Class 8  >  ML Aggarwal Solution: Class 8 Math  >  ML Aggarwal: Data Handling - 3

ML Aggarwal: Data Handling - 3 Notes | Study ML Aggarwal Solution: Class 8 Math - Class 8

Document Description: ML Aggarwal: Data Handling - 3 for Class 8 2022 is part of ML Aggarwal Solution: Class 8 Math preparation. The notes and questions for ML Aggarwal: Data Handling - 3 have been prepared according to the Class 8 exam syllabus. Information about ML Aggarwal: Data Handling - 3 covers topics like and ML Aggarwal: Data Handling - 3 Example, for Class 8 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for ML Aggarwal: Data Handling - 3.

Introduction of ML Aggarwal: Data Handling - 3 in English is available as part of our ML Aggarwal Solution: Class 8 Math for Class 8 & ML Aggarwal: Data Handling - 3 in Hindi for ML Aggarwal Solution: Class 8 Math course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: ML Aggarwal: Data Handling - 3 Notes | Study ML Aggarwal Solution: Class 8 Math - Class 8
Download, print and study this document offline
 Page 1


 
1. List the outcomes you can see in these experiments 
 
Solution: 
(i) The outcomes in spinning wheel = A, A, A, B, C, D 
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls 
of different colours = White, Red, Blue, Green, Yellow 
 
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 
Solution: 
Total outcomes of a die when rolled once: 
1, 2, 3, 4, 5, 6 = 6 
Page 2


 
1. List the outcomes you can see in these experiments 
 
Solution: 
(i) The outcomes in spinning wheel = A, A, A, B, C, D 
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls 
of different colours = White, Red, Blue, Green, Yellow 
 
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 
Solution: 
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 
Therefore, 
Probability P(E) = 


 = 



 
(ii) Multiple of 3 = 3, 6 
i.e., Favourable outcomes = 2 
Therefore, 
Probability P(E) = 


 = 



 
(iii) Not a multiple of 3 = 1, 2, 4, 5 
i.e. Favourable outcomes = 4 
Therefore, 
Probability P(E) = 


 = 


 
 
3. Two coins are tossed together. Find the probability of getting 
(i) Two tails 
(ii) At least one tail 
(iii) No tail 
Solution: 
The total outcomes, when two coins are tossed together = 2 × 2 = 4 
Therefore, outcomes are, 
HH, HT, TH, TT 
(i) Favourable outcomes of getting two tails = 1 
Page 3


 
1. List the outcomes you can see in these experiments 
 
Solution: 
(i) The outcomes in spinning wheel = A, A, A, B, C, D 
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls 
of different colours = White, Red, Blue, Green, Yellow 
 
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 
Solution: 
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 
Therefore, 
Probability P(E) = 


 = 



 
(ii) Multiple of 3 = 3, 6 
i.e., Favourable outcomes = 2 
Therefore, 
Probability P(E) = 


 = 



 
(iii) Not a multiple of 3 = 1, 2, 4, 5 
i.e. Favourable outcomes = 4 
Therefore, 
Probability P(E) = 


 = 


 
 
3. Two coins are tossed together. Find the probability of getting 
(i) Two tails 
(ii) At least one tail 
(iii) No tail 
Solution: 
The total outcomes, when two coins are tossed together = 2 × 2 = 4 
Therefore, outcomes are, 
HH, HT, TH, TT 
(i) Favourable outcomes of getting two tails = 1 
Hence, 
Probability P(E) = 



 
(ii) Favourable outcomes of getting at least one tail = TH, HT, TT = 3 
Hence, 
Probability P(E) = 


 
(iii) Favourable outcomes of getting no tail = HH = 1 
Hence, 
Probability P(E) = 



 
 
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 
Solution: 
Three coins are tossed together 
Hence, 
Total outcomes = 8 
= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT 
(i) Favourable outcomes of getting at least two heads = HHH, HHT, HTH, 
THH 
= 4 in numbers 
Therefore, 
Page 4


 
1. List the outcomes you can see in these experiments 
 
Solution: 
(i) The outcomes in spinning wheel = A, A, A, B, C, D 
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls 
of different colours = White, Red, Blue, Green, Yellow 
 
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 
Solution: 
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 
Therefore, 
Probability P(E) = 


 = 



 
(ii) Multiple of 3 = 3, 6 
i.e., Favourable outcomes = 2 
Therefore, 
Probability P(E) = 


 = 



 
(iii) Not a multiple of 3 = 1, 2, 4, 5 
i.e. Favourable outcomes = 4 
Therefore, 
Probability P(E) = 


 = 


 
 
3. Two coins are tossed together. Find the probability of getting 
(i) Two tails 
(ii) At least one tail 
(iii) No tail 
Solution: 
The total outcomes, when two coins are tossed together = 2 × 2 = 4 
Therefore, outcomes are, 
HH, HT, TH, TT 
(i) Favourable outcomes of getting two tails = 1 
Hence, 
Probability P(E) = 



 
(ii) Favourable outcomes of getting at least one tail = TH, HT, TT = 3 
Hence, 
Probability P(E) = 


 
(iii) Favourable outcomes of getting no tail = HH = 1 
Hence, 
Probability P(E) = 



 
 
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 
Solution: 
Three coins are tossed together 
Hence, 
Total outcomes = 8 
= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT 
(i) Favourable outcomes of getting at least two heads = HHH, HHT, HTH, 
THH 
= 4 in numbers 
Therefore, 
Probability P(E) = 
    !"!# $% 	
   &''(# $% 	
 
= 


 
= 



 
(ii) Favourable outcomes of getting at least one tail = HHT, HTH, HTT, 
TTT, THH, THT, TTH 
= 7 in numbers 
Therefore, 
Probability P(E)  
= (Number of favourable outcome) / (Number of possible outcome) 
= 


 
(iii) Favorable outcomes of getting at most one tail = HHH, HHT, HTH, 
THH 
= 4 in numbers 
Therefore, 
Probability P(E) = 
    !"!# $% 	
   &''(# $% 	
 
= 


 
= 



 
 
5. Two dice are rolled simultaneously. Find the probability of getting 
(i) The sum as 7 
(ii) The sum as 3 or 4 
Page 5


 
1. List the outcomes you can see in these experiments 
 
Solution: 
(i) The outcomes in spinning wheel = A, A, A, B, C, D 
(ii) The outcomes in drawing a ball from a bag containing 5 identical balls 
of different colours = White, Red, Blue, Green, Yellow 
 
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 
Solution: 
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 
Therefore, 
Probability P(E) = 


 = 



 
(ii) Multiple of 3 = 3, 6 
i.e., Favourable outcomes = 2 
Therefore, 
Probability P(E) = 


 = 



 
(iii) Not a multiple of 3 = 1, 2, 4, 5 
i.e. Favourable outcomes = 4 
Therefore, 
Probability P(E) = 


 = 


 
 
3. Two coins are tossed together. Find the probability of getting 
(i) Two tails 
(ii) At least one tail 
(iii) No tail 
Solution: 
The total outcomes, when two coins are tossed together = 2 × 2 = 4 
Therefore, outcomes are, 
HH, HT, TH, TT 
(i) Favourable outcomes of getting two tails = 1 
Hence, 
Probability P(E) = 



 
(ii) Favourable outcomes of getting at least one tail = TH, HT, TT = 3 
Hence, 
Probability P(E) = 


 
(iii) Favourable outcomes of getting no tail = HH = 1 
Hence, 
Probability P(E) = 



 
 
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 
Solution: 
Three coins are tossed together 
Hence, 
Total outcomes = 8 
= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT 
(i) Favourable outcomes of getting at least two heads = HHH, HHT, HTH, 
THH 
= 4 in numbers 
Therefore, 
Probability P(E) = 
    !"!# $% 	
   &''(# $% 	
 
= 


 
= 



 
(ii) Favourable outcomes of getting at least one tail = HHT, HTH, HTT, 
TTT, THH, THT, TTH 
= 7 in numbers 
Therefore, 
Probability P(E)  
= (Number of favourable outcome) / (Number of possible outcome) 
= 


 
(iii) Favorable outcomes of getting at most one tail = HHH, HHT, HTH, 
THH 
= 4 in numbers 
Therefore, 
Probability P(E) = 
    !"!# $% 	
   &''(# $% 	
 
= 


 
= 



 
 
5. Two dice are rolled simultaneously. Find the probability of getting 
(i) The sum as 7 
(ii) The sum as 3 or 4 
(iii) Prime numbers on both the dice. 
Solution: 
Two dice are rolled simultaneously, then 
Total outcomes = 6 × 6 = 36 
(i) Sum as 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6 
Therefore, 
Probability P(E) = 
)!"!# $% 	
*$!# $% 	
 
= 


 
= 



 
(ii) The sum as 3 or 4 = (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) = 5 
Therefore, 
Probability P(E) = 
)!"!# $% 	
*$!# $% 	
 
= 


 
(iii) Prime numbers on both the side  
= (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5) 
= 9 
Therefore, 
Probability P(E) = 
)!"!# $% 	
*$!# $% 	
 
= 


 
= 



 
Read More
Download as PDF

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

Related Searches

Exam

,

Viva Questions

,

shortcuts and tricks

,

mock tests for examination

,

Free

,

video lectures

,

pdf

,

MCQs

,

Extra Questions

,

ML Aggarwal: Data Handling - 3 Notes | Study ML Aggarwal Solution: Class 8 Math - Class 8

,

ppt

,

study material

,

past year papers

,

Previous Year Questions with Solutions

,

Sample Paper

,

Summary

,

practice quizzes

,

Objective type Questions

,

ML Aggarwal: Data Handling - 3 Notes | Study ML Aggarwal Solution: Class 8 Math - Class 8

,

Important questions

,

Semester Notes

,

ML Aggarwal: Data Handling - 3 Notes | Study ML Aggarwal Solution: Class 8 Math - Class 8

;