Let S be the sample space. Then, n(S) = 36.
(i) Let
E
1
E
be the event of getting the sum 5. Then,
E
1
=
{
(
1
,
4
)
,
(
2
,
3
)
,
(
3
,
2
)
,
(
4
,
1
)
}
⇒
n
(
E
1
)
=
4
.
E
∴
P
(
E
1
)
=
n
(
E
1
)
n
(
S
)
=
4
36
=
1
9
∴
⇒
⇒
odds in favour of
E
1
=
P
(
E
1
)
1
−
P
(
E
1
)
=
(
1
/
9
)
(
1
−
1
/
9
)
=
1
8
.
E
(ii) Let
E
2
E
be the event of getting the sum 6. Then,
E
2
=
{
(
1
,
5
)
,
(
2
,
4
)
,
(
3
,
3
)
,
(
4
,
2
)
,
(
5
,
1
)
}
⇒
n
(
E
2
)
=
5
.
E
∴
P
(
E
2
)
=
n
(
E
2
)
n
(
S
)
=
5
36
∴
⇒
⇒
odds against
E
2
=
1
−
P
(
E
2
)
P
(
E
2
)
=
(
1
−
5
36
)
(
5
/
36
)
=
31
5
.
E
