Let S be the sample space. Then, n(S) = 52.
Let
E
1
E
= event of getting a spade,
and
E
2
E
= event of getting a king.
Then,
(
E
1
∩
E
2
)
=
(
event of getting a king of spades.
Clearly,
n
(
E
1
)
=
13
,
n
(
E
2
)
=
4
and
n
(
E
1
∩
E
2
)
=
1
.
n
∴
P
(
E
1
)
=
n
(
E
1
)
n
(
S
)
=
13
52
=
1
4
,
P
(
E
2
)
=
n
(
E
2
)
n
(
S
)
=
4
52
=
1
13
.
∴
and
P
(
E
1
∩
E
2
)
=
n
(
E
1
∩
E
2
)
n
(
S
)
=
1
52
.
P
∴
∴
P(getting a spade or a king)
=
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 for two events]
=
(
1
4
+
1
13
−
1
52
)
=
16
52
=
4
13
.
=
Hence, the required probability is
4
13
4
.
