Conditional probability
A random event has been before determined as an event that can take place or not to take place for holding the set of conditions S. If there are no any other restrictions except for the conditions S for a calculation of the probability of an event then such a probability is unconditional; if there are other auxiliary conditions then the probability of an event is said to be conditional. For example, the probability of an event B is very often calculated with an auxiliary condition that an event A was happened. Observe that unconditional probability also, strictly speaking, is conditional since it is supposed that the conditions S hold.
The conditional probability PA(B) is the probability of the event B calculated in assumption that the event A has already happened.
Example. There are 3 white and 3 black balls in an urn. One takes out twice on one ball from the urn without replacement. Find the probability of appearance of a white ball at the second trial (the event B) if a black ball was extracted at the first trial (the event A).
Solution: There are 5 balls after the first trial, and 3 of them are white. The required probability is PA(B) = 3/5.
The conditional probability
of an event B
with the condition that an event A
has already happened is equal to:
where P (A) >
0.
Indeed, return to our example. The probability of
appearance of a white ball at the first trial P(A)
= 3/6 =
1/2. Find the probability P(AB) that
a black ball will be appeared at the first trial and a white ball –
at the second trial. The total number of events – a joint
appearance of two balls (indifferently from colour) is equal to the
number of allocations
The event AB
is favored 3
3 = 9 events from the total number. Consequently, P(AB)
= 9/30 = 3/10. Thus, PA(B)
= P(AB)/P(A) = (3/10)/(1/2) = 3/5.
Theorem of multiplication of probabilities
Theorem. The probability of joint appearance of two events is equal to the product of the probability of one of them on the conditional probability of another event calculated in assumption that the first event has already happened: P(AB) = P(A) PA(B)
Remark. P(BA) = P(B) PB(A).
Since the event BA does not differ from the event AB, P (AB) = P(B) PB(A).
Consequently, P(A) PA(B) = P(B) PB(A).
Corollary. The probability of joint appearance of several events is equal to the product of the probability of one of them on the conditional probabilities of all the rest events so that the probability of each subsequent event is calculated in assumption that all preceding events have already happened:
where
is the probability of the event An
calculated in assumption that the events A1,
A2,
…, An–1
have happened. In particular, for three events
Observe that the order of location of the events can be chosen any, i.e. it is indifferently which event is assumed the first, the second and et cetera.
Example. There are 3 conic and 7 elliptic cylinders at a collector. The collector has taken one cylinder, and then he has taken the second cylinder. Find the probability that the first taken cylinder is conic, and the second – elliptic.
Solution: The probability that the first cylinder will be conic (the event A) P(A) = 3/10. The probability that the second cylinder will be elliptic (the event B) calculated in assumption that the first cylinder is conic, i.e. the conditional probability PA (B) = 7/9. By theorem of multiplication, the required probability is P(AB) = P(A) PA(B) = (3/10) (7/9) = 7/30.
Example. There are 5 white, 4 black and 3 blue balls in an urn. Each trial consists in extracting at random one ball without replacement. Find the probability that a white ball will appear at the first trial (the event A), a black ball will appear at the second trial (the event B), and a blue ball will appear at the third trial (the event C).
Solution: The probability of appearance of a white ball at the first trial P(A) = 5/12. The probability of appearance of a black ball at the second trial calculated in assumption that a white ball has appeared at the first trial, i.e. the conditional probability PA(B) = 4/11. The probability of appearance of a blue ball at the third trial calculated in assumption that a white ball has appeared at the first trial, and a black ball has appeared at the second trial, i.e. the conditional probability PAB(C) = 3/10. The required probability is
