Vedantu’s NCERT Solutions for Class 11 Maths, Chapter 14 Probability (Ex. 14.1), Exercise 14.1?

If you have the correct study resources, getting all A’s in math shouldn't be too difficult. For a topic like mathematics, studying just the chapters is never sufficient; therefore, students need to practice the corresponding exercise questions. NCERT solutions have been developed by Vedantu's knowledgeable staff in accordance with CBSE standards. The solutions are correctly and thoroughly stated. Utilizing quick-response strategies will help you finish your exam on time, increase your scores, and become ready for competitive exams. The Vedantu website and mobile app both offer free downloads of all of these solutions.