When two dice are tossed together, there are
(
6
×
6
)
(
outcomes. Let S be the sample space. Then, n(S) = 36.
Let
E
1
=
E
event of getting a doublet,
and
E
2
=
E
event of getting a total of 6.
Then,
E
1
=
{
(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
,
(
5
,
5
)
,
(
6
,
6
)
}
E
and
E
2
=
{
(
2
,
4
)
,
(
3
,
3
)
,
(
4
,
2
)
}
.
E
∴
(
E
1
∩
E
2
)
=
{
(
3
,
3
)
}
.
∴
Thus,
n
(
E
1
)
=
6
,
n
(
E
2
)
=
3
and
n
(
E
1
∩
E
2
)
=
1
.
n
∴
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
∩
E
2
)
=
n
(
E
1
∩
E
2
)
n
(
S
)
=
1
36
.
P
∴
∴
P(getting a doublet or a total of 6)
=
P
(
E
1
or
E
2
)
=
P
(
E
1
∪
E
2
)
=
=
P
(
E
1
)
+
P
(
E
2
)
−
P
(
E
1
∩
E
2
)
=
[by the addition theorem]
=
(
1
6
+
1
12
−
1
36
)
=
8
36
=
2
9
.
=
Hence, the required probability is
2
9
.
2
