Probability of appearance of at least one event

Let as a result of a trial n events independent in union or some of them (in particular, only one or none) can appear, so that the probabilities of appearance of each of the events are known. How can we find the probability that at least one of these events will happen? For example, if as a result of a trial three events can appear, then an appearance of at least one of these events means an appearance of either one, or two, or three events. The answer on the posed question is given by the following theorem.

Theorem. The probability of appearance of at least one of the events A₁, A₂, …, A_n independent in union is equal to the difference between 1 and the product of the probabilities of the opposite events

P(A) = 1 – q₁q₂…q_n

where A is the appearance of at least one of the events A₁, A₂, …, A_n; .

Partial case. If the events A₁, A₂, …, A_n have the same probability which is equal to p then the probability of appearance of at least one of these events:

 

P(A) = 1 – qⁿ

where A is the appearance of at least one of the events A₁, A₂, …, A_n; .

Example. The probabilities of hit in a target at shooting by three guns are the following: p₁ = 0,8; p₂ = 0,7; p₃ = 0,9. Find the probability of at least one hit (the event A) at one shot by all three guns.

Solution: The probability of hit in the target by each of the guns doesn’t depend on results of shooting by other guns, therefore the considered events A ₁ (hit by the first gun), A ₂ (hit by the second gun) and A ₃ (hit by the third gun) are independent in union. The probabilities of events which are opposite to the events A ₁, A ₂ and A ₃ (i.e. the probabilities of misses) are equal respectively:

q₁ = 1 – p₁ = 1 – 0,8 = 0,2; q₂ = 1 – p₂ = 1 – 0,7 = 0,3;

q₃ = 1 – p₃ = 1 – 0,9 = 0,1.

The required probability P(A) = 1 – q₁q₂q₃ = 1 – 0,2 × 0,3 × 0,1 = 0,994.

Example. There are 4 flat-printing machines at typography. For each machine the probability that it works at the present time is equal to 0,9. Find the probability that at least one machine works at the present time (the event A).

Solution: The events «a machine works» and «a machine doesn’t work» (at the present time) are opposite, therefore the sum of their probabilities is equal to 1: p + q = 1. Consequently, the probability that a machine doesn’t work at the present time is equal to q = 1 – p = 1 – 0,9 = 0,1.

The required probability P(A) = 1 – q⁴ = 1 – (0,1)⁴ = 0, 9999.

 

Example. A student looks for one formula necessary to him in three directories. The probability that the formula is contained in the first, second and third directories, is equal to 0,6; 0,7 and 0,8 respectively. Find the probability that the formula is contained:

a) only in one directory (the event A);

b) only in two directories (the event B);

c) in all the directories (the event C);

d) at least in one directory (the event D);

e) neither of the directories (the event E).

Solution: Consider elementary events and their probabilities:

А ₁ – the formula is in the first directory,

А ₂ – the formula is in the second directory,

А ₃ – the formula is in the third directory,

 

Express all the events A–E by the elementary events and their negations, and apply the above-stated theorems:

a)

b)

c)

d)

e)

 

Glossary

independent events – независимые события

mutual – взаимный; independent in union – независимые в совокупности

flat-printing machine – плоскопечатная машина

directory – справочник

 

Exercises for Seminar 4

4.1. The probability that a shooter hit in a target at one shot is equal to 0,9. The shooter has made 3 shots. Find the probability that all 3 shots will strike the target.

 

4.2. A coin and a die are tossed. Find the probability of joint appearance of the following events: «the coin lands on heads» and «the die lands on 6».

 

4.3. What is the probability that at tossing three dice 6 aces will appear at least on one of the dice (the event А)?

The answer: 0,722.

 

4.4. There are 8 standard details in a batch of 10 details. Find the probability that there is at least one standard detail among two randomly taken details.

 

4.5. Two dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers?

 

4.6. The probability of hit in a target by the first shooter at one shot is equal to 0,8, and by the second shooter – 0,6. Find the probability that the target will be struck only with one shooter.

The answer: 0,44.

 

4.7. The probability to receive high dividends under shares at the first enterprise – 0,2; on the second – 0,35; on the third – 0,15. Determine the probability that a shareholder having shares of all the enterprises will receive high dividends:

a) only at one enterprise;

b) at least on one enterprise (a share – акция).

The answer: a) 0,4265; b) 0,564.

 

4.8. The first brigade has 6 tractors, and the second – 9. One tractor demands repair in each brigade. A tractor is chosen at random from each brigade. What is the probability that:

a) both chosen tractors are serviceable;

b) one of the chosen tractors demands repair (serviceable – исправный).

The answer: a) 20/27; b) 13/54.

 

Exercises for Homework 4

4.9. There are details in two boxes: in the first – 10 (3 of them are standard), in the second – 15 (6 of them are standard). One takes out at random on one detail from each box. Find the probability that both details will be standard.

The answer: 0,12.

 

4.10. There are 3 television cameras in a TV studio. For each camera the probability that it is turned on at present, is equal to p = 0,6. Find the probability that at least one camera is turned on at present (the event А).

The answer: 0,936.

 

4.11. What is the probability that at least one of a pair of dice lands on 6, given that the sum of the dice is 8?

The answer: 0,4.

 

4.12. 10 of 20 savings banks are located behind a city boundary. 5 savings banks are randomly selected for an inspection. What is the probability that among the selected banks appears inside the city:

a) 3 savings banks; b) at least one?

The answer: a) 0,348; b) 0,984.

 

4.13. There are 16 details made by the factory № 1 and 4 details of the factory № 2 at a collector. Two details are randomly taken. Find the probability that at least one of them has been made by the factory № 1.

The answer: 92/95.

 

4.14. Three buyers went in a shop.The probability that each buyer makes purchases is equal to 0,3. Find the probability that:

a) two of them will make purchases;

b) all three will make purchases;

c) only one of them will make purchases.

The answer: a) 0,189; b) 0,027; c) 0,441.

 

4.15. Three students pass an exam. The probability that the exam will be passed on "excellent" by the first student is equal to 0,7; by the second – 0,6; and by the third – 0,2. What is the probability that the exam will be passed on "excellent" by:

a) only one student; b) two students;

c) at least one; d) neither of the students?

The answer: a) 0,392; b) 0,428; c) 0,904; d) 0,096.

 

4.16. Three shots are made in a target. The probability of hit at each shot is equal to 0,6. Find the probability that only one hit will be in result of these shots.

The answer: 0,288.

 

 

L E C T U R E 5