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
Glossary
opposite events – противоположные события
to extract – извлекать; without replacement – без возвращения
conditional probability – условная вероятность
preceding – предшествующий
conic – конический; elliptic – эллиптический
cylinder – валик; collector – сборщик
Exercises for Seminar 3
3.1. In a cash–prize lottery 150 prizes and 50 monetary winnings are played on every 10000 tickets. What is the probability of a winning indifferently monetary or prize for an owner of one lottery ticket equal to?
3.2. The events A, B, C and D form a complete group. The probabilities of the events are those: P(A) = 0,1; P(B) = 0,4; P(C) = 0,3. What is the probability of the event D equal to?
3.3. The probability that a shooter will beat out 10 aces at one shot is equal to 0,1 and the probability to beat out 9 aces is equal to 0,3. Find the probabilities of the following events: A – the shooter will beat out 8 or less aces; B – the shooter will beat out no less than 9 aces.
3.4. There are 10 details in a box, and 2 of them are non-standard. Find the probability that in randomly selected 6 details appears no more than one non-standard detail.
Direction: If A is «there is no non-standard details» and B is «there is one non-standard detail» then P(A + B) = P(A) + P(B) =…
3.5. An enterprise produces 95% standard products, and 86% of them have the first grade. Find the probability that a randomly taken product made at the enterprise will be standard and the first grade (grade – сорт).
3.6. Two dice are thrown. Find the conditional probability that each die lands on 5 if it is known that the sum of aces is divided on 5.
3.7. If two dice are rolled, what is the conditional probability that the first one lands on 4 given that the sum of the dice is 8?
3.8. In a certain community, 36 percent of the families own a dog, and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is
(a) the probability that a randomly selected family owns both a dog and a cat;
(b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?
3.9. An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace (a pile – стопка; a deck – колода; an ace – туз).
3.10. A coin is tossed until it will not land on the same side 2 times in succession. Find the probability that the experiment will terminate before the sixth tossing (in succession – подряд).