When two dice are tossed together, there are `(6 xx 6)` outcomes. Let S be the sample space. Then, n(S) = 36.
Let `E_(1) =` event of getting a doublet,
and `E_(2) =` event of getting a total of 6.
Then, `E_(1) = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}`
and `E_(2) = {(2, 4), (3, 3), (4, 2)}.`
`therefore (E_(1) nn E_(2)) = {(3, 3)}.`
Thus, `n(E_(1)) = 6, n(E_(2)) = 3 and n(E_(1) nn E_(2)) = 1.`
`therefore P(E_(1)) = (n(E_(1)))/(n(S)) = 6/36 = 1/6, P(E_(2)) = (n(E_(2)))/(n(S)) = 3/36 = 1/12`
and `P(E_(1) nn E_(2)) = (n(E_(1) nn E_(2)))/(n(S)) = 1/36.`
`therefore` P(getting a doublet or a total of 6)
`= P(E_(1) or E_(2)) = P(E_(1) uu E_(2))`
`= P(E_(1)) + P(E_(2)) - P(E_(1) nn E_(2))" "` [by the addition theorem]
`= (1/6 + 1/12 - 1/36) = 8/36 = 2/9.`
Hence, the required probability is `2/9.`
