Bayes’ theorem in medicine_
stud
.doc
// This file prepares students for
// homework #3 and quizzes #4c, d, e.
// It continues the file
//
Conditional_probability_2_stud.doc
Feb 22’10
BAYES’ THEOREM
// See
handout #5
Suppose an event
A
can occur only with one of
k
mutually exclusive
events
B
1
,
B
2
, …,
B
k
that represent a set of
all
possible outcomes (a sample space) of an experiment. The
events
B
i
are called
hypotheses
associated with the occurrence of an event
A
.
M.V.:
The events B
i
represent actually our
hypotheses
about possible reasons due to
which an event A may occur
.
Since the events
AB
i
=
A∩B
i
and
AB
j
=
A∩B
j
are
mutually exclusive
at
i ≠ j
, the
event A is a
sum of events
A
1
=
A∩B
1
,
A
2
= A∩B
2
, …,
A
k
= A∩B
k
, and in accordance
with the
theorem (5.2) of addition of probabilities
we can write
k
k
P(A)
=
Σ P(A
i
)
=
Σ
P(A∩B
i
)
(5.6)
i =1
i =1
In accordance with the product rule (5.4a)
P(A∩B
i
) = P(B
i
)*P(AB
i
)
; thus, we can
represent (5.6) as
k
k
P(A)
=
Σ
P(A
i
)
=
Σ
P(B
i
)*P(AB
i
)
(5.7)
i =1
i =1
The relation (5.7) is called
a formula for total probability
. It allows us to determine the
probability
P(A)
of the event
A
, if the probability
P(B
i
)
of each hypothesis
B
i
and
conditional probability
P(AB
i
)
of an event
A
given event
B
i
are known.
When the probability
P(B
i
)
and conditional probabilities
P(AB
i
)
are known, we can
find the
conditional probabilities
P(B
i
A)
of hypotheses
B
i
which represent for us main
interest. By combining the relations (5.4a) and (5.4b) we obtain
P(
A
∩
B
i
) = P(
B
i
)*P(A
B
i
) = P(
A
)*P(
B
i

A
)
from which
P(
B
i

A
) = P(
B
i
)*P(A
B
i
)/P(
A
)
(5.8)
Now by substituting the relation (5.7) in the denominator of (5.8) we obtain the
Bayes’
formula
P(
B
i
)*P(
A

B
i
)
P(
B
i
)*P(
A

B
i
)
P(
B
i
)*P(
A

B
i
)
P(
B
i

A
) =  =  = 
(5.9)
P(
A
)
k
k
Σ
P(
B
i
)*P(
A

B
i
)
Σ P(
A
∩
B
i
)
i =1
i =1
1
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View Full DocumentIn a special case, when there are only two mutually exclusive
complementary
hypotheses
B
1
and
B
2
the formula (5.9) leads to the following expressions.
P(B
1

A
) = P(B
1
)*P(AB
1
) / [P(B
1
)*P(AB
1
) + P(B
2
)*P(AB
2
)]
(5.9#)
P(B
2

A
) = P(B
2
)*P(AB
2
) / [P(B
1
)*P(AB
1
)
+ P(B
2
)*P(AB
2
)]
(5.9##)
_
Sometimes it may be convenient to introduce new notations
B
1
= B
and
B
2
= B
in two
latter formulas. In such a case these formulas take the forms:
_
_
P(B
A
) = P(B)*P(
A
B)/P(
A
) = P(B)*P(
A
B) / [P(B)*P(
A
B) + P(B)*P(
A
B)]
(5.9a)
_
_
_
_
_
_
_
P(B
A
) = P(B)*P(
A
B)/P(A) = P(B)*P(
A
B) / [P(B)*P(
A
B) + P(B)*P(
A
B)]
(5.9b)
//////////////////////////////// END OF HANDOUT #5 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
//
See the file
Example of using of BAYES’ formula.doc
or beginning of the
homework #3
Using Tree Diagrams for Solving Bayes’ Theorem Problems
P(AB
1
)
P(B
1
)
/
A
P(
A
∩B
1
)
= P(B
1
)*P(AB
1
)

B
1
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 Spring '05
 Kzaer
 Conditional Probability, Probability, Probability theory, Type I and type II errors, *p

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