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Let’s check out your problem:
If
P
(
B
)
=
0.55
,
P
(
A
∣
B
)
=
0.80
,
P
(
B
′
)
=
0.45
P(B)=0.55, P(A \mid B)=0.80, P\left(B^{\prime}\right)=0.45
P
(
B
)
=
0.55
,
P
(
A
∣
B
)
=
0.80
,
P
(
B
′
)
=
0.45
, and
P
(
A
∣
B
′
)
=
0.40
P\left(A \mid B^{\prime}\right)=0.40
P
(
A
∣
B
′
)
=
0.40
, find
P
(
B
∣
A
)
P(B \mid A)
P
(
B
∣
A
)
View step-by-step help
Home
Math Problems
Algebra 2
Evaluate expression when a complex numbers and a variable term is given
Full solution
Q.
If
P
(
B
)
=
0.55
,
P
(
A
∣
B
)
=
0.80
,
P
(
B
′
)
=
0.45
P(B)=0.55, P(A \mid B)=0.80, P\left(B^{\prime}\right)=0.45
P
(
B
)
=
0.55
,
P
(
A
∣
B
)
=
0.80
,
P
(
B
′
)
=
0.45
, and
P
(
A
∣
B
′
)
=
0.40
P\left(A \mid B^{\prime}\right)=0.40
P
(
A
∣
B
′
)
=
0.40
, find
P
(
B
∣
A
)
P(B \mid A)
P
(
B
∣
A
)
Rephrase Problem:
First, let's rephrase the "Find the conditional
probability
P
(
B
∣
A
)
P(B|A)
P
(
B
∣
A
)
."
Apply Bayes' Theorem:
To find
P
(
B
∣
A
)
P(B|A)
P
(
B
∣
A
)
, we will use Bayes' theorem, which states that
P
(
B
∣
A
)
=
P
(
A
∣
B
)
⋅
P
(
B
)
P
(
A
)
P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}
P
(
B
∣
A
)
=
P
(
A
)
P
(
A
∣
B
)
⋅
P
(
B
)
.
Calculate
P
(
A
)
P(A)
P
(
A
)
:
We already have
P
(
A
∣
B
)
=
0.80
P(A|B) = 0.80
P
(
A
∣
B
)
=
0.80
and
P
(
B
)
=
0.55
P(B) = 0.55
P
(
B
)
=
0.55
. We need to find
P
(
A
)
P(A)
P
(
A
)
, which is the total probability of
A
A
A
occurring.
Use Law of Total Probability:
To find
P
(
A
)
P(A)
P
(
A
)
, we use the law of total probability:
P
(
A
)
=
P
(
A
∣
B
)
⋅
P
(
B
)
+
P
(
A
∣
B
′
)
⋅
P
(
B
′
)
P(A) = P(A|B) \cdot P(B) + P(A|B') \cdot P(B')
P
(
A
)
=
P
(
A
∣
B
)
⋅
P
(
B
)
+
P
(
A
∣
B
′
)
⋅
P
(
B
′
)
.
Calculate
P
(
A
)
P(A)
P
(
A
)
:
We have
P
(
A
∣
B
)
=
0.80
P(A|B) = 0.80
P
(
A
∣
B
)
=
0.80
,
P
(
B
)
=
0.55
P(B) = 0.55
P
(
B
)
=
0.55
,
P
(
A
∣
B
′
)
=
0.40
P(A|B') = 0.40
P
(
A
∣
B
′
)
=
0.40
, and
P
(
B
′
)
=
0.45
P(B') = 0.45
P
(
B
′
)
=
0.45
. Let's calculate
P
(
A
)
P(A)
P
(
A
)
.
\newline
P
(
A
)
=
(
0.80
×
0.55
)
+
(
0.40
×
0.45
)
P(A) = (0.80 \times 0.55) + (0.40 \times 0.45)
P
(
A
)
=
(
0.80
×
0.55
)
+
(
0.40
×
0.45
)
\newline
P
(
A
)
=
0.44
+
0.18
P(A) = 0.44 + 0.18
P
(
A
)
=
0.44
+
0.18
\newline
P
(
A
)
=
0.62
P(A) = 0.62
P
(
A
)
=
0.62
Calculate
P
(
B
∣
A
)
P(B|A)
P
(
B
∣
A
)
:
Now we can calculate
P
(
B
∣
A
)
P(B|A)
P
(
B
∣
A
)
using Bayes' theorem.
\newline
P
(
B
∣
A
)
=
P
(
A
∣
B
)
⋅
P
(
B
)
P
(
A
)
P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}
P
(
B
∣
A
)
=
P
(
A
)
P
(
A
∣
B
)
⋅
P
(
B
)
\newline
P
(
B
∣
A
)
=
(
0.80
⋅
0.55
)
0.62
P(B|A) = \frac{(0.80 \cdot 0.55)}{0.62}
P
(
B
∣
A
)
=
0.62
(
0.80
⋅
0.55
)
\newline
P
(
B
∣
A
)
=
0.44
0.62
P(B|A) = \frac{0.44}{0.62}
P
(
B
∣
A
)
=
0.62
0.44
Simplify Fraction:
Finally, we simplify the
fraction
to get the value of
P
(
B
∣
A
)
P(B|A)
P
(
B
∣
A
)
.
P
(
B
∣
A
)
=
0.44
0.62
≈
0.7097
P(B|A) = \frac{0.44}{0.62} \approx 0.7097
P
(
B
∣
A
)
=
0.62
0.44
≈
0.7097
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=
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(
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⋅
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i
)
=
\newline
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i
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+
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a
a
and
b
b
b
are real numbers.
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35
−
23
i
)
+
(
13
+
25
i
)
=
\newline
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i
)
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