Lecture 2. Probability of a random event
In this lecture, we are going to consider the concept of probability and give examples of probability functions. Historically, the first definition of probability was a statistical definition. Let us consider the ratio of N(A) to N (this ratio is the frequency of a random event). If, for N tending to infinity, there is a frequency limit equal to p, then the number p is the probability of a random event. The probability theory considers only statistically stable events, that is, those with a finite frequency limit, and the limit may not exist. Andrey Kolmogorov gave the modern definition of probability in 1923. This is the so-called axiomatic definition of a random event. We need to consider the notion of σ-algebra of events. Let us consider the Ω space of elementary outcomes of some experiment and the F set (this is the set of subsets of the space of elementary outcomes, but not necessarily all of them). The F set is called σ-algebra if the following conditions are met. First, the set of all elementary outcomes belongs to the F set. Second, if an event is from the F set, then its negation is also from the F set. Third, if a certain finite or countable set of events belongs to the set F, then their sum also belongs to the set F (the product of a countable set of sets can replace the third condition).
The probability or probability measure for a pair (Ω, F) is the function P that acts from the σ-algebra F to the set of real numbers and has the following properties:
1) the probability of any event is not negative;
2) for a countable set of pairwise incompatible events, the probability of their sum is equal to the sum of their probabilities (note that the set must consist of pairwise incompatible events, i.e. no two events can intersect or have common elementary outcomes);
3) the probability of Ω (i.e., a reliable event) is 1.
These three properties are the axioms of probability, and while considering any probability function, it is necessary to check that this function satisfies these axioms. The first example of a probability function is classical probability. If the set of elementary outcomes is finite, and all elementary outcomes are equally probable (we can often conclude that elementary outcomes are equally probable due to some symmetry observed in the experiment), then we can consider the classical probability. That is, the probability of an event A consisting of m elementary outcomes is equal to the ratio of the numbers m (the number of favorable outcomes) and N (the power of the set Ω, i.e. the total number of elementary outcomes of this experience). The classical probability can be considered only if the elementary outcomes are equally probable. Let us consider an example.
A dice is tossed. The set of elementary outcomes is {1, 2, 3, 4, 5, 6}, if we consider an ideal dice (that is, absolutely symmetric), then all these outcomes are equally probable, and we can consider is the classical probability, with the probability of each elementary outcome being equal to 1/6. If we consider the event A = {an even number of points falls out}, which consists of elementary outcomes 2, 4, and 6, then the probability of this event is 3/6 or 1/2. Let's consider another experiment. The black and white dice are tossed. In this experiment, the elementary outcome is a pair of numbers A and B, where A is the number on the black dice and B is the number on the white dice. In this case, there are 36 elementary outcomes. If we consider the event A when we have four on the white dice, then this event is favored by all elementary outcomes with the second value equal to four (that is, 1-4, 2-4, and so on, 6 - 4). There are 6 such elementary outcomes, which means that the probability of the event A is 6/36 or 1/6. In the same experiment, we can consider the event B = {in total, eight points fall out}. This event is favored by elementary outcomes B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, that is, five elementary outcomes out of 36. The probability of the event B is equal to 5/36.
Let us consider another experiment: let the shooter shoot at the target once. In this case, the set of elementary outcomes consists of two outcomes: Ω = {the shooter hit the target and the shooter missed}. Note that these elementary outcomes are not equally likely, i.e. the probability of a hit is not necessarily 1/2. So let's consider a more general case – this is a discrete probability space. If the set of all elementary outcomes is finite or countable, then each elementary outcome can correspond to a certain number pi from the interval [0, 1], so that the sum of all pi =1. In this case, we say that we consider a discrete probability space. Then the probability of the event A is equal to the sum of the probability of all elementary outcomes included in the set A.
Let us consider the following experiment: the shooter shoots at the target twice, then the set of elementary outcomes consists of four different outcomes: Ω = {(hit, hit), (hit, missed), (missed, hit), (missed, missed)}. Let's set the probabilities for these elementary outcomes. ω1 is the probability that the shooter hit twice = 0.49. ω2 is the probability that the shooter hit the first time and missed the second time = 0.21. ω3 is the probability that the shooter missed the first time and hit the second time = 0.21. The probability that he missed twice is 0.09.
It is easy to check that the sum of these four numbers is equal to 1, so we have a discrete probability space. For this experiment, we are going to consider the event A = {the shooter hit the target at least once}. This event consists of elementary outcomes {(hit both times), (first time hit, second time missed), (first time missed, second time hit)}. The probability of the event A is the sum of three numbers corresponding to elementary outcomes, i.e. 0,49 + 0,21 + 0,21 thus, the probability of the event A is 0.91.
Another experiment: the coin is tossed before the first heads falls. The elementary outcomes are the number of coin flips up to the first heads. These are numbers from 1 to ∞, that is, the set of elementary outcomes of this experiment coincides with the set of natural numbers. If we have the probabilities of events P(ωi) = 1/ 2i, it is easy to check that the sum of the probabilities of all elementary outcomes is 1 (this is the sum of the geometric progression). If it takes no more than 3 throws to get the event A in this experiment, then the probability of this event can be calculated as the sum of three probabilities, that is, 7/8. Another example of a probability function is geometric probability. We consider it if the space of elementary events of an experiment is a subset of Rn (i.e., we consider an experiment on a line or on a plane or in R3, and so on). The measure Ω is finite (the length is on a straight line; the square is on a plane and volume is in space R3). Then the event A is a subset of the set Ω.
Note that in this case, not all subsets are events, since there is an unmeasurable subset of Rn. If getting into any point of the set Ω is equally probable (for example, we say that we have a point on the line segment from 0 to 1, and its being at any point of this segment is equally probable), then we calculate the probability of the event A as the ratio of the measure of the set A to the measure of the set Ω (be it length, area, volume and so on). Let's consider some examples of geometric probability. Example 1: let's generate a random number from the interval [0, 1]. We consider the generator ideal, that is, we assume that any number from a given interval can be generated with the same probability. We are going to consider the event A = {the number is not equal to 1/2}. Note that the set A is a segment [0, 1] punctured at 1/2. The measure of this set is equal to one, the measure of the segment [0, 1] also equals one, then the probability of the event A is equal to one. Note that the event A is not reliable. When we introduced the axioms of probability, we said that the probability of a reliable event is equal to one. This example shows that the opposite is not true, i.e. if the probability of an event is equal to one, then the event does not have to be reliable.
You can also use a geometric probability if the problem doesn't have an explicit geometric basis. Let us consider a classical problem called the Problem of meeting. The young man and the girl are to meet at 12 o'clock, but each operates according to the following plan: he can come at any time from 12 to 13 o'clock, and they agreed that they wait for each other only for 20 minutes. That is, the one who comes first waits for 20 minutes, and if the second one does not come during these 20 minutes, then the first one leaves, and the date does not take place. We need to find the probability of the young man and the girl’s meeting. We can apply geometric probability here. Let's formalize the problem: let x be the time of the young man’s arrival, and y be the time of the girl’s arrival. Then we can mark a point with the coordinates (x, y) in a unit square in the coordinates x, y (see the video). If we measure it in minutes instead of hours, we get that each side of the square is 60 minutes and the area of this square is 3600 minutes squared. Let us consider the set A = {the point is in the shaded area}. The shaded area shows that the young man and the girl meet, that is, modulo (x – y) is not more than 20. Then the probability of the event A can be considered as the ratio of the set A, that is, the square of the shaded area, to the area of a large square. The area of the shaded area can be calculated as the area of the square minus the area of the white triangles, but the area of the square is 3600, then the probability of the event A is 5/9.
Let us consider some properties of a probability function or probability. First, the probability of negating an event is calculated as one minus the probability of that event. To prove this, it is enough to note that the event and its negation are incompatible events, which means that according to the 2 axiom of probability, the probability of their sum (i.e., a reliable event) is equal to the sum of their probabilities. Since according to axiom 3, the probability of a reliable event is equal to one, this equation follows. Second, the probability of an impossible event is zero. It is enough to use the third axiom and the previous properties (that is, instead of the event A, we substitute a reliable event, then the event non-A is an impossible event). Third, if the event A entails the event B (as we have already said, the set A is a subset of the set B), then the probability of the event A is no greater than the probability of the event B. In fact, the event A is favored by fewer elementary outcomes than the event B.
You can prove more strictly by representing the event B as the event A and the event B / non-A. These events are incompatible, so the probability of their sum is equal to the sum of their probabilities. The probability of their sum is the probability of the event B, but it is equal to the sum of the probability of A and the probability of the event b / non-A, which in turn is not negative. This means that the probability of the event B is not less than the probability of the event A. Fourth, it is the generalization of the probability of sums, that is, it is a general case. Axiom 2 of probability states that the probability of the sum of incompatible events is equal to the sum of probabilities; in the general case, it is not true. If we consider two events that are possibly compatible, the probability of their sum is equal to the sum of their probabilities minus the probability of their product. Let's introduce another relationship between events. Events are independent if the probability of their product is equal to the product of probabilities. In the next lecture, we are going to consider a second definition of event independence and show that this definition is equally probable. In simple terms, we can say the following. Events are independent if and only if the event A occurs or does not occur, it does not affect the probability of the event B occurring or not. Let's consider some examples.
Example 1: a dice is tossed twice. You need to find the probability that the first time you get six, and the second time you get an odd number. Let's give two events. Event A = {six falls out the first time}. This event corresponds to elementary outcomes, where the first number is six, and the second is any number (that is, out of 36 elementary outcomes, 5 outcomes are suitable for us). Thus, the probability of the event A is 1/6. Event B = {an odd number falls out the second time}. Similarly, the first time anything can fall out, and the second time we have three options out of six, that is, we have all pairs where the first number is any number (there are 6 options), and the second number is one of three, so a total of 18 options out of 36 are suitable for us. The probability of the event B is 1/2.
Let us consider the event C = {six falls out the first time, and the second time we have an odd number}. These events are the product of the events A and B. Since the tosses are independent of each other (i.e. what happens the first time does not affect what happens the second time), we write that the probability of the event C (this is the probability of the product of the events A and B) is equal to the product of the probabilities of the events A and B. That is, the probability that the first time you get six and the second time you get an odd number is 1/12.