The downloadable PDF of the Exercise 15.2 of NCERT Solutions for class 10 Maths Chapter 15- Probability is given here. This exercise is an optional exercise given by the NCERT, providing ample extra questions to the students, to test their preparation and understanding. There are 5 questions in this exercise that covers selected topics, that are very important, from the chapter Probability.

While solving the problems given in the NCERT textbooks, the subject experts at BYJUâ€™S make sure that it follows the NCERT Syllabus and guidelines. For students aiming to score great marks in Class 10 first and second term examination of CBSE, it is more than just important to practice the NCERT solutions of class 10 Maths.

### Download PDF of NCERT Solutions for Class 10 Maths Chapter 15- Probability Exercise 15.2

### Access Another Exercise Solutions of Class 10 Maths Chapter 15- Probability

Exercise 15.1 Solutions 25 Questions (1 MCQ, 21 Short Answer Questions, 2 Long Answer Question, 1 Main Question with 4 Sub-questions)

### Access Answers to NCERT Class 10 Maths Chapter 15 â€“ Probability Exercise 15.2

**1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on **

**(i) the same day? **

**(ii) consecutive days? **

**(iii) different days?**

**Solution:**

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5Ã—5 = 25

**(i)** The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = â…•

**(ii)** The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) = 8/25

**(iii)** P (both visiting on the different days) = 1-P (both visiting on the same day)

So, P (both visiting on the different days) = 1-(â…•) = â…˜

**2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:**

**What is the probability that the total score is**

**(i) even? **

**(ii) 6? **

**(iii) at least 6?**

**Solution:**

The table will be as follows:

+ | 1 | 2 | 2 | 3 | 3 | 6 |

1 | 2 | 3 | 3 | 4 | 4 | 7 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

2 | 3 | 4 | 4 | 5 | 5 | 8 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

3 | 4 | 5 | 5 | 6 | 6 | 9 |

6 | 7 | 8 | 8 | 9 | 9 | 12 |

So, the total number of outcomes = 6Ã—6 = 36

**(i)** E (Even) = 18

P (Even) = 18/36 = Â½

**(ii)** E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

**(iii)** E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

**3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball**

**is double that of a red ball, determine the number of blue balls in the bag.**

**Solution:**

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x+5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

âˆ´ P (drawing a blue ball) = [x/(x+5)] â€”â€”â€“(i)

Similarly,

P (drawing a red ball) = [5/(x+5)] â€”â€”â€“(i)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

**4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the**

**box, what is the probability that it will be a black ball?**

**If 6 more black balls are put in the box, the probability of drawing a black ball is now**

**double of what it was before. Find x**

**Solution:**

Total number of black balls = x

Total number of balls = 12

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (getting black balls) = x/12 â€”â€”â€”â€”â€”â€”-(i)

Now, when 6 more black balls are added,

Total balls become = 18

âˆ´ Total number of black balls = x+6

Now, P (getting black balls) = (x+6)/18 â€”â€”â€”â€”â€”â€”-(ii)

Itâ€™s given that, **the probability of drawing a black ball now is double of what it was before**

(ii) = 2 Ã— (i)

(x+6)/18 = 2 Ã— (x/12)

x + 6 = 3x

2x = 6

âˆ´ x = 3

**5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at**

**random from the jar, the probability that it is green is â…”. Find the number of blue balls**

**in the jar.**

**Solution:**

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24-x

P(getting green marble) = x/24

From the question, x/24 = â…”

So, the total green marbles = 16

And, the total blue marbles = 24-x = 8

Some of the topics that Exercise 15.2 of 10th standard Maths cover are as follows:

- The occurrence of an event: if any one of the elementary events corresponding to the event is the outcome of that event, then an event corresponding to a random experiment is said to occur.
- Impossible events: Impossible events are those events which do not occur ever. So, we say that the probability of an impossible event is always zero.
- Sure event: An event is said to be a sure event if it occurs.

When students of Class 10 CBSE board solve the problems given in exercise 15.2, the students will be able to understand the concepts mentioned above clearly. Same way, the NCERT solutions of all the chapters will help the students to understand the concepts explained in the chapters more profoundly, in turn improving their mathematical problem-solving ability along with enhancing their confidence.