Classical definition of probability
Basic concepts of probability theory
Hereinafter, instead of speaking «the set of conditions S holds» we shall speak briefly: «the trial has been made». Thus, an event will be considered as a result of a trial.
Example. A shooter shoots in a target subdivided into four areas. One shot is the trial. Hit in a certain area of the target is an event.
Example. There are colour balls in an urn. One takes at random one ball from the urn. An extracting a ball from the urn is the trial. An appearance of a ball of a certain colour is an event.
Events are incompatible if an appearance of one of them excludes an appearance of other events in the same trial. Otherwise, they are compatible.
Example. A detail is extracted at random from a box with details. An appearance of a standard detail excludes an appearance of a non-standard detail. The events «a standard detail has appeared» and «a non-standard detail has appeared» are incompatible.
Example. A coin is tossed. An appearance of «heads» excludes an appearance of «tails». The events «heads have appeared» («the coin lands on heads») and «tails have appeared» («the coin lands on tails») are incompatible.
For example, a landing two different prizes under only one ticket of a lottery are incompatible events, and a landing the same prizes under two tickets are compatible events. Obtaining marks «excellent», «good» and «satisfactory» by a student at an exam in one discipline are incompatible events and an obtaining the same marks at exams in three disciplines are compatible events.
Some events form a complete group if in result of a trial at least one of them will appear. In other words, the appearance of at least one of events of a complete group is a reliable event.
In particular, if events forming a complete group are pairwise incompatible then in result of a trial one and only one of these events will appear.
Example. Two tickets of a money-thing lottery have been bought. One necessarily will happen one and only one from the following events: «a landing a prize on the first ticket and a non-landing a prize on the second», «a landing a prize on both tickets», «a non-landing a prize on the first ticket and a landing a prize on the second», «a non-landing a prize on both tickets». These events form a complete group of pairwise incompatible events.
Example. A shooter has made one shot in a target. One necessarily will happen one from the following two events: hit, miss. These two incompatible events form a complete group.
Events are equally possible if there is reason to consider that none of them is more possible (probable) than other.
Example. An appearance of heads and an appearance of tails at tossing a coin are equally possible events. Appearances of «one», «two», «three», «four», «five» or «six» on a tossed die are equally possible events.
Several events are uniquely possible if at least one of them will necessarily happen as a result of a trial. For example, the events consisting in that a family with two children has: A – «two boys», B – «one boy and one girl» and C – «two girls» are uniquely possible.
Classical definition of probability
Example. Let an urn contain 6 identical, carefully shuffled balls, and 2 of them are red, 3 – blue and 1 – white. Obviously, the possibility to take out at random from the urn a colour ball (i.e. red or blue) is more than the possibility to extract a white ball.
Whether it is possible to describe this possibility by number? It appears it is possible. This number is said to be the probability of an event (appearance of a colour ball). Thus, the probability is the number describing the degree of possibility of an appearance of an event.
Let the event A be an appearance of a colour ball. We call each of possible results of a trial (the trial is an extracting a ball from the urn) by elementary event. We denote elementary events by w1, w2, w3 and et cetera. In our example the following 6 elementary events are possible: w1 – the white ball has appeared; w2, w3 – a red ball has appeared; w4, w5, w6 – a blue ball has appeared. These events form a complete group of pairwise incompatible events (it necessarily will be appeared only one ball) and they are equally possible (a ball is randomly extracted; the balls are identical and carefully shuffled).
We call those elementary events in which the event interesting for us occurs, as favorable to this event. In our example the following 5 events favor to the event A (appearance of a colour ball): w2, w3, w4, w5, w6. In this sense the event A is subdivided on some elementary events; an elementary event is not subdivided into other events. It is the distinction between the event A and an elementary event.
The ratio of the number of favorable to the event A elementary events to their total number is said to be the probability of the event A and it is denoted by P(A). In the considered example we have 6 elementary events; 5 of them favor to the event А. Therefore, the probability that the taken ball will be colour is equal to P(A) = 5/6. This number gives such a quantitative estimation of the degree of possibility of an appearance of a colour ball which we wanted to find.
The probability of the event A is the ratio of the number of favorable elementary events for this event to their total number of all equally possible incompatible elementary events forming a complete group.
Thus, the probability of the event A is determined by the formula:
where m is the number of elementary events favorable to A; n is the number of all possible elementary events of a trial. Here we suppose that elementary events are incompatible, equally possible and form a complete group.
The definition of probability implies the following its properties:
Property 1. The probability of a reliable event is equal to 1.
In fact, if an event is reliable then each elementary event of a trial favors to the event. In this case m = n and consequently P(A) = m/n = n/n = 1.
Property 2. The probability of an impossible event is equal to 0.
Indeed, if an event is impossible then none of elementary events of a trial favors to the event. In this case m = 0 and consequently P(A) = m/n = 0/n = 0.
Property 3. The probability of a random event is the positive number between 0 and 1.
In fact, a random event is favored only part of the total number of elementary events of a trial. In this case 0 < m < n; then 0 < m/n < 1 and consequently 0 < P(A) < 1.
Thus, the probability of an arbitrary event A satisfies the double inequality: 0 £ P(A) £ 1
The relative frequency (statistical probability) of an event is the ratio of the number of trials, in which the event has appeared, to the total number of actually made trials.
Thus, the relative frequency of the event A is defined by the formula:
where m is the number of appearances of the event, n is the total number of trials.
Comparing the definitions of probability and relative frequency, we conclude: the definition of probability does not demand that the trials should be made actually; the definition of relative frequency assumes that the trials were made actually. In other words, the probability is calculated before an experiment, and the relative frequency – after an experiment.
Example. The quality department has detected 3 non-standard details in a group consisting of 80 randomly selected details. The relative frequency of appearance of non-standard details is W(A) = 3/80.
Example. There have been made 24 shots in a target, and 19 hits were registered. The relative frequency of hit in the target is W(A) = 19/24.
The long observations have shown that if experiments are made in identical conditions, in each of which the number of trials is rather great, the relative frequency has a stability property. This property is that for different experiments the relative frequency is changed a little (the less changes, the more trials were made), oscillating about some constant number. There was found out that this constant number is the probability of appearance of the event.
Thus, if the relative frequency is established by a practical experiment, the obtained number can be accepted for approximate value of probability.
To overcome defect of the classical definition of probability consisting that it is inapplicable to trials with infinite number of events (outcomes) enter geometric probabilities – the probability of hit of a point in area (segment, part of a plane and etc.).
Let a segment l be a part of a segment L. A point is set (thrown) at random in the segment L. It means that the following suppositions hold: the thrown point can appear in any point of the segment L, the probability of hit of the point in the segment l is proportional to the length of this segment and does not depend on its disposition concerning the segment L. In these suppositions the probability of hit of the point in the segment l is determined by the equality P = the length of l / the length of L.
Example. A point B(x) is thrown at random in a segment OA of the length L of the numeric axis Ox. Find the probability that the smaller of the segments OB and BA has the length more than L/3. It is assumed that the probability of hit of a point in the segment is proportional to the length of the segment and does not depend on its disposition on the numeric axis.
Solution: Let's divide the segment OA by points C and D on three equal parts. The request of the problem will be executed if the point B(x) will hit in the segment CD of the length L/3. The required probability P = (L/3)/L = 1/3.
Let a flat figure g be a part of a flat figure G. A point is thrown at random in the figure G. It means that the following suppositions hold: the thrown point can appear in any point of the figure G, the probability of hit of the thrown point in the figure g is proportional to the area of this figure and does not depend on both its disposition concerning the figure G and the form of g. In these suppositions the probability of hit of the point in the figure g is determined by the equality P = the area of g / the area of G.
Example. Two concentric circles of which the radiuses are 5 and 10 cm respectively are drawn on the plane. Find the probability that the point thrown at random in the large circle will hit in the ring formed by the constructed circles. It is assumed that the probability of hit of a point in a flat figure is proportional to the area of this figure and does not depend on its disposition concerning the large circle.
Solution: The area of the ring (the figure g)
The area of the large circle (the figure G)
The required probability
probability theory – теория вероятностей; reliable event – достоверное событие
random event – случайное событие; vessel – сосуд; heads or tails? – орел или решка?
trial (experiment) – испытание (опыт, эксперимент); urn – урна; at random – наудачу
to land a prize – получить приз; complete group of events – полная группа событий
equally possible events – равновозможные события
uniquely possible events – единственно возможные события; ace – очко (при игре в кости)
die – кость (игральная); dice – игра в кости, кости; hit – попадание; miss – промах
to shuffle – перемешивать; relative frequency – относительная частота
favorable case – благоприятствующий (благоприятный) случай
mass homogeneous events – массовые однородные события
Exercises for Seminar 1
1.1. There are 50 identical details (and 5 of them are painted) in a box. Find the probability that the first randomly taken detail will be painted.
1.2. A die is tossed. Find the probability that an even number of aces will appear.
1.3. Participants of a toss-up pull a ticket with numbers from 1 up to 100 from a box. Find the probability that the number of the first randomly taken ticket does not contain the digit 5 (toss-up – жеребьевка; to pull – тянуть; ticket – жетон).
1.4. In a batch of 100 details the quality department has found out 5 non-standard details. What is the relative frequency of appearance of non-standard details equal to? (batch – партия)
1.5. At shooting by a rifle the relative frequency of hit in a target has appeared equal to 0,85. Find the number of hits if 120 shots were made (a rifle – винтовка).
1.6. One die is randomly taken from a carefully hashed full set of 28 dice of domino. Find the probability that the second randomly taken die can be put to the first if the first die: a) is a double;
b) is not double (to hash – перемешивать; double – дубль). The answer: a) 2/9; b) 4/9.
1.7. A point B(x) is randomly put in a segment OA of the length L of the numeric axis Ox. Find the probability that the smaller of the segments OB and BA has the length which is less than L /4. It is supposed that the probability of hit of a point in the segment is proportional to its length and does not depend on its location on the numeric axis.
1.8. Two persons have agreed to meet in a certain place between 18 and 19 o'clock and have agreed that a person come the first waits for another person within 15 minutes then leaves. Find the probability of their meeting if arrival of everyone within the specified hour can take place at any time and the moments of arrival are independent. The answer: 7/16.
1.9. Two dice are tossed. Find the probability that: a) the same number of aces will appear on both dice;
b) two aces will appear at least on one die; c) the sum of aces will not exceed 6.
1.10. A point is randomly taken inside a circle of radius 5. Find the probability that the point will be inside a proper (equilateral) triangle entered in the circle (a proper triangle – правильный треугольник; equilateral – равносторонний). The answer: .
Exercises for Homework 1
1.11. A die is tossed. Find the probability that the upper side of the die shows:
a) six aces; b) an odd number of aces; c) no less than four aces; d) no more than two aces.
1.12. An urn contains 12 balls: 3 white, 4 black and 5 red. Find the probability that a randomly taken ball will be black.
1.13. The first box contains 5 balls with numbers from 1 up to 5, and the second – 5 balls with numbers from 6 up to 10. It has been randomly extracted on one ball from each box. Find the probability that the sum of numbers of the extracted balls will be: 1) no less than 7; 2) equal to 11; 3) no more than 11.
The answer: 2) 0,2; 3) 0,6.
1.14. The relative frequency of workers of an enterprise having a higher education is equal to 0,18. Determine the number of workers having a higher education if the total number of workers of the enterprise is 350.
1.15. 78 seeds have germinated of 100 planted seeds. Find the relative frequency of germination of seeds (seed – семя; to germinate – прорастать).
1.16. Two dice are tossed. Find the probability that: a) the sum of aces equals 5, and the product equals 6;
b) the product of aces doesn’t exceed 6; c) the product of aces is divided on 6.
1.17. A point C is randomly appeared in a segment AB of the length 3. Determine the probability that the distance between C and B exceeds 1. The answer: 2/3.
1.18. A point is randomly thrown inside of a circle of the radius R. Find the probability that the point will be inside the square entered in the circle. It is supposed that the probability of hit of a point in the square is proportional to its area and does not depend on its location regarding the circle. The answer: 2/p.
1.19. A coin is tossed twice. Find the probability that the coin lands on heads in both times.
The answer: 1/4.
1.20. A point B(x) is randomly put in a segment OA of the length 6 of the numeric axis Ox. Find the probability that both the segments OB and BA have the length which is greater than 2. The answer: 1/3.
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