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Byju's Answer
Standard X
Mathematics
Addition Theorem of Probability for 2 Events
If P A = 12...
Question
If
P
(
A
)
=
1
2
,
P
(
B
)
=
1
3
,
P
(
C
)
=
1
2
a
n
d
P
(
A
∩
B
)
=
P
(
B
∩
C
)
=
P
(
C
∩
A
)
=
1
4
a
n
d
P
(
A
∩
B
∩
C
)
=
1
8
then find
P
(
A
∪
B
∪
C
)
.
Open in App
Solution
Given,
P
(
A
)
=
1
2
,
P
(
B
)
=
1
3
,
P
(
C
)
=
1
2
a
n
d
P
(
A
∩
B
)
=
P
(
B
∩
C
)
=
P
(
C
∩
A
)
=
1
4
a
n
d
P
(
A
∩
B
∩
C
)
=
1
8
.......(1)
Now, we have,
P
(
A
∪
B
∪
C
)
=
P
(
A
)
+
P
(
B
)
+
P
(
C
)
−
P
(
A
∩
B
)
−
P
(
B
∩
C
)
−
P
(
A
∩
C
)
+
P
(
A
∩
B
∩
C
)
=
1
2
+
1
3
+
1
2
−
3
×
1
4
+
1
8
=
1
+
1
3
−
3
4
+
1
8
=
1
+
1
3
−
5
8
=
1
−
7
24
=
17
24
.
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0
Similar questions
Q.
Assertion :If A, B, C are three events such that
P
(
A
)
=
1
4
,
P
(
B
)
=
1
6
&
P
(
C
)
=
2
3
then events A, B, C are mutually exclusive. Reason: If
P
(
A
∪
B
∪
C
)
=
P
(
A
)
+
P
(
B
)
+
P
(
C
)
then A, B, C are mutually exclusive events.
Q.
If
A
,
B
,
C
are three events, then show that
P
(
A
∪
B
∪
C
)
=
P
(
A
)
+
P
(
B
)
+
P
(
C
)
−
P
(
A
∩
B
)
−
P
(
B
∩
C
)
−
P
(
C
∩
A
)
+
P
(
A
∩
B
∩
C
)
Q.
If
A
,
B
,
C
are three events show that
P
(
A
∪
B
∪
C
)
=
P
(
A
)
+
P
(
B
)
+
P
(
C
)
−
P
(
A
∩
B
)
−
P
(
b
∩
C
)
−
P
(
C
∩
A
)
+
P
(
A
∩
B
∩
C
)
Q.
If
P
(
A
)
=
0.7
,
P
(
B
)
=
0.55
,
P
(
C
)
=
0.5
,
P
(
A
∩
B
)
=
x
,
P
(
A
∩
C
)
=
0.45
,
P
(
B
∩
C
)
=
0.3
and
P
(
A
∩
B
∩
C
)
=
0.2
, then
Q.
If events are mutually exclusive and
P
(
A
)
=
1
3
,
P
(
B
)
=
1
3
,
P
(
C
)
=
1
4
, then
P
(
A
′
∩
B
′
∩
C
′
)
is equal to
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