Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Data Handling-IV (Probability) - Exercise 26
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Page 1
1. The probability that it will rain tomorrow is 0.85. What is the probability that it
will not rain tomorrow?
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
? Probability of not raining = P (A) = 1 – 0.85
= 0.15
2. A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = Total prime numbers/Total number of outcomes
= 3/6
= 1/2
? Probability of getting a prime number = 1/2
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is Total numbers/Total number of outcomes
= 2/6
= 1/3
? Probability of getting 2 and 4 is 1/3
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes
= 4/6
= 2/3
? Probability of getting a multiple of 2 or 3 is 2/3
Page 2
1. The probability that it will rain tomorrow is 0.85. What is the probability that it
will not rain tomorrow?
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
? Probability of not raining = P (A) = 1 – 0.85
= 0.15
2. A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = Total prime numbers/Total number of outcomes
= 3/6
= 1/2
? Probability of getting a prime number = 1/2
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is Total numbers/Total number of outcomes
= 2/6
= 1/3
? Probability of getting 2 and 4 is 1/3
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes
= 4/6
= 2/3
? Probability of getting a multiple of 2 or 3 is 2/3
3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) An even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
Let us construct a table.
Here the first number denotes the outcome of first die and second number denotes the
outcome of second die.
(i) 8 as the sum
Total number of outcomes in the above table are 36
Number of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)
Therefore numbers of outcomes having 8 as sum are 5
Probability of getting numbers of outcomes having 8 as sum is = Total numbers/Total
number of outcomes
= 5/36
? Probability of getting numbers of outcomes having 8 as sum is 5/36
Page 3
1. The probability that it will rain tomorrow is 0.85. What is the probability that it
will not rain tomorrow?
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
? Probability of not raining = P (A) = 1 – 0.85
= 0.15
2. A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = Total prime numbers/Total number of outcomes
= 3/6
= 1/2
? Probability of getting a prime number = 1/2
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is Total numbers/Total number of outcomes
= 2/6
= 1/3
? Probability of getting 2 and 4 is 1/3
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes
= 4/6
= 2/3
? Probability of getting a multiple of 2 or 3 is 2/3
3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) An even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
Let us construct a table.
Here the first number denotes the outcome of first die and second number denotes the
outcome of second die.
(i) 8 as the sum
Total number of outcomes in the above table are 36
Number of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)
Therefore numbers of outcomes having 8 as sum are 5
Probability of getting numbers of outcomes having 8 as sum is = Total numbers/Total
number of outcomes
= 5/36
? Probability of getting numbers of outcomes having 8 as sum is 5/36
(ii) a doublet
Total number of outcomes in the above table are 36
Number of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
Number of outcomes as doublet are 6
Probability of getting numbers of outcomes as doublet is = Total numbers/Total number
of outcomes
= 6/36
= 1/6
? Probability of getting numbers of outcomes as doublet is 1/6
(iii) a doublet of prime numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of prime numbers are 3
Probability of getting numbers of outcomes as doublet of prime numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of prime numbers is 1/12
(iv) a doublet of odd numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of odd numbers are 3
Probability of getting numbers of outcomes as doublet of odd numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of odd numbers is 1/12
(v) a sum greater than 9
Total number of outcomes in the above table are 36
Number of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6,
5)
Number of outcomes having sum greater than 9 are 6
Probability of getting numbers of outcomes having sum greater than 9 is = Total
numbers/Total number of outcomes
= 6/36 = 1/6
? Probability of getting numbers of outcomes having sum greater than 9 is 1/6
Page 4
1. The probability that it will rain tomorrow is 0.85. What is the probability that it
will not rain tomorrow?
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
? Probability of not raining = P (A) = 1 – 0.85
= 0.15
2. A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = Total prime numbers/Total number of outcomes
= 3/6
= 1/2
? Probability of getting a prime number = 1/2
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is Total numbers/Total number of outcomes
= 2/6
= 1/3
? Probability of getting 2 and 4 is 1/3
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes
= 4/6
= 2/3
? Probability of getting a multiple of 2 or 3 is 2/3
3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) An even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
Let us construct a table.
Here the first number denotes the outcome of first die and second number denotes the
outcome of second die.
(i) 8 as the sum
Total number of outcomes in the above table are 36
Number of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)
Therefore numbers of outcomes having 8 as sum are 5
Probability of getting numbers of outcomes having 8 as sum is = Total numbers/Total
number of outcomes
= 5/36
? Probability of getting numbers of outcomes having 8 as sum is 5/36
(ii) a doublet
Total number of outcomes in the above table are 36
Number of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
Number of outcomes as doublet are 6
Probability of getting numbers of outcomes as doublet is = Total numbers/Total number
of outcomes
= 6/36
= 1/6
? Probability of getting numbers of outcomes as doublet is 1/6
(iii) a doublet of prime numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of prime numbers are 3
Probability of getting numbers of outcomes as doublet of prime numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of prime numbers is 1/12
(iv) a doublet of odd numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of odd numbers are 3
Probability of getting numbers of outcomes as doublet of odd numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of odd numbers is 1/12
(v) a sum greater than 9
Total number of outcomes in the above table are 36
Number of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6,
5)
Number of outcomes having sum greater than 9 are 6
Probability of getting numbers of outcomes having sum greater than 9 is = Total
numbers/Total number of outcomes
= 6/36 = 1/6
? Probability of getting numbers of outcomes having sum greater than 9 is 1/6
(vi) An even number on first
Total number of outcomes in the above table are 36
Number of outcomes having an even number on first are: (2, 1), (2, 2), (2, 3), (2, 4), (2,
5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6,
6)
Number of outcomes having an even number on first are 18
Probability of getting numbers of outcomes having an even number on first is = Total
numbers/Total number of outcomes
= 18/36
= 1/2
? Probability of getting numbers of outcomes having an even number on first is 1/2
(vii) An even number on one and a multiple of 3 on the other
Total number of outcomes in the above table are 36
Number of outcomes having an even number on one and a multiple of 3 on the other are:
(2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6)
Number of outcomes having an even number on one and a multiple of 3 on the other are
6
Probability of getting an even number on one and a multiple of 3 on the other is = Total
numbers/Total number of outcomes
= 6/36
= 1/6
? Probability of getting an even number on one and a multiple of 3 on the other is 1/6
(viii) Neither 9 nor 11 as the sum of the numbers on the faces
Total number of outcomes in the above table are 36
Number of outcomes having 9 nor 11 as the sum of the numbers on the faces are: (3, 6),
(4, 5), (5, 4), (5, 6), (6, 3) and (6, 5)
Number of outcomes having neither 9 nor 11 as the sum of the numbers on the faces are 6
Probability of getting 9 nor 11 as the sum of the numbers on the faces is = Total
numbers/Total number of outcomes
= 6/36
= 1/6
Probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P (E) =
1/6
?Probability of getting neither 9 nor 11 as the sum of the numbers on the faces is 1/6
Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is
Page 5
1. The probability that it will rain tomorrow is 0.85. What is the probability that it
will not rain tomorrow?
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
? Probability of not raining = P (A) = 1 – 0.85
= 0.15
2. A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = Total prime numbers/Total number of outcomes
= 3/6
= 1/2
? Probability of getting a prime number = 1/2
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is Total numbers/Total number of outcomes
= 2/6
= 1/3
? Probability of getting 2 and 4 is 1/3
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes
= 4/6
= 2/3
? Probability of getting a multiple of 2 or 3 is 2/3
3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) An even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
Let us construct a table.
Here the first number denotes the outcome of first die and second number denotes the
outcome of second die.
(i) 8 as the sum
Total number of outcomes in the above table are 36
Number of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)
Therefore numbers of outcomes having 8 as sum are 5
Probability of getting numbers of outcomes having 8 as sum is = Total numbers/Total
number of outcomes
= 5/36
? Probability of getting numbers of outcomes having 8 as sum is 5/36
(ii) a doublet
Total number of outcomes in the above table are 36
Number of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
Number of outcomes as doublet are 6
Probability of getting numbers of outcomes as doublet is = Total numbers/Total number
of outcomes
= 6/36
= 1/6
? Probability of getting numbers of outcomes as doublet is 1/6
(iii) a doublet of prime numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of prime numbers are 3
Probability of getting numbers of outcomes as doublet of prime numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of prime numbers is 1/12
(iv) a doublet of odd numbers
Total number of outcomes in the above table are 36
Number of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)
Number of outcomes as doublet of odd numbers are 3
Probability of getting numbers of outcomes as doublet of odd numbers is = Total
numbers/Total number of outcomes
= 3/36
= 1/12
? Probability of getting numbers of outcomes as doublet of odd numbers is 1/12
(v) a sum greater than 9
Total number of outcomes in the above table are 36
Number of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6,
5)
Number of outcomes having sum greater than 9 are 6
Probability of getting numbers of outcomes having sum greater than 9 is = Total
numbers/Total number of outcomes
= 6/36 = 1/6
? Probability of getting numbers of outcomes having sum greater than 9 is 1/6
(vi) An even number on first
Total number of outcomes in the above table are 36
Number of outcomes having an even number on first are: (2, 1), (2, 2), (2, 3), (2, 4), (2,
5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6,
6)
Number of outcomes having an even number on first are 18
Probability of getting numbers of outcomes having an even number on first is = Total
numbers/Total number of outcomes
= 18/36
= 1/2
? Probability of getting numbers of outcomes having an even number on first is 1/2
(vii) An even number on one and a multiple of 3 on the other
Total number of outcomes in the above table are 36
Number of outcomes having an even number on one and a multiple of 3 on the other are:
(2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6)
Number of outcomes having an even number on one and a multiple of 3 on the other are
6
Probability of getting an even number on one and a multiple of 3 on the other is = Total
numbers/Total number of outcomes
= 6/36
= 1/6
? Probability of getting an even number on one and a multiple of 3 on the other is 1/6
(viii) Neither 9 nor 11 as the sum of the numbers on the faces
Total number of outcomes in the above table are 36
Number of outcomes having 9 nor 11 as the sum of the numbers on the faces are: (3, 6),
(4, 5), (5, 4), (5, 6), (6, 3) and (6, 5)
Number of outcomes having neither 9 nor 11 as the sum of the numbers on the faces are 6
Probability of getting 9 nor 11 as the sum of the numbers on the faces is = Total
numbers/Total number of outcomes
= 6/36
= 1/6
Probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P (E) =
1/6
?Probability of getting neither 9 nor 11 as the sum of the numbers on the faces is 1/6
Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is
given by P (E) = 1 – 1/6 = (6-1)/5 = 5/6
? Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is
5/6
(ix) A sum less than 6
Total number of outcomes in the above table are 36
Number of outcomes having a sum less than 6 are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2,
2), (2, 3), (3, 1), (3, 2), (4, 1)
Number of outcomes having a sum less than 6 are 10
Probability of getting a sum less than 6 is = Total numbers/Total number of outcomes
= 10/36
= 5/18
? Probability of getting sum less than 6 is 5/18
(x) A sum less than 7
Total number of outcomes in the above table are 36
Number of outcomes having a sum less than 7 are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2,
1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)
Number of outcomes having a sum less than 7 are 15
Probability of getting a sum less than 7 is = Total numbers/Total number of outcomes
= 15/36
= 5/12
? Probability of getting sum less than 7 is 5/12
(xi) A sum more than 7
Total number of outcomes in the above table are 36
Number of outcomes having a sum more than 7 are: (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4,
6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Number of outcomes having a sum more than 7 are 15
Probability of getting a sum more than 7 is = Total numbers/Total number of outcomes
= 15/36
= 5/12
? Probability of getting sum more than 7 is 5/12
(xii) At least once
Total number of outcomes in the above table 1 are 36
Number of outcomes for at least once are 11
Probability of getting outcomes for at least once is = Total numbers/Total number of
outcomes
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The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Data Handling-IV (Probability) - Exercise 26 on the App
Students of Class 8 can study Data Handling-IV (Probability) - Exercise 26 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Data Handling-IV (Probability) - Exercise 26,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Data Handling-IV (Probability) - Exercise 26 is prepared as per the latest Class 8 syllabus.
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