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RD Sharma Solutions Class 8 Mathematics Solutions for Data Handling Probability Exercise 26.1 in Chapter 26 - Data Handling Probability
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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 than9
(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 a number other than 5 on any dice.
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
= 11/36
∴ Probability of getting outcomes for at least once is 11/36
(xiii) A number other than 5 on any dice.
Total number of outcomes in the above table 1 are 36
Number of outcomes having 5 on any die are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)
Number of outcomes having outcomes having 5 on any die are 15
Probability of getting 5 on any die is = Total numbers/Total number of outcomes
= 11/36
∴ Probability of getting 5 on any die is 11/36 Probability of not getting 5 on any die P (E) = 1 – P (E)
= 1 – 11/36
= (36-11)/36
= 25/36
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