NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Data Handling |

Exercise |
Ex 5.3 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3

**Ex 5.3 Class 8 Maths Question 1.**

List the outcomes you can see in these experiments.

**
(a)** Spinning a wheel,

**(b)**Tossing two coins together.

**Solution:**

**(a)**List of outcomes of spinning the given wheel are A, B, C and D.

**(b)**When two coins are tossed together, the possible outcomes of the experiment are HH, HT, TH and TT.

**Ex 5.3 Class 8 Maths Question 2.**

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.

**Solution:**

**(i)** In a throw of die, list of the outcomes 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 and 4.

**Ex 5.3 Class 8 Maths Question 3.**

Find the

**
(a)** Probability of the pointer stopping on D in (Question 1 (a)).

**(b)**Probability of getting an ace from a well shuffled deck of 52 playing cards.

**(c)**Probability of getting a red apple (see adjoining figure).

**Solution:**

**(a)**Out of 5 sectors, the pointer can stop dt any of sectors in 5 ways.

∴ Total number of elementary events = 5. There is only one ‘D’ on the spinning wheel.

∴ Favourable number of outcomes = 1

∴ Required probability = \(\frac { 1 }{ 5 } \)

**(b)** Out of 52 cards, one card can be drawn in 52 ways.

∴ 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.

∴ Favourable number of cases = 4

So, the required probability = \(\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)

**(c)** Out of 7 apples, one apple can be drawn in 7 ways.

∴ 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.

∴ Favourable number of case = 4

So, the required probability = \(\frac { 4 }{ 7 } \)

**Ex 5.3 Class 8 Maths Question 4.**

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 chosen 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?

**Solution:**

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

∴ Required probability = \(\frac { 1 }{ 10 } \)

**(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

∴ 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

∴ 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

∴ Required probability =\(\frac { 9 }{ 10 } \)

**Ex 5.3 Class 8 Maths Question 5.**

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?

**Solution:**

Out of 5 sectors, the pointer can stop at any of sectors in 5 ways.

∴ Total number of outcomes = 5

There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways.

∴ Favourable number of outcomes = 3

So, the required probability = \(\frac { 3 }{ 5 } \)

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 = \(\frac { 4 }{ 5 } \)

**Ex 5.3 Class 8 Maths Question 6.**

Find the probabilities of the events given in Question 2.

**Solution:**

In a single throw of a die, we can get any one of the six numbers 1, 2, 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.

∴ Favourable number of outcomes = 3

Hence, \(P\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)

**(ii)** Let A denote the event “not getting a prime number”. Clearly, event A occurs if we obtain 1, 4, 6 as an outcome.

∴ Favourable number of outcomes = 3

Hence, \(P\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)

**(iii)** The event “getting a number greater than 5” will occur if we obtain the number 6.

∴ Favourable number of outcomes = 1

Hence, required probability = \(\frac { 1}{ 6 } \)

**(iv)** The event “getting a number not greater than 5” will occur if we obtain one of the numbers 1, 2, 3, 4, 5.

∴ Favourable number of outcomes = 5

Hence, required probability = \(\frac { 5 }{ 6 } \)

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