# ML Aggarwal Data Handling Exe-19.3 Class 8 ICSE Ch-19 Maths Solutions

ML Aggarwal Data Handling Exe-19.3 Class 8 ICSE Ch-19 Maths Solutions. We Provide Step by Step Answer of Exe-19.3 Questions for Data Handling as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Data Handling Exe-19.3 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-19 | Data Handling |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-19.3 Questions |

Edition | 2023-2024 |

**Data Handling Exe-19.3**

ML Aggarwal Class 8 ICSE Maths Solutions

**Page-338**

**Question 1. List the outcomes you can see in these experiments.**

**Answer:**

(i) Outcomes in spinning wheel = A, A, A, B, C, D.

(ii) Outcomes in drawing a ball from a bag containing 5 identical balls

of different colours, says Red, black, green, blue, yellow.

**Question 2. A die is rolled once. Find the probability of getting**

(i) an even number

(ii) a multiple of 3

(iii) not a multiple of 3

**Answer:**

Total outcomes of a die when rolled once:

1, 2, 3, 4, 5, 6 = 6

**(i) An even number: 2, 4, 6**

i.e, Favourable outcomes = 3

Probability P(E) = 3 / 6

= 1 / 2

**(ii) Multiple of 3 = 3, 6**

i.e, Favourable outcomes = 2

Probability P(E) = 2 / 6

= 1 / 3

**(iii) Not a multiple of 3 = 1, 2, 4, 5**

i.e Favourable outcomes = 4

Probability P(E) = 4 / 6

= 2 / 3

**Question 3. Two coins are tossed together. Find the probability of getting**

(i) two tails

(ii) at least one tail

(iii) no tail

**Answer:**

When two coins are tossed together, then

Total outcomes = 2 × 2 = 4

i.e. HH, HT, TH, TT

(i) Favourable outcomes of getting two tails = 1

∴ Probability P(E) = 1/4

(ii) Favourable outcomes of getting at least one tail

TH, HT, TT = 3

∴ Probability P(E) = 3/4

(iii) Favourable outcomes of getting No tail: HH = 1

∴ Probability P(E) = 1/4

**Question 4. Three coins are tossed together. Find the probability of getting**

(i) at least two heads

(ii) at least one tail

(iii) at most one tail.

**Answer:**

Three coins are tossed together.

Total outcomes = 8

= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT

**(i) Favourable outcomes of getting atleast two heads = HHH, HHT, HTH, THH**

= 4 in numbers

Probability P(E) = (Number of favourable outcomes) / (Number of possible outcomes)

= 4 / 8

= 1 / 2

**(ii) Favourable outcomes of getting atleast one tail = HHT, HTH, HTT, TTT, THH, THT, TTH**

= 7 in numbers

Probability P(E) = (Number of favourable outcomes) / (Number of possible outcomes)

= 7 / 8

**(iii) Favorable outcomes of getting atmost one tail = HHH, HHT, HTH, THH**

= 4 in numbers

Probability P(E) = (Number of favourable outcomes) / (Number of possible outcomes)

= 4 / 8

= 1 / 2

**Question 5. A box contains 600 screws, one tenth are rusted. One screw is taken out at random from the box. Find the probability that it is**

(i) a rusted screw

(ii) not a rusted screw

**Answer:**

Rusted screw = (1 / 10) of 600

= (1 / 10) × 600

= 60 seconds

**(i) Favourable outcomes of picking rusted screw = 60**

Probability P(E) = 60 / 600

= 1 / 10

**(ii) Probability (of not rusted screw) = 1 – Probability (of rusted screw)**

= 1 – 1 / 10

= (10 – 1) / 10

= 9 / 10

**Question 6. A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel?**

**Answer:**

‘TRIANGLE’

Total number of outcomes = 8

Vowels : I, A, E = 3

∴ Probability P(E) = 3/8

**Question 7. A bag contains 5 red, 6 black and 4 white balls. A ball is drawn at random from the bag, find the probability the ball is drawn is**

(i) white

(ii) not black

(iii) red or black

(iv) neither red nor black

**Answer:**

In a bag, there are 5 red, 6 black and 4 white balls.

Total number of outcomes = 5 + 6 + 4=15

(i) Probability of white ball = 4/15

(ii) Probability of not black (5 + 4 = 9) balls = (9/15) = (3/5)

(iii) Probability of red or black ball (5 + 6 = 11)= 11/15

(iv) Probability of ball which is neither red nor black, white ball (4) P(E) = 4/15

**Question 8. A box contains 17 cards numbered 1, 2, 3, ……….,17 and are mixed thoroughly. A card is drawn at random from the box. Find the probability that the number on the card is**

(i) odd

(ii) even

(iii) prime

(iv) divisible by 3

(v) divisible by 2 and 3 both

**Answer:**

A box contains 17 cards numbered 1 to 17

Total number of outcomes = 17

**(i) Card bearing odd number (1, 3, 5, 7, 9, 11, 13, 15, 17) = 9**

∴Probability P(E) = 9/17

**(ii) Even number (2, 4, 6, 8, 10, 12, 14, 16) = 8**

∴Probability P(E) = 8/17

**(iii) Prime numbers {2, 3, 5, 7, 11, 13, 17) = 7**

∴Probability P(E) = 7/17

**(iv) Numbers divisible by 3 = 3, 6, 9, 12, 15 = 5**

∴Probability P(E) = 5/17

**(v) Number divisible by 2 and 3 both 6, 12 = 2**

∴Probability P(E) = 2/17

**Question 9. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is:**

(i) an ace

(ii) a red card

(iii) neither a king nor a queen

(iv) a red face card or an ace

(v) a card of spade

(vi) non-face card of red colour.

**Answer:**

Total number of playing cards = 52

One card is drawn

**(i) An ace = 4**

Probability P(E) = 4 / 52

= 1 / 13

**(ii) A red card = 13 + 13 = 26**

Probability P(E) = 26 / 52

= 1 / 2

**(iii) Neither a king nor a queen**

Number of cards = 52 – (4 + 4)

= 52 – 8

= 44

Probability P(E) = 44 / 52

= 11 / 13

**(iv) A red face card = 6**

Probability P(E) = 6 / 52

= 3 / 26

**(v) A card of spade or an ace = 13 + 3**

= 16

Probability P(E) = 16 / 52

= 4 / 13

**(vi) Non-face card of red colour = 26 – 6**

= 20

Probability P(E) = 20 / 52

= 5 / 13

**Data Handling Exe-19.3**

ML Aggarwal Class 8 ICSE Maths Solutions

**Page-339**

**Question 10. In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.**

**Answer:**

Number of prized tickets = 5

Number of blank tickets = 995

Total number of tickets = 5 + 995 = 1000

Probability of prized ticket

Probability of prized ticket P(E) = (Number of favourable outcomes) / (Number of possible outcomes)

= 5 / 1000

= 1 / 200

**The probability of his winning the prize is 1 / 200.**

— End of Data Handling **Exe-19.3 **Class 8 ICSE Maths Solutions :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

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