NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.2) Exercise 13.2

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Click on the given link to download the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.2) Exercise 13.2

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Exercise 13.2 Class 12th is from Unit Probability. Probability means the possibility of the given event. A random event’s occurrence is the subject of this area of Mathematics. The range of the value is 0 to 1. Probability has been applied to Mathematics to predict the likelihood of different events. Probability generally refers to the degree to which something is likely to occur. This fundamental theorem of Probability, which also applies to the probability distribution, can help students comprehend the possible outcomes of a random experiment. Before one can determine the probability that a certain event will occur, one must first know the total number of outcomes.

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Access NCERT Solutions for Class 12 Maths Chapter 13 – Probability

It is possible to solve all the questions of the Probability chapter with the help of  NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

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NCERT Solutions for Class 12 Maths PDF Download

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Q.1 If P(A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.

Ans

Since, P(A∩B)=P(A).P(B)  [As A and B are independent events.]    =35.15    =325Thus, P(A∩B)=325.

Q.2 Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Ans

Total number of cards=52Number of black cards=26Let​ A=A black card in first draw        B=A black card in second draw∴P(A)=2652=12     P(B)=2551Thus, probability of getting both the cards black      =P(A).P(B)     =12×2551     =25102

Q.3 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Ans

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.∴     Probability that first drawn orange is good, P(A)=1215Probability that second drawn orange is good, P(B)=1114⇒Probability that third drawn orange is good, P(C)=1013The box is approved for sale, if all the three oranges are good.   Thus, probability of getting all the oranges good=P(A).P(B).P(C)   =1215×1114×1013   =4491Therefore, the probability that the box is approved for sale is  4491.

Q.4 A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Ans

Sample space, when a fair coin and an unbiased die are tossed,                   S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}                   A=Head appears on the coin                  ={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}                   B=3 on the die                  ={(H,3),(T,3)}           A∩B={(H,3)}     P(A∩B)=112P(A)×P(B)=612×212    =112Since,      P(A∩B)=P(A)×P(B)Therefore, A and B are independent events.

Q.5 A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, the number is even, and B be the event, the number is red. Are A and B independent?

Ans

Sample space, when an unbiased die are tossed,                   S={1,2,3,4,5,6}                   A=The number is even                  ={2,  4,  6}                   B=The number is red                  ={4,5,6}           A∩B={4,6}     P(A∩B)=26=13P(A)×P(B)=36×36    =14Since,      P(A∩B)≠P(A)×P(B)Therefore, A and B are not independent events.

Q.6 Let E and F be events with P (E) = 3/5, P(F) = 3/10 and P (E ∩ F) = 1/5 . Are E and F independent?

Ans

E and F be events with P(E)=35, P(F)=310  and P (E∩F)=1/5P(E)×P(F)=35×310    =950Since, P (E∩F)≠P(E)×P(F)Therefore, A and B are not independent events.

Q.7 Given that the events A and B are such that
P(A) = 1/2 , P (A ∪ B) = 3/5 and
P(B) = p. Find p if they are
(i) mutually exclusive (ii) independent.

Ans

Given that the events A and B are such thatP(A)=12, P(A∪B)=35  and P(B)=p(i) When the events A and B are mutually exclusive i.e., P(A∩B)=0     P(A∪B)=P(A)+ P(B)−P(A∩B)35=12+p−0⇒p=35−12    =6−510⇒p=110(ii) When the events A and B are independent i.e., P(A∩B)=P(A). P(B)∴     P(A∪B)=P(A)+ P(B)−P(A∩B)      P(A∪B)=P(A)+ P(B)−P(A). P(B)⇒35=12+p−12×p⇒  35−12=p(1−12)⇒  6−510=12pp=110×21=15

Q.8 Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A ∩ B) (ii) P(A ∪ B)
(iii) P(A|B) (iv) P (B|A)

Ans

A and B be independent events with P(A)=0.3 and P(B)=0.4(i)  P(A∩B)=P(A)×P(B)     =0.3×0.4     =0.12(ii)P(A∪B)=P(A)+P(B)−P(A)×P(B)     =0.3+0.4−0.12     =0.7−0.12     =0.58(iii)P(A|B)=P(A∩B)P(B)     =0.120.4     =0.3(iv)P(B|A)=P(A∩B)P(A)     =0.120.3     =0.4

Q.9 If A and B are two events such that P(A) =1/4, P(B)= 1/2 and P(A∩B)= 1/8, find P(not A and not B).

Ans

A and B are two events such that P(A)=14, P(B)=12  and P(A∩B)=18P(not A and not B)=P(A‘∩B‘) =P(A∪B)‘[By De Morgan‘s Law(A‘∩B‘)=(A∪B)‘] =1−P(A∪B) =1−{P(A)+P(B)−P(A)P(B)} =1−{14+12−14×12} =1−(14+12−18) =1−(2+4−18) =1−58 =8−58 =38Thus, P(not A and not B) is  38.

Q.10 Events A and B are such that P (A) =1/2, P(B)=7/12 and P(not A or not B) = 1/4. State whether A and B are independent?

Ans

Events A and B are such that P(A)=12,  P(B)=712  and P(not A or not B) =14P(not A or not B)=P(A‘∪B‘)       14 =P(A∩B)‘[By De Morgan‘s Law(A‘∩B‘)=(A∪B)‘]       14=1−P(A∩B)  P(A∩B)=1−14  =34P(A).P(B)=12×712  =724So,   P(A∩B)≠P(A).P(B)Thus,A and B are independent events.

Q.11 Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B) (ii) P(A and not B)
(iii) P(A or B) (iv) P(neither A nor B)

Ans

Given two independent events A and B such that P(A)=0.3, P(B)=0.6(i) P(A and B)=P(A∩B)  =P(A).P(B)  =0.3×0.6  =0.18(ii)P(A and not B)  =P(A∩B‘)  =P(A)−P(A∩B)  =0.3−0.18  =0.12(iii) P(A or B)=P(A∪B)  =P(A)+P(B)−P(A∩B)  =0.3+0.6−0.18  =0.9−0.18  =0.72(iv)P(neither A nor B)  =P(A‘∩B‘)  =[P(A∪B)‘][By De Morgan‘s Law   A‘∩B‘=(A∪B)‘]  =1−P(A∪B)  =1−0.72  =0.28

Q.12 A die is tossed thrice. Find the probability of getting an odd number at least once.

Ans

 Probability of getting an odd number in a throw of a die=36=12Probability of getting an even number in a throw of a die=36=12     Probability of getting an even number three times=12.12.12        =18Therefore, probability of getting an odd number at least once = 1−Probability of getting an odd number in none of the throws = 1−Probability of getting an even number thrice       =1−18       =78

Q.13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.

Ans

Two balls are drawn at random with  replacement from a box containing 10 black  and 8 red balls. Total balls=10+8=18Black balls=10    Red balls=8(i)        Probability of getting a red ball in the first draw=818=49       The ball is replaced after the first draw. Probability of getting second red ball in the first draw=818=49Probability of getting   both the red balls=49×49         =1681(ii)   Probability of getting a black ball in the first draw=1018       The ball is replaced after the first draw. Probability of getting second red ball in the first draw=818Probability of getting first ball as black and second ball as red        =1018×818        =2081(iii)                Probability of getting first ball as red=818=49The ball is replaced after the first draw.      Probability of getting second ball as black=1018=59∴Probability of getting first ball as red and second ball as black=49×59        =2081Therefore, probability of getting first ball as black and second ball as red=49×59        =2081Therefore, probability that one of them is black and other is red      = 2081+2081        =4081

Q.14 Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.

Ans

        Probability of solving specific problem by A, P(A)=12         Probability of solving specific problem by B, P(B)=13Probability of not solving specific problem by A, P(A‘)=1−12=12 Probability of not solving specific problem by B, P(B‘)=1−13=23(i)Probability of Problem solved by both independetly,P(A∩B)=12×13=16Probability that Problem is solved,P(A∪B)=P(A)+P(B)−P(A)P(B)=12+13−12×13=12+13−16=3+2−16=46=23(ii) Probability that exactly one of them solves the problem=P(A)P(B‘)+P(B)P(A‘)=12×23+13×12=13+16=2+16=36=12

Q.15 One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’.

Ans

Total cards in a well shuffled deck=52(i) E:‘the card drawn is a spade’           P(E)=452          =113 F:‘the card drawn is an ace’            P(F)=1352 =14     P(E∩F)=152  P(E). P(F)=113×14   =152∴P(E∩F)=P(E). P(F)Thus, the events E and F are independent.(ii) E:‘the card drawn is black’           P(E)=1352          =14 F:‘the card drawn is a king’            P(F)=452   =113  P(E). P(F)=14×113   =152∴P(E∩F)=P(E). P(F)Thus, the events E and F are independent.(iii) E:‘the card drawn is a king or queen’           P(E)=852=213 F:‘the card drawn is a queen or jack’            P(F)=852   =213  P(E). P(F)=213×213   =4169∴P(E∩F)≠P(E). P(F)Thus, the events E and F are not independent.

Q.16 In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.

Ans

Let H=the students who read Hindi newspaper and         E=the students who read English newspaper. It is given that,         P(H)=60%=60100=0.6        P(E)=40%=40100=0.4P(H∩E)=20%=20100=0.2(i) The probability that she reads neither Hindi nor English news papers=P(H∪E)‘        =1−P(H∪E)        =1−{P(H)+P(E)−P(H∩E)}        =1−(0.6+0.4−0.2)        =1−0.8        =0.2(ii)If she reads Hindi news paper, the probability that she reads English news paper        =P(E|H)        =P(E∩H)P(H)        =0.20.6=13(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper        =P(H|E)        =P(H∩E)P(E)        =0.20.4=12

Q.17 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Ans

When two dice are rolled, the number of outcomes is 36.The only even prime number is 2. Let E be the event of getting an even prime number on each die.
So, E = {(2, 2)}
P (E) = (1/36)
Thus, the correct option is D.

Q.18 Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A′B′) = [1 – P(A)][1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1

Ans

And, P(A)P(B)=m.n≠0∴          P(A∩B)≠P(A)P(B)So, A and B are not independent events.Therefore, option A is not correct.(B)Two events A and B are said to be independent, if P(AB)=P(A)×P(B) Consider the result given in alternative B.               P(A‘B‘)=[1−P(A)][1−P(B)]⇒             P(A‘∩B‘)=1−P(A)−P(B)+P(A)P(B)⇒             P(A∪B)‘=1−P(A)−P(B)+P(A)P(B)[By De​ Morgan‘s Law]⇒ 1−P(A∪B)=1−P(A)−P(B)+P(A)P(B)⇒           P(A∪B)=P(A)+P(B)−P(A)P(B)P(A)+P(B)−P(A∩B)=P(A)+P(B)−P(A)P(B)⇒            P(A∩B)=P(A)P(B)This implies that A and B are independent, ifP(A‘B‘)=[1−P(A)][1−P(B)]Therefore, option B is correct.(C)Let A: Event of getting an odd number on throw of a die={1, 3, 5}B: Event of getting an even number on throw of a die={2, 4, 6}∴P(A)=36=12​ and P(B)=36=12​So,  A∩B=Ï•⇒P(A∩B)=0  P(A)P(B)=12.12=14≠0⇒P(A∩B)≠ P(A)P(B)So, A and B are not independent events.Therefore, option C is not correct.(D)If P(A)+P(B)=1,  then, P(A∩B)≠P(A)P(B)So, A and B are not independent events.Therefore, option D is not correct.Thus, option B is correct.

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FAQs (Frequently Asked Questions)

1. Are the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 difficult?

The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 are not difficult.

Furthermore, with regular practice and proper guidance from Extramarks, students can easily understand the concepts of the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

2. Is it necessary to practice all the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2?

Yes, students should practice all the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2, as having a clear understanding of the basics and a fast calculation speed are

crucial to scoring well on the examination. Also, every question of the exercise contains a different concept, therefore students should practice all the questions.

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Students can refer to the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 provided by the Extramarks’ website to clear their doubts. Also, the website offers students with various learning tools like live doubt-solving classes, K12 live classes programs and much more so that students can resolve their doubts easily and score well in the board examination.

4. Are the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 available on Extramarks?

Extramarks provides students with the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2. Along with this, it provides students with K12 study material for their boards and live classes with the experts to provide detailed solutions to all the questions of students. Also, the learning app provides students with curriculum mapping, complete syllabus coverage, gamified learning experience and much more so that students can have a systematic and enjoyable learning process. Extramarks focuses on the holistic development of students. Therefore, students can choose Extramarks as their learning partner to study NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

5. How can NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 help students succeed in their board examinations?

Extramarks advises studying from the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2. Nearly all the examination questions are covered in the NCERT solutions.

6. Do NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 aid in completing the examination question paper in the allotted time?

Students can finish their question paper in an examination within the allotted time by regularly practising NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

7. How does Mathematics assist in resolving issues pertaining to other subjects?

Mathematics is employed in higher-level studies like Computer Engineering, as well as to solve Chemistry and Physics equations.

The  NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 provides a summary of the chapter. These solutions are curated by the expert educationalists of Extramarks and are detailed in a step-by-step approach. Students can easily understand the concepts of the chapter as these steps help them understand the logic behind the question. It is essential for students to have an in-depth understanding of the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 as they structure the basic concepts of students. Once they practice these questions, they will be easily able to solve any complicated problems that they may encounter during the examination. Along with the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2, Extramarks also provides young learners with sample papers, past years’ papers, revision notes, and important questions about all the subjects and classes.  Sample papers and past years’ papers are the best tools for students to prepare for their examinations. They provide students with a model for writing their question papers. By practising sample papers, students can gain confidence and perform well in their board examinations.

To some students, Mathematics is boring because of all the numbers and calculations, which are complex and confusing. The Extramarks’ website makes it interesting by making the content’s graphics and animations more engaging to divert students’ interest toward Mathematics. The NCERT solutions for all of the core courses are covered by Extramarks, and the experts consistently work to make the topics as simple as possible so that students can learn them easily.

Click on the given link to download the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.