NCERT Solutions Class 8 Maths Chapter 4 Data Handling Exercise 4.3

The NCERT Solutions for Class 8 Maths Chapter 4 Exercise 4.3 examines the ideas of chance and probability. This exercise introduces simple ideas of predicting outcomes based on possible results and demonstrates the difference between occasions that are certain, likely, or impossible. These are good ideas not only in Maths but also in science and everyday thinking. The questions are entertaining and simple, containing real-life examples to explain how probability works.

All questions are based on the latest NCERT syllabus for Class 8 standard. These NCERT Solutions will show you how to solve questions step-by-step, enabling you to do well in your exams. You will develop a good knowledge base to prepare for higher classes, but also to enter competitive exams such as Olympiads.

1.0Download NCERT Solutions Class 8 Maths Chapter 4 Data Handling Exercise 4.3: Free PDF

Exercise 4.3 explains the concept of chance and probability in a fun and simple way. These NCERT Solutions for Class 8 Maths Chapter 4 will help you solve all the relevant questions step by step. Download from below the PDF for free and start preparing for your examination!

2.0Key Concepts in Exercise 4.3 of Class 8 Maths Chapter 4

Exercise 4.3 explains the idea of chance in daily life using basic probability. The key concepts include:

• Understanding events that are sure, likely, or impossible
• Listing all possible outcomes in an event
• Finding the chance or probability of an event happening
• Using simple examples like coins, dice, or cards to explain probability

3.0NCERT Class 8 Maths Chapter 4: Other Exercises

4.0NCERT Class 8 Maths Chapter 4 Exercise 4.3: Detailed Solutions

• List the outcomes you can see in these experiments. (i) Spinning a wheel,
  
  (ii) Tossing two coins together. Sol. (i) List of outcomes of spinning the given wheel are A, B, C and D. (ii) When two coins are tossed together, the possible outcomes of the experiment are HH, HT, TH and TT.
• When a die is thrown, list the outcomes of an event of getting. (i) (a) A prime number, (b) Not a prime number. (ii) (a) A number greater than 5, (b) A number not greater than 5. Sol. (i) In a throw of die, list of the outcome of an event of getting (a) A prime number are 2,3 and 5. (b) Not a prime number are 1, 4 and 6. (ii) In a throw of die, list of the outcomes of an event of getting (a) A number greater than 5 is 6 . (b) A number not greater than 5 are 1, 2, 3,4 and 5 .
• Find the (i) Probability of the pointer stopping on $D$ in (Q. 1 (i)). (ii) Probability of getting an ace from a well shuffled deck of 52 playing cards. (iii) Probability of getting a red apple (see adjoining figure).
  
  Sol. (i) Out of 5 sectors, the pointer can stop at any of sectors in 5 ways. $\therefore$ Total number of elementary events $=$5. There is only one ' D ' on the spinning wheel. $\therefore$ Favourable number of outcomes $=1$ $\therefore$ Required probability $=1/5$ (ii) Out of 52 cards, one card can be drawn in 52 ways. $\therefore$ Total number of outcomes $=52$ There are 4 aces in a pack of 52 cards, out of which one ace can be drawn in 4 ways. $\therefore$ Favourable number of cases $=4$. So, the required probability $=\frac{4}{52}=\frac{1}{13}$ (iii) Out of 7 apples, one apple can be drawn in 7 ways. $\therefore$ Total number of outcomes $=7$ There are 4 red apples, in a bag of 7 apples, out of which 1 red apple can be drawn in 4 ways. $\therefore$ Favourable number of cases $=4$ So, the required probability $=4/7$
• Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is choosen from the box without looking into it. What is the probability of (i) Getting a number 6? (ii) Getting a number less than 6? (iii) Getting a number greater than 6 ? (iv) Getting a 1-digit number? Sol. Out of 10 slips, 1 slip can be drawn in 10 ways. So, the total number of outcomes $=10$ (i) An event of getting a number 6, i.e., if we obtain a slip having number 6 as an outcome. So favourable number of outcomes $=1$ $\therefore$ Required probability $=1/10$ (ii) An event of getting a number less than 6, i.e., if we obtain a slip having any of numbers $1,2,3,4,5$ as an outcome. So, favourable number of cases $=5$ $\therefore$ Required probability $=\frac{5}{10}=\frac{1}{2}$ (iii) An event of getting a number greater than 6, i.e., if we obtain a slip having any of numbers 7, 8, 9, 10 as an outcome. So, favourable number of cases $=4$ $\therefore$ Required probability $=\frac{4}{10}=\frac{2}{5}$ (iv) An event of getting a one-digit number, i.e., if we obtain a slip having any of numbers $1,2,3,4,5,6,7,8,9$ as an outcome. So, favourable number of cases $=9$. $\therefore$ Required probability $=9/10$.
• If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector? Sol. Out of 5 sectors, the pointer can stop at any of sectors in 5 ways. $\therefore$ Total number of outcomes $=5$ There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways. $\therefore$ Favourable number of outcomes $=3$ So, the required probability $=3/5$ Further, there are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways. So, the required probability $=4/5$.
• Find the probabilities of the events given in question 2. Sol. In a single throw of a die, we can get any one of the six numbers $1,2,\dots ,6$ marked on its six faces. Therefore, the total number of outcomes $=6$. (i) Let A denote the event "getting a prime number". Clearly, event A occurs if we obtain $2,3,5$ as an outcome $\therefore$ Favourable number of outcomes $=3$. Hence, $P(A)=\frac{3}{6}=\frac{1}{2}$ (ii) Let A denote the event "not getting a prime number". Clearly, event A occurs if we obtain $1,4,6$ as an outcomes $\therefore$ Favourable number of outcomes $=3$. Hence, $P(A)=\frac{3}{6}=\frac{1}{2}$ (iii) The event "getting a number greater than 5 " will occur if we obtain the number 6 . $\therefore$ Favourable number of outcomes $=1$. Hence, required probability $=1/6$ (iv) The event "getting a number not greater than 5 " will occur if we obtain one of the numbers $1,2,3,4,5$. $\therefore$ Favourable number of outcomes $=5$. Hence, required probability $=5/6$

5.0Key Features and Benefits of Class 8 Maths Chapter 4 Exercise 4.3

• The exercise addresses simple concepts of probability in line with the current NCERT syllabus. 
• Uses everyday situations to introduce chance and outcomes, and its implications. 
• Enables students to clearly identify sure events, likely events and impossible events. 
• Regular practice develops a strong knowledge base for solving probability based questions in board and Olympiad exams. 
• They also help in developing logical thinking and real world problem solving skills.