NCERT Solutions Class 10 Maths Chapter 14 Probability
NCERT Solutions for Class 10 Maths Chapter 14 Probability is an important chapter in class 10 Maths. This chapter will introduce the students to fundamental concepts such as random experiments, outcomes, events, and probability calculations.
These NCERT Solutions provide step-by-step explanations for even complex probability problems so that students feel confident while solving exercises. Since questions from Probability will surely appear in the CBSE Class 10 Maths board exams, applying to real-life situations, it is vital that students understand these problems.
In this blog, NCERT Solutions for Class 10 Maths Chapter 14 Probability are given in PDF, helping students to prepare effectively for their exams. We also ensure that students can study anytime, anywhere.
1.0Download NCERT Solutions Class 10 Maths Chapter 14 Probability PDF Online
Students will study the fundamentals of probability in Class 10 Maths Chapter 14, including experimental probability, theoretical probability, the probability of certain and impossible events, and elementary/complementary events to solve a variety of problems. We’ve provided a link below for free access to NCERT questions and solutions for Chapter 14 Probability, allowing you to practice comfortably at your own pace.
2.0NCERT Exercise-wise Solutions Class 10 Maths Chapter 14 Probability
The NCERT Solutions Class 10 Maths Chapter 14 are a very important set of study materials because they enable the understanding of the concept itself along with the practical application. This description makes complex ideas simpler and easier to understand. The detailed answers of the NCERT for every exercise are given below:
4.0Importance of NCERT Solutions Class 10 Maths Chapter 14 Probability
- Practising the solutions for Class 10 Maths is essential not only for exam preparation but also for understanding its real-life applications.
- Important ideas that serve as the basis for more complex mathematical topics are covered in this chapter.
- Students who practice these solutions will have a stronger understanding of probability and will have developed a methodical approach to problem-solving.
- The detailed solutions of NCERT help you improve your problem-solving skills and gain the maximum score in the exam.
- These exercises not only help in understanding the basics but also build the mathematical foundation needed for tackling more complex topics in higher studies.
5.0Sample NCERT Solutions Class 10 Maths Chapter 14 Probability Exercise 14.1
1. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes- two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes-an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Sol.
(i) Incorrect. The given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So the total outcomes is 4.
∴ Probability = 1/4.
(ii) Correct. Because the two outcomes are possible. Total outcomes = 6 and odd numbers = 3 and even numbers = 3. So, favorable outcomes = 3 (in both the cases even or odd).
∴ Probability = 3/6 = 1/2
2. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Sol. Possible outcomes are:
HHH, TTT, HHT, HTH, THH, TTH, THT, HTT
i.e., in all 8 outcomes.
Out of these, 6 outcomes other than HHH and TTT favor the event E than Hanif will lose the game.
Therefore, P (Hanif will lose the game) = 6/8 = 3/4
3. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:
(i) She will buy it
(ii) She will not buy it?
Solution:
Total number of ball pens = 144
=> Total number of possible outcomes = 144
(i) Since there are 20 defective pens
=> Number of good pens = 144 - 20 = 124
=> Number of favourable outcomes = 124
=> Probability that she will buy it = 124/144 = 31/36
(ii)Probability that she will not buy it = 1 - [Probability that she will buy it]
= 1 - 31/36
= (36 - 31) / 36
= 5/36.
4. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears.
(i) a two-digit number.
(ii) a perfect square number.
(iii) a number divisible by 5.
Sol. We have the total number of discs = 90.
Therefore, the total number of possible outcomes = 90.
(i) Since, two-digit numbers are 10, 11, 12, ..., 90.
Therefore, the number of two-digit numbers = 90 - 9 = 81.
Number of favorable outcomes = 81.
Therefore, P(Two-digit number) = (Number of favorable outcomes) / (Total number of possible outcomes) = 81/90 = 9/10
(ii) Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81.
Therefore, the number of perfect squares = 9.
Number of favorable outcomes = 9.
Therefore, P(Perfect square) = (Number of favorable outcomes) / (Total number of possible outcomes) = 9/90 = 1/10
(iii) Numbers divisible by 5 from 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.
i.e., There are 18 numbers from (1 to 90) which are divisible by 5.
Therefore, the number of favorable outcomes = 18.
P(Numbers divisible by 5) = (Number of favorable outcomes) / (Total number of possible outcomes) = 18/90 = 1/5
5. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Sol. We have the number of good pens = 132 and the number of defective pens = 12.
Therefore, the total number of pens = 132 + 12 = 144.
Total possible outcomes = 144.
There are 132 good pens.
Therefore, the number of favorable outcomes = 132.
P(good pens) = (Number of favorable outcomes) / (Total number of possible outcomes) = 132/144 = 11/12
6.0Benefits of Class 10 Maths Chapter 14 Probability
The Probability chapter in Class 10 Maths offers several key benefits:
- Real-Life Applications: It helps in understanding and predicting the likelihood of events, useful in fields like gaming, weather forecasting, and risk assessment.
- Decision-Making Skills: Students learn to assess chances and make informed decisions based on probability.
- Problem-Solving Abilities: It enhances logical thinking and the ability to solve problems involving uncertainty and chance.
- Foundation for Future Studies: Probability lays the groundwork for more advanced topics in statistics, economics, and data science.
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