13_p4

13.4 The concept of probability

Trials that we can repeat many times are called repeatable. For example, the tossing of a coin trial or the trial of measuring the size of some object can be repeated as many times as we wish. Please note that not all the trials have this feature. A good example is a trial of measuring the "life time" of a bulb - once it is broken we cannot use it for this trial any more.

Let's assume that the considered trials are repeatable and that their results are mutually independent. We can then observe the frequency of a certain event. For example, we can roll a die repeatedly and check how often 6 appears. If during 100 rolls 6 appears 20 times, we say that the frequency of 6 appearing in this trial is 20/100.

Generally, if during n trials the event A appeared m times, then the frequency of the event A is m/n, see Figure 13.4.1. It is noticeable that if the number of trials increases, the frequency of the event A tends to a certain number. This number is the probability of the event A.

Fig. 13.4.1 Event A = "the sun is shining" appeared 40 times during 60 days of observations, i.e. the frequency of the event A in this trial is 40/60.

Probability has risen from the notion of frequency and it is a theoretical equivalent of frequency. So it is natural that we want some of the basic properties of frequency to be transferred to probability. One of such properties is the dependence of the frequency of an event which is a union of mutually exclusive events A and B, upon the frequency of the event A and the frequency of the event B.

If during n trials the event A appeared n_A times, and the event B - n_B times, and additionally, the events A and B are exclusive, then the frequency of the event A ∪ B is

(n_A+n_B)/n = n_A/n + n_B/n,
so it is a sum of the frequency of the event A and the frequency of the event B.

The definition of probability by Kolmogorov (1933) is presented below.

Definition 13.4.1

Let Ω be a finite sample space. Probability is the function P defined on events such that

(1) P(A) ≥ 0 for any event A,

(2) P(A ∪ B) = P(A) + P(B) for any exclusive events A, B,

(3) P( Ω) = 1.

Definition 13.4.1 shows that the probability is always a positive number less or equal to 1, the probability of the certain event is 1, and the probability of a union of events is a sum of their probabilities, as long as they are exclusive.

Question 13.4.1: What is the probability of an impossible event? Justify your answer.

Check the answer

The following statement lets us express the probability of certain composite events by the probabilities of more simple events and it is a natural generalization of the rule (2) used in the probability definition.

Lemma 13.4.1

If the events A₁, A₂, ..., A_n, specified in a certain sample space Ω are pairwise disjoint, than

P(A₁ ∪ ... ∪ A_n) = P(A₁) + P(A₂) + ...+ P(A_n).

Proof by induction with respect to n.

For n=2, the statement arises from the probability definition.

Inductive assumption: P(A₁ ∪ ... ∪ A_k) = P(A₁) + P(A₂) + ...+ P(A_k) for a certain k and the sequence of k pairwise disjoint events.

Let's consider the sequence of (k+1) elements A₁, A₂, ..., A_k, A_k+1 of events with the property A_i ∩ A_j= ∅ for i ≠ j. First of all, let's notice that the events A = A₁ ∪ A₂ ∪ ... ∪ A_k and A_k+1 are exclusive. Indeed,

A ∩ A_k+1 = (A₁ ∩ A_k+1) ∪ (A₂ ∩ A_k+1) ∪ ... ∪ (A_k ∩ A_k+1) = ∅,
because each summand is an empty set according to the assumption.

Hence, we have, according to the condition (2) of the probability definition,

P(A ∪ A_k+1) = P(A₁ ∪ ... ∪ A_k ) + P(A_k+1).

From that and from the inductive assumption, we have a thesis in the instance of the sequence of (k+1) events. By the terms of induction, the statement is true for any sequence of events of n elements. ♦

The statement presented above is very practical, if we assume that the elementary events from the sample space Ω are equally probable and that the sample space is finite.

Let Ω = {w₁, w₂, ..., w_n} and let P(w_i) = p for a certain p, i.e. all the elementary events are equally probable.

By the terms of the previous statement we have:

P( Ω) = P({w₁, w₂, ...w_n}) = P({w₁} ∪ {w₂} ∪ ... ∪ {w_n} ) = P(w₁) + P(w₂) + ... +P(w_n) = n*p = 1

Hence p = 1/n.

Let's have any event A = {w_i1, w_i2, ...w_ik}, which has k favorable elementary events of the sample space Ω . Then, by the terms of the definition 13.4.1 and previous consideration, P({w_i}) = 1/n for all i =1,2,..n, and

P(A) = P({w_i1, w_i2, ...w_ik}) = P({w_i1} ∪ {w_i2} ∪ ... ∪ {w_ik} ) = P(w_i1) + P( w_i2) + ... +P(w_ik).

Finally P(A) = k/n.

Theorem 13.4.1

Probability of the event A ⊆ Ω is a quotient of the number of elementary events favorable to the event A and the number of elementary events of the considered finite sample space Ω, if the elementary events are equally probable, i.e.

The formula included in the above theorem is called the classical definition of probability, and it was formulated by Pierre Laplace.

The use of this formula will be demonstrating on examples. In all cases we assume that the elementary events are equally probable (i.e. for example during the roll of a die we are using only a "fair" die) .

Example 13.4.1

Let Ω be the sample space in rolling a pair of dice trial. The sample space consists of 36 events, see Figure 13.2.1. Let's consider the events A, B, C: A ="6 received at least once", B = "the sum of pips is 7", C = "product of pips is an even number". As the event A has 11 favorable elementary events, the event B has 6 favorable elementary events, the event C has 27 favorable elementary events, then

P(A)= 11/36, P(B) = 6/36, P(C)= 27/36.

Fig. 13.4.2 The arrangement of 9 people at a round table.

Example 13.4.2

9 people {a,b,c,..g,h,i} sit at a round table on nine chairs marked with numbers 1,2,3...9. What is the probability that two persons, a and b, sit next to each other?

The sample space is a set of all possible arrangements of 9 people on 9 chairs. It is the same number as the number of permutation of the set of 9 elements, so 9! If we choose a place for person a, then person b has to take an adjacent seat on the left or on the right. There are 9 possible arrangements for the situation (ab) and 9 arrangements for the situation (ba). The rest of the people can sit anywhere, i.e. 7! possible arrangements for the remaining 7 people. Finally, the probability is 2(9*7!/9!)=1/4.

Example 13.4.3

There are 9 balls in the box marked with numbers from 1 to 9. We draw two balls (sampling without replacement). The first one stands for the number of units, the second one stands for the number of dozens. What is the probability of the event A = "an even number obtained"?

If for the first time we draw 4, and 6 for the second, then our result is 6*10 + 4 = 64, and if for the first time we draw 1, and 2 for the second, then our result is 21.

The sample space Ω = {(k,l) : k,l ∈ {1,2,...9} and k ≠ l}. | Ω| = 9*8 =72.

All those situations are favorable to the event A, where the drawn number of units is even, which means those where the first number drawn is even. Hence A consists of the elementary events

(2,x), for x ≠ 2, (there are 8 such events),

(4,x), for x ≠ 4, (there are 8 such events)

(6,x), for x ≠ 6, (there are 8 such events)

(8,x), for x ≠ 8 (there are 8 such events).

So P(A)= 4*8/(9*8) = 4/9.

Example 13.4.4

We toss a coin 10 times. What is the probability that during those 10 tosses tails will be obtained exactly 4 times?

The sample space is a set of sequences with the values of T - tails and H - heads. There are as many such events as there are different functions from the set of 10 elements into the set of 2 elements (i.e. as many as different binary sequences of ten elements), which is 2¹⁰. All the events where tails appears 4 times in a sequence and heads - the remaining 6 times are favorable to the event A. The position of those four tails can be any of the set of 10 elements. There are as many of them as the four-elements subsets in a set of 10 elements, i.e. (10 nad 4). Hence, the probability of the considered event is 210/1024.

Question 13.4.2: Let's consider the trial described in the example 13.4.3. What is the probability that the number drawn is divisible by 3?

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