NCERT Solutions Class 12 Maths Exercise 13.1

The National Council of Educational Research and Training (NCERT) is a government agency that was established to help India’s schools improve their quality of education. The Government of India created it to preserve the uniformity of the curriculum since different books were being recommended to students in the same standard. The responsibility for creating content for various NCERT textbooks belongs to this body.  The Central Board of Secondary Education (CBSE) and many other State boards follow the NCERT pattern. NCERT Solutions is preferred by CBSE.These questions are predominant in Board examinations, therefore, students appearing for the board examinations are advised to revise the NCERT Solutions thoroughly for better results.

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NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.1) Exercise 13.1

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Class 12 Mathematics Chapter 13 is titled Probability. Probability is a measure of the uncertainty of events in a random experiment. The chapter is reportedly interesting, and arriving at the right answer calls for logical reasoning. The NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1 requires thinking and reasoning.

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NCERT Solutions For Class 12 Maths Chapter 13 Probability Exercise 13.1

Human intellect and logic are fundamentally based on Mathematics, which is essential to understanding the world. Mathematics is a powerful tool for developing mental discipline and promoting logical thinking. Probability is a branch of Mathematics that enables one to foresee the outcomes of future events through calculations. The chapter on Probability is fascinating. One of the most important exercises is Class 12 Maths Chapter 13 Exercise 13.1. The NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1. deal with basic calculations for a better understanding of the concept. To make studying easier and more effective Extramarks offers notes on Exercise 13.1 Class 12th Maths, past years’ papers, and much more.

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Topics Covered in Class 12 NCERT Maths of Chapter 13 Exercise 13.1

Exercise 13.1 Class 12 deals with Probability equations. Many things are challenging to predict with complete certainty. Using it, predictions can be made about how probable an event is to happen, or its chance of happening. Solving the NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1 is advisable for students’ preparation as it covers topics that could appear in the examination related to NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1. Extramarks is a website that consists of NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1. It not only provides NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1 but also offers sample papers from past years’ papers to help students build confidence by giving them an idea of how their exams can look. Extramarks covers all the topics of every chapter of Class 12 Mathematics and describes every solution precisely. Exercise 13.1 Class 12, each question is explained very specifically by subject matter experts in Mathematics in their respective fields.

What is Conditional Probability?

The possibility of an event or outcome being contingent on a prior event or outcome is known as Conditional Probability. The probability of the prior event is multiplied by the current likelihood of the subsequent, or conditional, occurrence to determine the conditional probability. The term is present in  NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.1.

Chapter wise NCERT Solutions for Class 12 Maths

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Access Other Exercises of Class 12 Maths Chapter 13

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The NCERT Solutions Class 10, NCERT Solutions Class 11, and NCERT Solutions Class 12 are available on Extramarks. Students of Class 10, Class 11 and Class 12 cannot take risks with their examinations, hence, these solutions make them ready for their board examination.

Q.1

Given that E and F are events such that P( E )=0.6,P( F )=0.3 and P(EF)=0.2, find P( E|F ) and P( F|E ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=DeacaWFPbGaa8NDaiaa=vgacaWFUbGaa8hiaiaa=rhacaWFObGaa8xyaiaa=rhacaWFGaGaa8xraiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa8Nraiaa=bcacaWFHbGaa8NCaiaa=vgacaWFGaGaa8xzaiaa=zhacaWFLbGaa8NBaiaa=rhacaWFZbGaa8hiaiaa=nhacaWF1bGaa83yaiaa=HgacaWFGaGaa8hDaiaa=HgacaWFHbGaa8hDaiaa=bcacaWFqbWaaeWaaeaacaWFfbaacaGLOaGaayzkaaGaa8xpaiaa=bdacaWFUaGaa8Nnaiaa=XcacaWFqbWaaeWaaeaacaWFgbaacaGLOaGaayzkaaGaa8xpaiaa=bdacaWFUaGaa83maiaa=bcaaeaacaWFHbGaa8NBaiaa=rgacaWFGaGaa8huaiaa=HcacaWFfbGaeyykICSaa8Nraiaa=LcacaWF9aGaa8hmaiaa=5cacaWFYaGaa8hlaiaa=bcacaWFMbGaa8xAaiaa=5gacaWFKbGaa8hiaiaa=bfadaqadaqaaiaa=veacaWF8bGaa8NraaGaayjkaiaawMcaaiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa8NzamaabmaabaGaa8Nraiaa=XhacaWFfbaacaGLOaGaayzkaaGaa8Nlaaaaaa@87C4@

Ans

Since, P(E|F)=P(EF)P(F)    P(E|F)=0.20.3  =23And  P(F|E)=P(EF)P(E)  =0.20.6  =26=13

Q.2

Compute P(A|B), if P(B) = 0.5and P(A∩B) =0.32.

Ans

It is given that P(B) = 0.5 and P(AB) = 0.32 So, P(A|B)=P(AB)P(B)    P(A|B)=0.320.5  =1625

Q.3

If P(A)=0.8,P(B)=0.5and P(B|A)=0.4, find(i)P(AB)(ii)P(A|B)(iii)P(AB)

Ans

ItisgiventhatP(A)=0.8,P(B)=0.5,andP(B|A)=0.4(i)P(B|A)=0.4P(AB)P(A)=0.4P(AB)0.8=0.4P(AB)=0.32(ii)P(A|B)=P(AB)P(B)  =0.320.5  =0.64(iii)P(AB)=P(A)+P(B)P(AB)  =0.8+0.50.32  =0.98

Q.4 E

valuate P(AB), if 2P( A )=P( B )= 5 13 and P( A|B )= 2 5 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWFfbGaa8NDaiaa=fgacaWFSbGaa8xDaiaa=fgacaWF0bGaa8xzaiaa=bcacaWFqbGaa8hkaiaa=feacqGHQicYcaWFcbGaa8xkaiaa=XcacaWFGaGaa8xAaiaa=zgacaWFGaGaa8Nmaiaa=bfadaqadaqaaiaa=feaaiaawIcacaGLPaaacaWF9aGaa8huamaabmaabaGaa8NqaaGaayjkaiaawMcaaiaa=1dadaWcaaqaaiaa=vdaaeaacaWFXaGaa83maaaacaWFGaGaa8xyaiaa=5gacaWFKbGaa8hiaiaa=bkacaWFqbWaaeWaaeaacaWFbbGaa8hFaiaa=jeaaiaawIcacaGLPaaacaWF9aWaaSaaaeaacaWFYaaabaGaa8xnaaaacaWFUaaaaa@616E@

Ans

It is given that 2P(A)=P(B)=513P(A)=526,P(B)=513andP(A|B)=25.P(A|B)=P(AB)P(B)      25=P(AB)(513)P(AB)=25×513    =213P(AB)=P(A)+P(B)P(AB)    =526+513213    =1526213    =15426=1126

Q.5

If P(A)=611,P (B)=511and (AB =711,find (i)P(AB)(ii) P(A|B) (iii)P(B|A).

Ans

It is given that P(A)=611, P(B)=511 and P(AB)=711(i)P(AB)=P(A)+P(B)P(AB)      711=611+511P(AB)  =1111P(AB)  P(AB)=1111711=411(ii)Since,  P(A|B)=P(AB)P(B)  =(411)(511)  =45(iii)Since,  P(B|A)=P(AB)P(A)=(411)(611)  =46=23

Q.6 A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(ii) E: at least two heads, F: at most two heads
(iii) E: at most two tails, F: at least one tail
Determine P (E|F).

Ans

If a coin is tossed three times, then the sample space S is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} It can be seen that the sample space has 8 elements. (i)     E={HHH, HTH, THH, TTH}           F={HHH, HHT} EF={HHH}So, P(E)=48=12,​ P(F)=28=14 and P(EF)=18P(E|F)=P(EF)P(F)=(18)(14)=12(ii))   E={HHH, HHT, HTH, THH} F={HHT, HTH, HTT, THH, THT, TTH, TTT} EF={HHT, HTH, THH}So, ​ P(F)=78 and P(EF)=38 P(E|F)=P(EF)P(F)=(38)(78)=37(iii)      E={HHH, HHT, HTT, HTH, THH, THT, TTH}     F={HHT, HTT, HTH, THH, THT, TTH, TTT}EF={HHT,HTT,HTH,THH,THT,TTH}      P(F)=78 and P(EF)=68So,​ P(E|F)=P(EF)P(F)    =(68)(78)=67

Q.7 Two coins are tossed once, where
(i) E: tail appears on one coin,
F: one coin shows head
(ii) E: no tail appears, F : no head appears
Determine P (E|F).

Ans

If two coins are tossed once, then the sample space S is       S = {HH, HT, TH, TT} (i) E = {HT, TH}       F = {HT, TH} EF={HT, TH}P(F)=24=12 and P(EF)=24=12P(E|F)=P(EF)P(F)=(12)(12)=1(ii)  E={HH},F={TT}and EF=Ï•    P(F)=14,P(EF)=0P(E|F)=P(EF)P(F)      =0(14)=0

Q.8 A die is thrown three times,
E: 4 appears on the third toss,
F: 6 and 5 appears respectively on first two tosses
Determine P (E|F).

Ans

If a die is thrown three times, then the number of elements in the sample space will be 6× 6 × 6 = 216

            E={(1,1,4),(1,2,4),(1,3,4),...  (1,6,4)(2,1,4),(2,2,4),(2,3,4),...(2,6,4)(3,1,4),(3,2,4),(3,3,4),...(3,6,4)(4,1,4),(4,2,4),(4,3,4),...(4,6,4)(5,1,4),(5,2,4),(5,3,4),...(5,6,4)(6,1,4),(6,2,4),(6,3,4),...(6,6,4)}            F={(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}EF={(6,5,4)}So, P(F)=6216 and P(EF)=1216Then,  P(E|F)=P(EF)P(F)        =(1216)(6216)=16

Q.9 Mother, father and son line up at random for a family picture
E: son on one end,
F: father in middle
Determine P (E|F).

Ans

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be S={MFS, MSF, FMS, FSM, SMF, SFM} E={MFS, FMS, SMF, SFM}F={MFS, SFM} So, EF={MFS, SFM}P(F)=26 and P(EF)=26P(E|F)=P(EF)P(F)=(26)(26)=1

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

Ans

Let the first observation be from the black die and second from the red die.When two dice (one black and another red) are rolled,the sample space S has 6 × 6=36 number of elements. (a) Let A: Obtaining a sum greater than 9 ={(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}B: Black die results in a 5. ={(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}         AB={(5, 5), (5, 6)}  P(B)=636 and P(AB)=236 The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).   P(A|B)=P(AB)P(B) =(236)(636) =26=13(b)) E: Sum of the observations is 8. = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} F: Red die resulted in a number less than 4. ={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)(3,1),(3,2),(3,3),(4,1),(4,2),(4,3)(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)}andEF={(5,3),(6,2)}      P(F)=1836 and P(EF)=236The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, P (E|F)=P(EF)P(F)=2361836=19

Q.11 A fair die is rolled. Consider events
E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find
(i) P(E|F) and P (F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Ans

When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6} then given events are E={1,3,5}, F={2,3} and G={2,3,4,5}So, EF={3}, then P(EF)=16, P(E)=36, P(F)=26 and P(G)=46(i)P(E|F)=P(EF)P(F)=(16)(26)=12    P(F|E)=P(EF)P(E)=(16)(36)=13(ii)          EG={5}              P(EG)=16              P(E|G)=P(EG)P(G)    =(16)(46)=14            P(G|E)=P(EG)P(E)=(16)(36)=13(iii)          (EF)={1,2,3,5}(EF)G={2,3,5}    P((EF)G)=36=12    P{(EF)|G}=P((EF)G)P(G)    =(12)(46)=34(EF)G={3}  P((EF)G)=16P((EF)|G)=P((EF)G)P(G)=(16)(46)=14

Q.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, (ii) at least one is a girl?

Ans

Let B and G represent the boy and the girl child respectively. If a family has two children, the sample space will be S = {(B, B), (B, G), (G, B), (G, G)} Let A= both children are girls={(G, G)}(i) Let B be the event that the youngest child is a girl, then   B={(B, G),(G, G)}AB={(G, G)}P(B)=24=12 and P(AB)=14The conditional probability that both are girls, given that the youngest child is a girl=P (A|B)=P(AB)P(B)=(14)(12)=12Thus, the required probability is 12.(ii) Let E be the event that at least one child is a girl.             E={(B,G),(G,B),(G,G)}Then,  AE={(G,G)}P(AE)=14 and P(E)=34The conditional probability that both are girls, given that at least one child is a girl= P(A|E)        =P(AE)P(E)        =(14)(34)=13Thus, the required probability is 13.

Q.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?

Ans

Number of easy True/False questions = 300
Number of easy multiple choice questions = 500
Number of difficult True/False questions = 200
Number of difficult multiple choice questions = 400
Total number of multiple choice questions = 900
Total questions = 1400
Let the notations of questions are as follows:
E = Easy questions, D = Difficult questions, M = Multiple choice questions and T = True/False questions.

So, Probability of easy and multiple choice questionsP(EM)=5001400        =514          Probability of multiple choice question,P(M)=9001400        =914Probability of a randomly selected an easy question, given that it is a multiple choice question=P(E|M)        =P(EM)P(M)        =(514)(914)=59Thus, required probability is 59.

Q.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’.

Ans

When two dice are thrown, sample space observations = 6 × 6 = 36
Let A = the sum of the numbers on the dice is 4 and
B = the two numbers appearing on throwing the two dice are different.

A={(1,3),(3,1)}    B={(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5)}Then,AB={(1,3),(3,1)}P(B)=3036=56 and P(AB)=230=118The probability that the sum of the numbers on the dice is 4,given that the two numbers appearing on throwing the two diceare different=P(A|B)=P(AB)P(B)=(118)(56)=115

Q.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’.

Ans

The sample space of the experiment is given below:S={(1,H),(1,T),(2,H),(2,T),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,H),(4,T),(5,H),(5,T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),}Let A = the coin shows a tail ={(1,T),(2,T),(4,T),(5,T)}and B=at least one die shows 3={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)}AB=Ï•P(AB)=0P(B)=P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}+P{(3,1)}=136+136+136+136+136+136+136=736Probability of the event that the coin shows a tail, given that at least one die shows 3=P(A|B)=P(AB)P(B)=0(736)=0Thus, the required probability is 0.

Q.16 If P (A) =1/2, P (B) = 0, then P (A|B) is
(A) 0 (B) 1/2
(C) not defined (D) 1

Ans

Given that P(A)=12, P(B)=0, then P(A|B)=P(AB)P(B)      =P(AB)0Therefore, P (A|B) is not defined. Thus, the correct answer is C.

Q.17 If A and B are events such that P (A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B
(C) A ∩ B = Φ (D) P (A) = P(B)

Ans

It is given that    P(A|B)=P(B|A)  P(AB)P(B)=P(AB)P(A)          1P(B)=1P(A)          P(A)=P(B)Thus, option D is correct.

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