Explaining Conditional Probability in Everyday Situations
Statistics and Probability · Flashcards · Application
STATISTICS AND PROBABILITY · FLASHCARDS
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QUESTION
Find
P
(
A
∣
B
)
P(A\mid B)
P
(
A
∣
B
)
given
P
(
A
∩
B
)
=
0.12
P(A\cap B)=0.12
P
(
A
∩
B
)
=
0.12
and
P
(
B
)
=
0.30
P(B)=0.30
P
(
B
)
=
0.30
.
QUESTION
Compute
P
(
A
∩
B
)
P(A\cap B)
P
(
A
∩
B
)
given
P
(
A
∣
B
)
=
0.25
P(A\mid B)=0.25
P
(
A
∣
B
)
=
0.25
and
P
(
B
)
=
0.60
P(B)=0.60
P
(
B
)
=
0.60
.
QUESTION
Find
P
(
A
)
P(A)
P
(
A
)
if
A
A
A
and
B
B
B
are independent,
P
(
A
∣
B
)
=
0.30
P(A\mid B)=0.30
P
(
A
∣
B
)
=
0.30
, and
P
(
B
)
>
0
P(B)>0
P
(
B
)
>
0
.
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ANSWER
0.30
0.30
0.30
. Independence means
P
(
A
∣
B
)
=
P
(
A
)
P(A\mid B)=P(A)
P
(
A
∣
B
)
=
P
(
A
)
.
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