Question

Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be

पर्याय

•  13/15

•  13/18

•  1/9

•  8/9

   

Answer 1

13/18

When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space is given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
      (2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
      (3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
      (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
      (5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
      (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ n(S) = 36
Let E be the event of getting the digits which are neither equal nor give a total of 9.
Then E' = event of getting either a doublet or a total of 9
Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
i.e. n(E') = 10
P(E') = \[\frac{n\left( E' \right)}{n\left( S \right)} = \frac{10}{36} = \frac{5}{18}\]

Hence, required probability P(E) = 1- P(E')
                                                   = \[1 - \frac{5}{18} = \frac{13}{18}\]