In thrown 3 dice together, the number of all possible outcomes is
(
6
×
6
×
6
)
=
216
(
.
Let E = event of getting a total of at least 6.
Then,
¯¯¯
E
=
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
(
¯¯¯
E
)
=
10
⇒
P
(
not
E
)
=
P
(
¯¯¯
E
)
=
n
(
¯¯¯
E
)
n
(
S
)
=
10
216
=
5
108
.
n
⇒
P
(
E
)
=
1
−
P
(
not
E
)
=
(
1
−
5
108
)
=
103
108
.
⇒
Hence, the required probability is
103
108
103
.