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Statistics

Notes 3.2 Baye’s Theorem

Theorem 1 (Bayes’ Theorem) : If B , B , . . . , B are n events which constitute a partition of  and
1 2 n
A is an event of positive probability, then
P(B )P(A|B )
P(B | A) = r r
r n
 P(B )P(A|B )
j
j
j 1

foranyr, 1  r  n .
Proof: Observe that by definition,
P(A  B )
P(Br | A) = r
P(A)

P(B )P(A|B )
= r r , by (10)
P(A)
P(B )P(A|B )
= r r , by (12)
n
 P(B )P(A|B )
j
j

j 1
The proof is complete.
In the examples that follow, you will see a variety of situations in which Bayes’ theorem is
useful.

Example 16: It is known that 25 per cent of the people in a community suffer from TB. A
test to diagnose this’disease is such that the probability is 0.99 that a person suffering from it will
show a positive result indicating its presence. The same test has probability 0.20 that a person
not suffering from TB has a positive test result. If a randomly selected person from the community
has positive test result, let us find the probability that he has TB.

c
Let B denote the event that a randomly selected person has TB. Let B = B . Then from the given
1 2 j
information, P(B ) = 0.25, P(B ) = 0.75. Let A denote the event that the test for the randomly
1 2
selected person yields a positive result. Then P(A | B ) = 0.99 and P(A | B ) = 0.20. We need to
1 2
obtain P(B1 | A). By applying Bayes’ theorem we get
P(B )P(A|B )
P(B1 | A) = 1 1
P(B )P(A|B ) P(B )P(A|B )

1
1
2
2
Example 17: We have three boxes, each containing two covered compartments. The first
box has a gold coin in each compartment. The second box has a gold coin in one compartment
and a silver coin in the other. The third box has a silver coin in each of its compartments. We
choose a box at random and open a drawer at random. It contains a gold coin. We would like to
know the probability that the other compartment also has a gold coin.
Let B , B , B , respectively, denote the events that Box 1, Box 2 and Box 3 are selected. It is easy to
1 2 3
see that B , B , B constitute a partition of the sample space of the experiment.
1 2 3
Since the boxes are selected at random, we have
P(B ) = P(B ) = P(B ) = 1/3.
1 2 3

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