Let S be the sample space. Then, n(S) = 52.
Let E1,E2andE3
be the event of getting a king, a heart and a red card repsectively. Then, n(E1)=4,n(E2)=13andn(E2)=13andn(E3)=26.
(E1∩E2)=
event of getting a king of hearts,
(E2∩E3)=
event of getting a heart [∵
a heart is a red cards also],
(E3∩E1)=
event of getting a red king,
and (E1∩E2∩E3)=
event of getting a king of hearts.
∴n(E1∩E2)=1,n(E2∩E3)=13,n(E3∩E1)=2
and n(E1∩E2∩E3)=1.
∴P(E1)=
n(E1)
n(S)
=
4
52
=
1
13
,P(E2)=(n
E2
n(S)
=
13
52
=
1
4
,
P(E3)=
n(E3)
n(S)
=
26
52
=
1
2
,P(E1∩E2)=
n(E1∩E2)
n(S)
=
1
52
,
P(E2∩E3)=
n(E2∩E3)
n(S)
=
2
52
=
1
26
and P(E1∩E2∩E3)=
n(E1∩E2∩E3)
n(S)
=
1
52
.
∴
P(getting a king or a heart or a red card)
=P(E1 or E2 or E3)=P(E1∪E2∪E3)
=P(E1)+P(E2)+P(E3)-P(E1∩E2)-P(E2∩E3)-P(E3∩E1)+P(E1∩E2∩E3)
=
P
(
E
1
)
+
P
(
E
2
)
+
P
(
E
3
)
−
P
(
E
1
∩
E
2
)
−
P
(
E
2
∩
E
3
)
−
P
(
E
3
∩
E
1
)
+
P
(
E
1
∩
E
2
∩
E
3
)
=(
1
13
+
1
4
+
1
2
-
1
52
-
1
4
-
1
26
+
1
52
)=
28
52
=
7
13
.
Hence, the required probability is
7
13
.
