# Excercise 5.3 Data Handling- NCERT Solution Class 8

Data Handling

Exercise 5.3

## Chapter 5 Ex.5.3 Question 1

List the outcomes you can see in these experiments.

(a) Spinning a wheel

(b) Tossing two coins together

**Solution**

**Video Solution**

**What is known?**

Two different condition.

**What is unknown?**

Outcomes of the given conditions.

**Reasoning: **

All the possibilities of an event are known as its outcomes.

**Steps:**

(a) There are four letters \(A, B, C\) and \(D\) in a spinning wheel. So, there are \(4\) outcomes.

(b) When two coins are tossed together there are four possible outcomes

\(HH, HT, TH, TT.\)

(Here \(HT\) means head on first coin and tail on second coin and so on.)

## Chapter 5 Ex.5.3 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**

**Video Solution**

**What is known?**

A situation in which a die is thrown.

When a dice is thrown,the possible outcomes are \(1,2,3,4,5\) and \(6.\)

**What is unknown?**

Outcomes of the situation under given conditions.

**Reasoning: **

All the possibilities of an event are known as its outcomes.

**Steps:**

(i)

(a) Outcomes of event of getting a prime number are \(2, 3\) and \(5.\)

(b) Outcomes of event of not getting a prime number are \(1, 4 \) and \(6.\)

(ii)

(a) Outcomes of event of getting a number greater than \(5\) is \(6.\)

(b) Outcomes of event of not getting a number greater than \(5\) are \(1, 2, 3, 4\) and \(5.\)

## Chapter 5 Ex.5.3 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 figure alongside)

**Solution**

**Video Solution**

**What is known?**

(i) Spinning wheel image

(ii) Deck of \(52\) playing cards

(iii) Image depicting Red and Green apples

**What is unknown?**

(i) Probability of the pointer stopping on \(D.\)

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

(iii) Probability of getting a red apple

**Reasoning: **

Probability

\[\begin{align}= \! \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}\]

**Steps:**

(a) In a spinning wheel, there are five pointers \(A, A, B, C, D.\) So, there are five outcomes. Pointer stops at \(D\) which is \(1\) outcome. So, the probability of the pointer stopping on \(\begin{align}D = \frac{1}{5}\end{align}\)

(b) There are \(4\) aces in a deck of \(52\) playing cards. So, there are \(4\) events of getting an ace.

So, probability of getting an ace \(\begin{align} = \frac{4}{{52}} = \frac{1}{{13}}\end{align}\)

(c) Total number of apples \(= 7\)

Number of red apples \(= 4\)

So, probability of getting red apple \(\begin{align} = \frac{4}{7}\end{align}\)

## Chapter 5 Ex.5.3 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 number less than \(6?\)

(v) getting a \(1-\)digit number?

**Solution**

**Video Solution**

**What is known?**

Numbers \(1\) to \(10\) are written on ten separate slips (one number on one slip), kept in a box and mixed well

**What is unknown?**

(i) Probability of getting a number \(6\)

(ii) Probability of getting a number less than \(6 \)

(iii) Probability of getting a number greater than \(6\)

(iv) Probability of getting a number less than \(6\)

(v) Probability of getting a \(1-\)digit number

**Reasoning: **

Probability\(\begin{align} = \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}\)

**Steps:**

(i) Outcome of getting a number \(6\) from ten separate slips is \(1\). Therefore, probability of getting a number \(\begin{align}6 = \frac{1}{{10}}\end{align}\)

(ii) Numbers less than \(6\) are \(1, 2, 3, 4\) and \(5\). So, there are \(5\) outcomes. Therefore, probability of getting a number less than

\[\begin{align} 6 = \frac{5}{{10}} = \frac{1}{2} \end{align}\]

(iii) Number greater than 6 are \(7, 8, 9, 10.\) So there are \(4\) possible outcomes. Therefore, probability of getting a number greater than

\[\begin{align} 6 = \frac{4}{{10}} = \frac{2}{5} \end{align}\]

(iv) One-digit numbers are

\(1, 2, 3, 4, 5, 6, 7, 8, 9\) out of ten.

Therefore, probability of getting a \(1\)-digit number

\[\begin{align} = \frac{9}{{10}} \end{align}\]

## Chapter 5 Ex.5.3 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**

**Video Solution**

**What is known?**

Spinning wheel with \(3\) green sectors, \(1\) blue sector and \(1\) red sector

**What is unknown?**

i) probability of getting a green sector

ii) probability of getting a non-blue sector

**Reasoning: **

Probability

\(\begin{align}= \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}\)

**Steps:**

Total number of sectors \(= 5\)

Total number of green sectors \(= 3\)

Therefore, probability of getting a green sector\(\begin{align} = \frac{3}{5} \end{align}\)

Total number of blue sectors \(= 1\)

Total number of non-blue sectors \(= 5 – 1 = 4\)

Therefore, probability of getting a non-blue sector \(\begin{align} = \frac{4}{5}\end{align}\)

## Chapter 5 Ex.5.3 Question 6

Find the probability of the events given in Question \(2.\)

**Solution**

**Video Solution**

**What is known?**

A situation in which a die is thrown.

**What is unknown?**

Probability of the events

**Reasoning: **

Probability

\(\begin{align} =\frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} \end{align}\)

**Steps:**

When a die is thrown, there are total six outcomes, i.e., \(1, 2, 3, 4, 5\) and \(6.\)

(i) (a) \(2, 3, 5\) are prime numbers. So, there are \(3\) outcomes out of \(6.\)

Therefore,probability of getting a prime number

\(\begin{align} = \frac{3}{6} = \frac{1}{2} \end{align}\)

(b) \(1, 4, 6\) are not the prime numbers. So, there are \(3\) outcomes out of \(6.\)

Therefore, probability of getting a prime number

\(\begin{align} = \frac{3}{6} = \frac{1}{2} \end{align}\)

(ii) (a) Only \(6\) is greater than \(5.\) So, there is one outcome out of \(6.\)

Therefore, probability of getting a number greater than

\(\begin{align} 5 = \frac{1}{6} \end{align}\)

(b) Numbers not greater than \(5\) are \(1, 2, 3, 4\) and \(5.\) So there are \(5\) outcomes out of \(6.\)

Therefore, probability of not getting a number greater than

\(\begin{align} 5 = \frac{1}{6} \end{align}\)