In thrown 3 dice together, the number of all possible outcomes is `(6 xx 6 xx 6) = 216`.
Let E = event of getting a total of at least 6.
Then, `bar(E) =` event of getting a total of less than 6.
= event of getting a total of 3 or 4 or 5.
`= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1(}`
Now, `n(bar(E)) = 10 rArr P("not "E) = P(bar(E)) = (n(bar(E)))/(n(S)) = 10/216 = 5/108.`
`rArr P(E) = 1 - P("not " E) = (1 - 5/108) = 103/108.`
Hence, the required probability is `103/108`.
