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NCERT Exemplar Solution for Class 10 Mathematics Chapter 13: Exercise 13.3

Mar 19, 2018 17:24 IST
    NCERT Exemplar Solution for Class 10 Maths Chapter 13
    NCERT Exemplar Solution for Class 10 Maths Chapter 13

    Here you get the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 13, Statistics and Prbability. This part includes solutions for problems given in exercise 13.3 of class 10 Maths NCERT Exemplar Book. It is consisted of only the short answer type questions framed from various important topics discussed in the chapter. Each question is provided with a detailed yet simple and step-wise solution to help students learn the concept and logic working behind that particular question.

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as for other competitive exams.

    CBSE Class 10 Maths NCERT Exemplar Solution: Exercise 13.2

    Few questions from the Class 10 Maths NCERT Exemplar Exercise 13.3 are given below:

    Q. Two dice are thrown at the same time. Find the probability of getting

    (i) same number on both dice.

    (ii) different number on both dice.

    Sol.

    As the two dice are thrown at the same time,                                

    So, total number of possible outcomes = 62 = 36

    (i) In case of same numbers on both dice the possible outcomes are

          (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

    ∴ Number of possible outcomes = 6

    So, required probability = 6/36 = 1/6

    (ii) In case of different numbers on both dice the possible outcomes are

          = 36 – Number of possible outcomes for same number on both dice

          = 36 – 6 = 30

    ∴ Required probability = 30/36 = 5/6

    Q. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

    (i) 7?                           

    (ii) a prime number ?  

    (iii) 1?

    Sol.

    As the two dice are thrown at the same time,                                

    So, total number of possible outcomes = 62 = 36

    (i) Sum of the numbers appearing on the dice is 7.

    Here, the possible outcomes are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1),

    Thus, the number of possible outcomes = 6

    ∴  Required probability = 6/36 = 1/6.

    (ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

    Here, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3,4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

    Thus, number of possible outcomes = 15

    \ Required probability = 15/36 = 5/12.

    (iii) Sum of the numbers appearing on the dice is 1.

    It is an impossible event, so its probability is zero.

    Q. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

    (i) 6                            

    (ii) 12                         

    (iii) 7

    Sol.

    As the two dice are thrown at the same time,                                

    So, total number of possible outcomes = 62 = 36

    (i) Product of the numbers on the top of the dice is 6.

    Here, the possible ways are (1, 6), (2, 3), (3, 2) and (6, 1).

    Thus, the number of possible outcomes = 4

    ∴ Required probability = 4/36 = 1/9.

    (ii) Product of the numbers on the top of the dice is 12.

    Here, the possible ways are (2, 6), (3, 4), (4, 3) and (6, 2).

    Thus, the number of possible outcomes = 4

    ∴ Required probability = 4/36 = 1/9.

    (iii) Product of the numbers on the top of the dice is 7.

    It is an impossible event.

    So, its probability is zero.

    Q. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

    Sol.

    As the two dice are thrown at the same time,

    Number of total outcomes = 36

    For the product of numbers appearing on them to be less than 9, the possible outcomes are

    (1, 6), (1,5) (1,4), (1,3), (1,2), (1, 1),(2, 2), (2, 3), (2, 4), (3,2), (4, 2), (4, 1), (3,1), (5,1), (6, 1) and (2,1).

    Thus, the number of possible outcomes = 16

    ∴ Required probability = 16/36 = 4/9

    To get the solutions to remaining questions of the exercise, visit the following link:

    CBSE Class 10 Maths NCERT Exemplar Solution: Exercise 13.3

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