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