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Byju's Answer
Standard XII
Mathematics
Intersection of Sets
For any 2 eve...
Question
For any 2 events A and B in a sample space which one of the following is incorrect?
A
P
(
A
/
B
)
≥
P
(
A
)
+
P
(
B
)
−
1
P
(
B
)
,
P
(
B
)
≠
0
, is always true
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B
P
(
A
∩
¯
¯¯
¯
B
)
=
P
(
A
)
−
P
(
A
∩
B
)
doen't hold
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C
P
(
A
∪
B
)
=
1
−
P
(
¯
¯¯
¯
A
)
P
(
¯
¯¯
¯
B
)
if A & B are independent
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D
P
(
A
∪
B
)
=
1
−
P
(
¯
¯¯
¯
A
)
P
(
¯
¯¯
¯
B
)
, if A & B are disjoint.
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Solution
The correct option is
A
P
(
A
/
B
)
≥
P
(
A
)
+
P
(
B
)
−
1
P
(
B
)
,
P
(
B
)
≠
0
, is always true
Consider
2
events
A
ξ
B
P
(
A
B
)
≥
P
(
A
)
+
P
(
B
)
−
1
P
(
B
)
,
P
(
B
)
≠
0
,
is always true.
This condition is not correct.
P
(
A
B
)
=
N
u
m
b
e
r
o
f
e
l
e
m
e
n
t
a
r
y
e
v
e
n
t
s
f
a
v
o
r
a
b
l
e
A
∩
B
N
u
m
b
e
r
o
f
e
l
e
m
e
n
t
a
r
y
e
v
e
n
t
s
f
a
v
o
r
a
b
l
e
t
o
B
P
(
A
B
)
=
n
(
A
∩
B
)
n
(
B
)
=
n
(
A
∩
B
)
n
(
s
)
n
(
B
)
n
(
s
)
∴
P
(
A
B
)
=
(
A
∩
B
)
(
B
)
Similarly
P
(
B
A
)
=
(
A
∩
B
)
(
A
)
Hence, the answer is
P
(
A
B
)
≥
P
(
A
)
+
P
(
B
)
−
1
P
(
B
)
,
P
(
B
)
≠
0
,
is always true.
Suggest Corrections
0
Similar questions
Q.
Assertion :If A & B are independent with probabilities
P
(
A
)
=
1
2
,
P
(
B
)
=
1
5
then
P
(
A
∪
B
)
=
3
5
. Reason: For independent events A & B the result
P
(
A
∩
B
)
=
P
(
A
)
P
(
B
)
holds true.
Q.
Let
A
and
B
be two events such that
P
(
A
∪
B
)
=
P
(
A
)
+
P
(
B
)
−
P
(
A
)
P
(
B
)
. If
0
<
P
(
A
)
<
1
and
0
<
P
(
B
)
<
1
, then
P
(
A
∪
B
)
′
is equal to
Q.
If A and B are two events such that
P
(
A
∪
B
)
=
P
(
A
∩
B
)
,
then the true relation is
Q.
If A and B are any two events such that P (A) + P (B) − P (A and B) = P (A), then (A) P (B|A) = 1 (B) P (A|B) = 1 (C) P (B|A) = 0 (D) P (A|B) = 0
Q.
Let
0
<
P
(
A
)
<
1
,
0
<
P
(
B
)
<
1
and
P
(
A
∪
B
)
=
P
(
A
)
+
P
(
B
)
−
P
(
A
)
P
(
B
)
, then
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