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NCERT Solutions for Exercise 13.1 Class 12 Maths Chapter 13 - Probability

NCERT Solutions for Exercise 13.1 Class 12 Maths Chapter 13 - Probability

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:15 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1

NCERT Solutions for Exercise 13.1 Class 12 Maths Chapter 13 Probability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 13.1 Class 12 Maths chapter 13 is very important for the students who wish to perform well in the 12th board exam. NCERT book exercise has covered topics like the probability of events, conditional probability, properties of conditional property, etc. Probability has a lot of real-life applications like forecasting the weather, sports, online gaming, insurance, politics lottery tickets, playing cards, statistics, etc. There area total of 17 questions covered in the exercise 13.1 Class 12 Maths which are mostly related to conditional probability. You are advised to solve all these NCERT syllabus problems on your own. You can go through NCERT solutions for Class 12 Maths chapter 13 exercise 13.1 If you are facing difficulties while solving these problems.

12th class Maths exercise 13.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Probability Class 12 Chapter 13-Exercise: 13.1

Question:1 Given that E and F are events such that P(E)=0.6,P(F)=0.3 and p(E\cap F)=0.2, find P(E\mid F) and P(F\mid E)

Answer:

It is given that P(E)=0.6,P(F)=0.3 and p(E\cap F)=0.2,

P ( E | F ) = \frac{p(E\cap F)}{P(F)}= \frac{0.2}{0.3}= \frac{2}{3}

P ( F | E ) = \frac{p(E\cap F)}{P(E)}= \frac{0.2}{0.6}= \frac{1}{3}

Question:2 Compute P(A\mid B), if P(B)=0.5 and P(A\cap B)=0.32

Answer:

It is given that P(B)=0.5 and P(A\cap B)=0.32

P ( A | B ) = \frac{p(A\cap B)}{P(B)}= \frac{0.32}{0.5}=0.64

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(i) P(A\cap B)

Answer:

It is given that P(A)=0.8, P(B)=0.5 and P(B|A)=0.4

P ( B | A ) = \frac{p(A\cap B)}{P(A)}

0.4 = \frac{p(A\cap B)} {0.8}

p(A\cap B) = 0.4 \times 0.8

p(A\cap B) = 0.32

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(ii) P(A\mid B)

Answer:

It is given that P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4,

P(A\cap B)=0.32

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

P ( A | B ) = \frac{0.32}{0.5}

P ( A | B ) = \frac{32}{50}=0.64

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(iii) P(A\cup B)

Answer:

It is given that P(A)=0.8,P(B)=0.5

P(A\cap B)=0.32

P(A\cup B)=P(A)+P(B)-P(A\cap B)

P(A\cup B)=0.8+0.5-0.32

P(A\cup B)=1.3-0.32

P(A\cup B)=0.98

Question:4 Evaluate P(A\cup B), if 2P(A)=P(B)=\frac{5}{13} and P(A\mid B)=\frac{2}{5}

Answer:

Given in the question 2P(A)=P(B)=\frac{5}{13} and P(A\mid B)=\frac{2}{5}

We know that:

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

\frac{2}{5} = \frac{p(A\cap B)}{\frac{5}{13}}

\frac{2\times 5}{5\times 13} = p(A\cap B)

p(A\cap B)=\frac{2}{ 13}

Use, p(A\cup B)=p(A)+p(B)-p(A\cap B)

p(A\cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}

p(A\cup B)=\frac{11}{26}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}. , find

(i) P(A\cap B)

Answer:

Given in the question

P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}.

By using formula:

p(A\cup B)=p(A)+p(B)-p(A\cap B)

\frac{7}{11}=\frac{6}{11}+\frac{5}{11}-p(A\cap B)

p(A\cap B)=\frac{11}{11}-\frac{7}{11}

p(A\cap B)=\frac{4}{11}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}, find

(ii) P(A\mid B)

Answer:

It is given that - P(A)=\frac{6}{11},P(B)=\frac{5}{11}

p(A\cap B)=\frac{4}{11}

We know that:

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

P ( A | B ) = \frac{\frac{4}{11}}{\frac{5}{11}}

P ( A | B ) = \frac{4}{5}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}, find

(iii) P(B\mid A)

Answer:

Given in the question-

P(A)=\frac{6}{11},P(B)=\frac{5}{11} and p(A\cap B)=\frac{4}{11}

Use formula

P ( B | A ) = \frac{p(A\cap B)}{P(A)}

P ( B | A ) = \frac{\frac{4}{11}}{\frac{6}{11}}

P ( B | A ) = \frac{4}{6}=\frac{2}{3}

Question:6 A coin is tossed three times, where

(i)E : head on third toss ,F : heads on first two tosses

Answer:

The sample space S when a coin is tossed three times is

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

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E: head on third toss, F: heads on first two tosses

E=\left \{ {HHH},{TTH},{HTH},{THH} \right \}

F=\left \{ {HHH},{HHT} \right \}

E\cap F =HHH

P(F)=\frac{2}{8}=\frac{1}{4}

P(E\cap F)=\frac{1}{8}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{8}}{\frac{1}{4}}

P(E| F)=\frac{4}{8}=\frac{1}{2}

Question:6 A coin is tossed three times, where

(ii)E : at least two heads ,F : at most two heads

Answer:

The sample space S when a coin is tossed three times is

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

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E : at least two heads , F : at most two heads

E=\left \{ {HHH},{HTH},{THH},{HHT}\right \}=4

F=\left \{ {HTH},{HHT},{THH},{TTT},{HTT},{TTH},{THT} \right \}=7

E\cap F =\left \{ {HTH},THH,HHT\right \}=3

P(F)=\frac{7}{8}

P(E\cap F)=\frac{3}{8}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{3}{8}}{\frac{7}{8}}

P(E| F)=\frac{3}{7}

Question:6 A coin is tossed three times, where

(iii)E : at most two tails ,F : at least one tail

Answer:

The sample space S when a coin is tossed three times is

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

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E: at most two tails, F: at least one tail

E=\left \{ {HHH},{TTH},{HTH},{THH},THT,HTT,HHT \right \}=7

F=\left \{ {TTT},{TTH},{HTH},{THH},THT,HTT,HHT \right \}=7

E\cap F=\left \{ {TTH},{HTH},{THH},THT,HTT,HHT \right \}=6

P(F)=\frac{7}{8}

P(E\cap F)=\frac{6}{8}=\frac{3}{4}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{3}{4}}{\frac{7}{8}}

P(E| F)=\frac{6}{7}

Question:7 Two coins are tossed once, where

(i) E : tail appears on one coin, F : one coin shows head

Answer:

E : tail appears on one coin, F : one coin shows head

Total outcomes =4

E=\left \{ HT,TH \right \}=2

F=\left \{ HT,TH \right \}=2

E\cap F=\left \{ HT,TH \right \}=2

P(F)=\frac{2}{4}=\frac{1}{2}

P(E\cap F)=\frac{2}{4}=\frac{1}{2}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{2}}{\frac{1}{2}}

P(E| F)=1

Question:7 Two coins are tossed once, where

(ii)E : no tail appears,F : no head appears

Answer:

E : no tail appears, F : no head appears

Total outcomes =4

\\E={HH}\\F={TT}

E\cap F=\phi

n(E\cap F)=0

P(F)=1

P(E\cap F)=\frac{0}{4}=0

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{0}{1}=0

Question:8 A die is thrown three times,

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Answer:

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Total outcomes =6^{3}=216

E=\left \{ 114,124,134,144,154,164,214,224,234,244,254,264,314,324,334,344,354,364,414,424,434,454,464,514,524,534,544,554,564,614,624,634,644,654,664 \right \} n(E)=36

F=\left \{ 651,652,653,654,655,656 \right \}

n(F)=6

E\cap F=\left \{ 654 \right \}

n(E\cap F)=1

P(E\cap F)=\frac{1}{216}

P( F)=\frac{6}{216}=\frac{1}{36}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{216}}{\frac{1}{36}}

P(E| F)=\frac{1}{6}

Question:9 Mother, father and son line up at random for a family picture

E : son on one end, F : father in middle

Answer:

E : son on one end, F : father in middle

Total outcomes =3!=3\times 2=6

Let S be son, M be mother and F be father.

Then we have,

E= \left \{ SMF,SFM,FMS,MFS \right \}

n(E)=4

F=\left \{ SFM,MFS \right \}

n(F)=2

E\cap F=\left \{ SFM,MFS \right \}

n(E\cap F)=2

P(F)=\frac{2}{6}=\frac{1}{3}

P(E\cap F)=\frac{2}{6}=\frac{1}{3}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{3}}{\frac{1}{3}}

P(E| F)=1

Question:10 A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5.

Answer:

A black and a red dice are rolled.

Total outcomes =6^{2}=36

Let the A be event obtaining a sum greater than 9 and B be a event that the black die resulted in a 5.

A=\left \{ 46,55,56,64,65,66 \right \}

n(A)=6

B=\left \{ 51,52,53,54,55,56 \right \}

n(B)=6

A\cap B=\left \{ 55,56 \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P( B)=\frac{6}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}

Question:10 A black and a red dice are rolled.

(b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4 .

Answer:

A black and a red dice are rolled.

Total outcomes =6^{2}=36

Let the A be event obtaining a sum 8 and B be a event thatthat the red die resulted in a number less than 4 .

A=\left \{ 26,35,53,44,62, \right \}

n(A)=5

Red dice is rolled after black dice.

B=\left \{ 11,12,13,21,22,23,31,32,33,41,42,43,51,52,53,61,62,63 \right \}

n(B)=18

A\cap B=\left \{ 53,62 \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P( B)=\frac{18}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(i) P(E\mid F) and P(F\mid E)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \}

E\cap F=\left \{ 3\right \}

n(E\cap F)=1

n( F)=2

n( E)=3

P( E)=\frac{3}{6} P( F)=\frac{2}{6} and P(E\cap F)=\frac{1}{6}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{6}}{\frac{2}{6}}

P(E| F)=\frac{1}{2}

P(F| E)=\frac{P(F\cap E)}{P(E)}

P(F| E)=\frac{\frac{1}{6}}{\frac{3}{6}}

P(F| E)=\frac{1}{3}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F=\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(ii) P(E\mid G) and P(G\mid E)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \} , G=\left \{ 2,3,4,5 \right \}

E\cap G=\left \{ 3,5\right \}

n(E\cap G)=2

n( G)=4

n( E)=3

P( E)=\frac{3}{6} P( G)=\frac{4}{6} P(E\cap F)=\frac{2}{6}

P(E| G)=\frac{P(E\cap G)}{P(G)}

P(E| G)=\frac{\frac{2}{6}}{\frac{4}{6}}

P(E| G)=\frac{2}{4}=\frac{1}{2}

P(G| E)=\frac{P(G\cap E)}{P(E)}

P(G| E)=\frac{\frac{2}{6}}{\frac{3}{6}}

P(G| E)=\frac{2}{3}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F=\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(iii) P((E\cup F)\mid G) and P((E\cap F)\mid G)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \}

E\cap G=\left \{ 3,5 \right \} , F\cap G=\left \{ 2,3\right \}

(E\cap G)\cap G =\left \{ 3 \right \}

P[(E\cap G)\cap G] =\frac{1}{6} P(E\cap G) =\frac{2}{6} P(F\cap G) =\frac{2}{6}

P((E\cup F)|G) = P(E|G)+P(F|G) - P[(E\cap F)|G]

=\frac{P(E\cap G)}{P(G)}+\frac{P(F\cap G)}{P(G)}-\frac{P((E\cap F)\cap G)}{P(G)}

=\frac{\frac{2}{6}}{\frac{4}{6}}+\frac{\frac{2}{6}}{\frac{4}{6}}-\frac{\frac{1}{6}}{\frac{4}{6}}

=\frac{2}{4}+\frac{2}{4}-\frac{1}{4}

=\frac{3}{4}

P((E\cap F)|G)=\frac{P((E\cap F)\cap G)}{P(G)}

P((E\cap F)|G)=\frac{\frac{1}{6}}{\frac{4}{6}}

P((E\cap F)|G)=\frac{1}{4}

Question:12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(i) the youngest is a girl,

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1\, \, \, and \, \, \,G2 respectively also first and second boy are denoted by B1\, \, \, and \, \, \,B2

If a family has two children, then total outcomes =2^{2}=4 =\left \{ (B1B2),(G1G2),(G1B2),(G2B1)\right \}

Let A= both are girls =\left \{(G1G2)\right \}

and B= the youngest is a girl = =\left \{(G1G2),(B1G2)\right \}

A\cap B=\left \{(G1G2)\right \}

P(A\cap B)=\frac{1}{4} P( B)=\frac{2}{4}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{1}{4}}{\frac{2}{4}}

P(A| B)=\frac{1}{2}

Therefore, the required probability is 1/2

Question:12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(ii) at least one is a girl?

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1\, \, \, and \, \, \,G2 respectively also first and second boy are denoted by B1\, \, \, and \, \, \,B2

If a family has two children, then total outcomes =2^{2}=4 =\left \{ (B1B2),(G1G2),(G1B2),(G2B1)\right \}

Let A= both are girls =\left \{(G1G2)\right \}

and C= at least one is a girl = =\left \{(G1G2),(B1G2),(G1B2)\right \}

A\cap B=\left \{(G1G2)\right \}

P(A\cap B)=\frac{1}{4} P( C)=\frac{3}{4}

P(A| C)=\frac{P(A\cap C)}{P(C)}

P(A| C)=\frac{\frac{1}{4}}{\frac{3}{4}}

P(A| C)=\frac{1}{3}

Question:13 An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer:

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.

Total number of questions =300+200+500+400=1400

Let A = question be easy.

n(A)= 300+500=800

P(A)=\frac{800}{1400}=\frac{8}{14}

Let B = multiple choice question

n(B)=500+400=900

P(B)=\frac{900}{1400}=\frac{9}{14}

A\cap B = easy multiple questions

n(A\cap B) =500

P(A\cap B) =\frac{500}{1400}=\frac{5}{14}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{5}{14}}{\frac{9}{14}}

P(A| B)=\frac{5}{9}

Therefore, the required probability is 5/9

Question:14 Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer:

Two dice are thrown.

Total outcomes =6^2=36

Let A be the event ‘the sum of numbers on the dice is 4.

A=\left \{ (13),\left ( 22 \right ),(31) \right \}

Let B be the event that two numbers appearing on throwing two dice are different.

B=\left \{ (12),(13),(14),(15),(16),(21)\left ( 23 \right ),(24),(25),(26,)(31),(32),(34),(35),(36),(41),(42),(43),(45),(46),(51),(52),(53),(54),(56) ,(61),(62),(63),(64),(65)\right \} n(B)=30

P(B)=\frac{30}{36}

A\cap B=\left \{ (13),(31) \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{30}{36}}

P(A| B)=\frac{2}{30}=\frac{1}{15}

Therefore, the required probability is 1/15

Question:15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin.

Total outcomes

=\left \{ (1H),(1T),(2H),(2T),(31),(32),(33),(34),(35),(36),(4H),(4T),(5H),(5T),(61),(62),(63),(64),(65),(66)\right \}

Total number of outcomes =20

Let A be a event when coin shows a tail.

A=\left \{ ((1T),(2T),(4T),(5T)\right \}

Let B be a event that ‘at least one die shows a 3’.

B=\left \{ (31),(32),(33),(34),(35),(36),(63)\right \}

n(B)=7

P(B)=\frac{7}{20}

A\cap B= \phi

n(A\cap B)= 0

P(A\cap B)= \frac{0}{20}=0

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{0}{\frac{7}{20}}

P(A| B)=0

Question:16 In the following Exercise 16 choose the correct answer:

If P(A)=\frac{1}{2},P(B)=0, then P(A\mid B) is

(A) 0

(B) \frac{1}{2}

(C) not\; defined

(D) 1

Answer:

It is given that

P(A)=\frac{1}{2},P(B)=0,

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{P(A\cap B)}{0}

Hence, P(A\mid B) is not defined .

Thus, correct option is C.

Question:17 In the following Exercise 17 choose the correct answer:

If A and B are events such that P(A\mid B)=P(B\mid A), then

(A) A\subset B but A\neq B

(B) A=B

(C) A\cap B=\psi

(D) P(A)=P(B)

Answer:

It is given that P(A\mid B)=P(B\mid A),

\Rightarrow \frac{P(A\cap B)}{P(B)} =\frac{P(A\cap B)}{P(A)}

\Rightarrow P(A)=P(B)

Hence, option D is correct.

More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1:-

Exercise 13.1 Class 12 Maths solutions are consist of questions related to conditional probability and different properties of conditional probability. There are 15 short answer type questions and 2 multiple choice types questions are given in this exercise. There are 7 solved examples given in the NCERT textbook before this exercise that is related to conditional probability. You can solve these examples before solving exercise problems, so it will be easy for you to solve exercise problems.

Also Read| Probability Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1:-

  • Class 12th Maths chapter 13 exercise 13.1 solutions are helpful for the students to get conceptual clarity about conditional probability and its applications.
  • These exercise 13.1 Class 12 Maths solutions are designed by the subject matter experts who know how best to write the answer in the board exams in order to get good marks.
  • These Class 12 Maths chapter 13 exercise 13.1 solutions could be be used for revision before the exam.
  • These are some examples given in the Class 12 NCERT book before exercise 13.1 which could be solved before solving exercise 13.1.
  • As most of the questions in the 12th board exams are directly asked from NCERT textbooks, you are advised to be thorough with the NCERT textbook.
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Key Features Of NCERT Solutions for Exercise 13.1 Class 12 Maths Chapter 13

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.1 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 13.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 13.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 13.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 13.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 13.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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Frequently Asked Questions (FAQs)

1. What is weightage of probability in CBSE Class 12 Maths board exam?

Probability has 10 marks weightage in the CBSE Class 12 Maths final board exam.

2. How many exercises are there in the NCERT Class 12 Maths chapter 13 Probability ?

There are 6 exercises including a miscellaneous exercise in the NCERT Class 12 Maths chapter 13 Probability.

3. How many questions are covered in Probability Exercise 13.1 ?

There are 17 questions in NCERT Class 12 Maths chapter 13 exercise 13.1.

4. Where can I get NCERT Exemplar solutions for Class 12 Maths probability?
5. What is Conditional probability?

The conditional probability of an event is the probability of that event given that the other event happened

6. What do you mean by Independent Events ?

Two events are called independent events if they exist such that the probability of occurrence of one event is not dependent on the occurrence of another event.

7. What do you mean by Null Event ?

If the probability of occurrence of an event is zero then such event is called a null event.

8. What do you mean by Certain Event ?

An event is said to be a certain event when all possible outcomes are favorable to the event.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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October 23, 2024 - 04:01 PM IST :
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