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Joint Entrance Examination
Mathematics
Probability
Question 251
Question 251 of 417
Medium
MCQ
P
(
A
∪
B
)
=
P
(
A
∩
B
)
P(A\cup B)=P(A\cap B)
P
(
A
∪
B
)
=
P
(
A
∩
B
)
. Find the INCORRECT statement.
A
A
,
B
A,B
A
,
B
equally likely
B
P
(
A
∩
B
′
)
=
0
P(A\cap B')=0
P
(
A
∩
B
′
)
=
0
C
P
(
A
′
∩
B
)
=
0
P(A'\cap B)=0
P
(
A
′
∩
B
)
=
0
D
P
(
A
)
+
P
(
B
)
=
1
P(A)+P(B)=1
P
(
A
)
+
P
(
B
)
=
1
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Official solution
From
P
(
A
∪
B
)
=
P
(
A
∩
B
)
P(A\cup B)=P(A\cap B)
P
(
A
∪
B
)
=
P
(
A
∩
B
)
, it follows
P
(
A
)
=
P
(
B
)
=
P
(
A
∩
B
)
=
P
(
A
∪
B
)
P(A)=P(B)=P(A\cap B)=P(A\cup B)
P
(
A
)
=
P
(
B
)
=
P
(
A
∩
B
)
=
P
(
A
∪
B
)
, so A, B, C hold, but D need not hold (e.g. if
P
(
A
)
=
P
(
B
)
=
0.5
P(A)=P(B)=0.5
P
(
A
)
=
P
(
B
)
=
0.5
works, but the equality doesn't force sum to 1 in general).
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