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

## Data Handling

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.

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.

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 = $\frac { 4 }{ 7 }$

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 =$\frac { 9 }{ 10 }$

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 = $\frac { 4 }{ 5 }$

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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