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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 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. Select the correct answer from above options

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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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