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) × P_A(B)

 

Remark. P(BA) = P(B) × P_B(A).

Since the event BA does not differ from the event AB, P(AB) = P(B) × P_B(A).

Consequently, P(A) × P_A(B) = P(B) × P_B(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 A _n calculated in assumption that the events A ₁, A ₂, …, A _n–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 P_A(B) = 7/9. By theorem of multiplication, the required probability is

P(AB) = P(A) × P_A(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 P_A(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 P_AB(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).

Exercises for Homework 3

3.11. By the statistical data of a repair shop 20 stops of a lathe are on the average: 10 – for change of a cutter; 3 – because of malfunction of a drive; 2 – because of delayed submission of details. The rest stops occur for other reasons. Find the probability of stop of the lathe for other reasons (repair shop – ремонтная мастерская; lathe – токарный станок; cutter – резец; malfunction – неисправность; drive – привод).

The answer: 0,25.

 

3.12. There are 30 TVs in a shop, and 20 of them are import. Find the probability that no less than 3 import TVs will be among 5 TVs sold for one day, assuming that the probabilities of purchase of TVs of different marks are identical.

The answer: 0,81.

 

3.13. If two dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is 11?

The answer: 0,5.

 

3.14. An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

The answer: 0,066.

 

3.15. Fifty-two percent of the students at a certain university are females. Five percent of the students in this university are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that

(a) this student is female, given that the student is majoring in computer science;

(b) this student is majoring in computer science, given that the student is female.

The answer: a) 0,4; b) 0,038.

 

3.16. Celine is undecided as to whether to take a French course or a chemistry course. She estimates that her probability of receiving an A grade would be 1/2 in a French course and 2/3 in a chemistry course. If Celine decides to base her decision on the flip of a coin, what is the probability that she gets an A in chemistry (grade – оценка)?

The answer: 0,33.

 

3.17. Suppose that an urn contains 8 red and 4 white balls. We draw 2 balls from the urn without replacement. If we assume that at each draw each ball in the urn is equally likely to be chosen, what is the probability that both balls drawn are red (to draw – тянуть)?

The answer: 0,424.

 

3.18. An urn contains 10 white, 15 black, 20 blue and 25 red balls. A ball is taken at random from the urn. Find the probability that the taken ball is:

a) white or black; b) blue or red.

 

3.19. Two shooters shoot in a target. Find the probability that the target will be struck at least one of the shooters if it is known that the probability of miss by both shooters is equal to 0,27.

 

3.20. Two cards are randomly selected from a pack of 36 playing cards. Find the probability that both cards are the same color (a pack – колода).

The answer: 0,486.

 

 

L E C T U R E 4

Independent events

 

An event B is said to be independent from an event A if appearance of the event A does not change the probability of the event B, i.e. if the conditional probability of the event B is equal to its unconditional probability:

 

P_A(B) = P(B) (*)

 

Since P(A) × P_A(B) = P(B) × P_B(A), by using (*) we have P(A) × P(B) = P(B) × P_B(A) and consequently P_B(A) = P(A), i.e. the event A does not depend from the event B.

Thus, if an event B does not depend from an event A then the event A does not depend from the event B; this means that the property of independence is mutual.

 

For independent events theorem of multiplication P(AB) = P(A) × P_A(B) has the following form:

P(AB) = P(A) × P(B) (**)

 

i.e. the probability of joint appearance of two independent events is equal to the product of the probabilities of these events. The equality (**) is accepted as a definition of independent events.

Two events are independent if the probability of their joint appearance is equal to the product of the probabilities of these events; otherwise they are dependent.

At practice one concludes on independence of events on the sense of a problem. For example, the probability of hit in a target by each of two guns does not depend on that whether another gun has hit in the target, therefore the events «the first gun has hit in the target» and «the second gun has hit in the target» are independent.

Example. Find the probability of joint hit in a target by two guns if the probability of hit in the target by the first gun (the event A) is equal to 0,8; and by the second gun (the event B) – 0,7.

Solution: The events A and B are independent, therefore by theorem of multiplication, the required probability P(AB) = P(A) × P(B) = 0,8 × 0,7 = 0,56.

Remark. If events A and B are independent, the events A and , and B, and are also independent.

Proof of Remark:

Then we have i.e. the events A and are independent. ð

 

Several events are pairwise independent if each two of them are independent. For example, events A, B and C are pairwise independent if the events A and B, A and C, B and C are independent.

Several events are independent in union (or just independent) if each two of them are independent and each event and all possible products of the rest events are independent. For example, if events A ₁, A ₂ and A ₃ are independent in union then the events A ₁ and A ₂, A ₁ and A ₃, A ₂ and A ₃, A ₁ and A ₂ A ₃, A ₂ and A ₁ A ₃, A ₃ and A ₁ A ₂ are independent.

Let's underline that a pairwise independence of several events does not imply their independence in union in general, i.e. the demand of independence of events in union is stronger than the demand of their pairwise independence.

Example. There are 4 coloured balls in an urn: one ball is coloured in red colour (A), one ball is coloured in blue colour (B), one ball is coloured in black colour (C) and one ball – in all these three colours (ABC). Find the probability that a ball extracted at random from the urn has red colour. Since two of four balls have red colour, P(A) = 2/4 = 1/2. Reasoning analogously, we have P(B) = 1/2, P(C) = 1/2. Assume now that the taken ball has blue colour, i.e. the event B has already happened. Is the probability that the extracted ball has red colour changed, i.e. is the probability of the event A changed? One ball of two balls having blue colour has also red colour, therefore the probability of the event A is still equal to 1/2. In other words, the conditional probability of the event A calculated in assumption that the event B has happened is equal to its unconditional probability. Consequently, the events A and B are independent. By analogy we have that the events A and C, B and C are independent. Thus, the events A, B and C are pairwise independent. Are these events independent in union? Let the extracted ball has two colours, for example, blue and black. What is the probability that this ball has also red colour? Since only one ball is coloured in all three colours, therefore the taken ball has also red colour. Thus, assuming that the events B and C have happened, we have the event A will necessarily happen. Consequently, this event is reliable and its probability is equal to 1. In other words, the conditional probability P_BC(A) = 1 of the event A is not equal to its unconditional probability P(A) = 1/2. Thus, the pairwise independent events A, B and C are not independent in union.

Corollary. The probability of joint appearance of several events that are independent in union is equal to the product of the probabilities of these events:

 

P(A₁A₂…A_n) = P(A₁) × P(A₂) × …× P(A_n).

Remark. If events A₁, A₂, …, A_n are independent in union, then opposite to them events , …, are also independent in union.

Example. Find the probability of a joint appearance of heads at tossing of two coins.

Solution: The probability of appearance of heads on the first coin (the event A) P(A) = 1/2. The probability of appearance of heads on the second coin (the event B) P(B) = 1/2. The events A and B are independent; therefore the required probability by theorem of multiplication is equal to:

P(AB) = P(A)× P(B) = 1/2 × 1/2 = 1/4.

Example. There are 3 boxes containing 10 details each. There are 8 standard details in the first box, 7 – in the second and 9 – in the third box. One takes at random on one detail from each box. Find the probability that all three taken details will be standard.

Solution: The probability that a standard detail has been taken from the first box (the event A) P(A) = 8/10 = 0,8. The probability that a standard detail has been taken from the second box (the event B) P(B) = 7/10 = 0,7. The probability that a standard detail has been taken from the third box (the event C) P(C) = 9/10 = 0,9. Since the events A, B and C are independent in union, the required probability (by theorem of multiplication) is equal to:

P(ABC) = P(A) × P(B) × P(C) = 0,8 × 0,7 × 0,9 = 0,504.