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Conditional probability

PROOF. Let n be the total number of uniquely possible, equiprobable and incompatible outcomes; m a
number of outcomes leading to the occurrence of the event A; l the number of outcomes leading to the
occurrence of the event B and k the number of outcomes leading to the occurrence of both A and B. Then

k km m k
P(AB) = = = · ,
n nm n n

but m = P(A), k = P A (B) and hence
n m

P(AB) = P(A) · P A (B).

From (3.1) we obtain
P(A ∩ B)
P(A|B) = (3.2)
P(B)

and
P(B ∩ A)
P(B|A) = . (3.3)
P(A)
In terms of Venn diagrams, we may think of P(B|A) as the probability of B in the reduced sample
space defined by A. Thus, if two events A and B are mutually exclusive then

P(A|B) = 0 = P(B|A). (3.4)

It follows from (3.1) that the probability of joint appearance of any number of events is equal
to the probability of one of them multiplied by: the conditional probability of another relative
to the first event, the conditional probability of the third event relative to the intersection of two
first events, etc. by the conditional probability of the last event relative to the intersection of all
preceding ones:

P(A 1 A 1 . . . A n ) = P(A 1 )P(A 2 /A 1 )P(A 3 /A 1 A 2 ) . . . P(A n /A 1 A 2 . . . A n−1 ). (3.5)

It is easy to see that conditional probabilities have properties analogous to those of ordinary
probabilities. For example,

1. 0 ≤ P A (B) ≤ 1

2. If A and B are incompatible, so that AB = ∅, then P A (B) = ∅.

3. If A implies B, so that A ⊂ B then P A (B) = 1.

If the events A and B are independent then the probability for their product is defined by the
formula
P(AB) = P(A)P(B). (3.6)

Example 3.2. The break in an electric circuit occurs when at least one out of
three elements connected in series is out of order. Compute the probability that
the break in the circuit will not occur if the elements may be out of order with
the respective probabilities 0.3, 0.4 and 0.6. How does the probability change if
the first element in never out of order? ,

Solution. The required probability equals the probability that all three elements in series are
working. Let A k (k = 1, 2, 3) denote the event that the k-th element functions. Then p =
p(A) = P(A 1 A 2 A 3 ). Since the events may be assumed independent, p = P(A 1 ) = P(A 2 ) =
P(A 3 ) = 0.7·0.6·0.4 = 0.168. If the first element is not out of order, then p = p(A 2 A 3 ) = 0.24.

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