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Conditional probability

Bayes Formula

In practice we are often interested in a total group of incompatible events H 1 , H 2 , ..., H n whose
probabilities P(H i ) (i = 1, 2, . . . , n) are known. These events are not observable but one may
(A) (i = 1, 2, . . . , n) are known.
observe some event A whose conditional probabilities P H i
Assume that a trial was performed resulting in the appearance of the event A. Using this result
of the trial it is required to make some inferences about the events H 1 , H 2 , . . . , H n , namely to
determine their probabilities after the trial. In other words, it is necessary to find the conditional
probabilities of the events H 1 , H 2 , . . . , H n with respect to the event A.
From the probability multiplication theorem (3.1) follows
(A)
P(AH k ) = P(A)P A (H k ) = P(H k )P H k
whence
(A)
P(H k )P H k
P A (H k ) = .
P(A)
Substituting the expression of the probability of the event A from the formula of total probability
(3.7) we obtain
(A)
P(H k )P H k
P A (H k ) = ∑ n (k = 1, 2, . . . , n). (3.8)
i=1 P(H i )P H i (A)
This formula which solves the problem is called Bayes formula.
The probabilities P(H k ) (k = 1, 2, . . . , n) of the events H 1 , H 2 , . . . , H n before the trial are
usually calledprior probabilities, from the Latina priori, which means “primary” i.e. in our case
(A) (k = 1, 2, . . . , n) of the same events after
before the trial was performed. The probabilities P H k
the trial are called posterior probabilities, from the Latin a posteriori, which means “after”, i.e.
after the trial was performed.
Example 3.5. A telegraphic communications system transmits the signals dot and
dash. Assume that the statistical properties of obstacles are such that an average
of 2 of the dots and 1 of the dashes are changed. Suppose that the ratio between
5 3
the transmitted dots and the transmitted dashes is 5 : 3. What is the probability
that a received signal will be the same as the transmitted signal if
1. the received signal is dot,
2. the received signal is a dash. ,

Solution. Let A be the event that a dot is received, and B that a dash is received.
One can make two hypotheses: H 1 that the transmitted signal was a dot; and H 2 that the
P(H 1 ) 5
transmitted signal was a dash. By assumption, = . Moreover, P(H 1 ) + P(H 2 ) = 1.
P(H 2 ) 3
3
5
Therefore P(H 1 ) = , P(H 2 ) = .
8 8
One knows that
3 1 2 2
(B) = .
P H 1 (A) = , P H 2 (A) = , P H 1 (B) = , P H 2
5 3 5 3
The probabilities of A and B are determined from the total probability formula:
5 3 3 1 1 5 2 3 2 1
P(A) = · + · = , P(B) = · + · = .
8 5 8 3 2 8 5 8 3 2
The required probabilities are:
(A) 5 · 3 3
P(H 1 )P H 1 8 5
P A (H 1 ) = = = ;
P(A) 1 4
2
(A) 3 · 2 1
P(H 2 )P H 2 8 3
P A (H 2 ) = = = ;
P(A) 1 2
2

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