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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.

2 Answers

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Best answer

(i) 8 as the sum Total number of outcomes when a pair of die is thrown simultaneously is: 

Here the first number denotes the outcome of first die and second number the outcome of second die.

First/second die 1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6

Total number of outcomes in the above table are 36 

Numbers 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 

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{5}{36}\) 

(ii) a doublet Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) 

Therefore Numbers of outcomes as doublet are 6

Probability of getting numbers of outcomes as doublet is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

Therefore Probability of getting numbers of outcomes as doublet is = \(\frac{1}{6}\)

(iii) a doublet of prime numbers Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)

Therefore Numbers of outcomes as doublet of prime numbers are 3 

Probability of getting numbers of outcomes as doublet of prime numbers is

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\) 

Therefore Probability of getting numbers of outcomes as doublet of prime numbers is = \(\frac{1}{12}\)

(iv) a doublet of odd numbers Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)

Therefore Numbers of outcomes as doublet of odd numbers are 3 

Probability of getting numbers of outcomes as doublet of odd numbers is

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

(v) a sum greater than 9 Total numbers of outcomes in the above table 1 are 36

Numbers of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6, 5) 

Therefore Numbers of outcomes having sum greater than 9 are 6 

Probability of getting numbers of outcomes having sum greater than 9 is

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

Therefore Probability of getting numbers of outcomes having sum greater than 9 is = \(\frac{1}{6}\)

(vi) An even number on first 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers of outcomes having an even number on first are 18 

Probability of getting numbers of outcomes having An even number on first is

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{18}{36}\) = \(\frac{1}{2}\) 

(vii) an even number on one and a multiple of 3 on the other 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers 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

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

+1 vote
by (26.3k points)

 (viii) neither 9 nor 11 as the sum of the numbers on the faces 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers 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 

\(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

The probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P(E) = \(\frac{1}{6}\)

Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is given by P(\(\text{E}\) ) = \(1-\frac{1}{6}=\frac{6-1}{6}=\frac{5}{6}\)

Therefore probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces = P (\(\text{E}\)) = \(\frac{5}{6}\) 

Therefore Probability of getting neither 9 nor 11 as the sum of the numbers on the faces is = \(\frac{1}{6}\)

(ix) a sum less than 6 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers of outcomes having a sum less than 6 are 10

Probability of getting a sum less than 6 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\) 

Therefore Probability of getting sum less than 6 is = \(\frac{5}{18}\)

(x) a sum less than 7 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers of outcomes having a sum less than 7 are 15

Probability of getting a sum less than 7 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\) 

Therefore Probability of getting sum less than 7 is = \(\frac{5}{12}\)

(xi) a sum more than 7 Total numbers of outcomes in the above table 1 are 36

Numbers 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) 

Therefore Numbers of outcomes having a sum more than 7 are 15

Probability of getting a sum more than 7 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

Therefore Probability of getting sum more than 7 is = \(\frac{5}{12}\)

(xii) at least once 

Total numbers of outcomes in the above table 1 are 36 

Therefore Numbers of outcomes for atleast once are 11 

Probability of getting outcomes for atleast once is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{11}{36}\) 

Therefore Probability of getting outcomes for atleast once is = \(\frac{11}{36}\)

(xiii) a number other than 5 on any dice. 

Total numbers of outcomes in the above table 1 are 36 

Numbers 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) 

Therefore Numbers of outcomes having outcomes having 5 on any die are 15

Probability of getting 5 on any die is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{11}{36}\) = \(\frac{11}{36}\) 

Therefore Probability of getting 5 on any die is = \(\frac{11}{36}\)

Probability of not getting 5 on any die P( ) = 1 –P (E)

P(\(\text{E}\)) = \(1-\frac{11}{36}\) = \(\frac{36-11}{36}\) = \(\frac{25}{36}\)

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