Practical lesson 1. Classical probability
To begin with, let us examine a few standard examples on this topic.
Let me remind you that if we have a space of elementary events "Omega" and a set A, which is a subset of "omega" As will be some event, all elementary events of " omega "space as a result of some of the assumptions may be considered equally probable.
Equal probability may arise either from symmetry, or of any special conditions of the experiment, it is possible to use the classical probability.
Let number N be the number of elements.
The number of elementary events of "omega"set and number mA - is the number of elements of elementary events in set A.
There are many favorable outcomes of event A.
Then the probability of event A can be calculated as the ratio of the number of favorable outcomes to the number of all outcomes.
This is often referred to as mA / N.
This is the formula that we will use in the following examples.
The first example: Suppose there are 10 numbered balls inside the urn.
The experiment is to extract one ball from this urn, then elementary outcome of this experiment is the number of the extracted ball.
That is, an elementary outcome is a ball number.
Then the set of "Omega", it will consist of numbers from 1 to 10.
Numbers of extracted balls.
That is, there are10 elementary outcomes.
Why can we assume that all outcomes are equally probable in this case?
We say that we take out a ball from the urn by chance, that is, with equal probability, we can take out any ball.
Thus, the classical probability can be used here.
The task is as follows: list elementary events favoring event A –taking out the ball with an odd number.
If we write a set of elementary outcomes favorable toevent A, there will be one of 1, 3, 5, 7, 9; 5 such outcomes, that mA is equal to 5, and thus the probability of event A is equal to 5/10 or ½.
The second event, event B –taking out the ball with an even number, you can either list elementary outcomes that are favorable to the event, or see that event B can be written as a negation of event A.
That is not true, that anodd-numbered ball according to the properties of probability, we can find the probability of event B: one minus the probability of event A, that is, 1- ½, i.e. ½
Well, event C –taking out the ball, with a number over three.
Accordingly, set C is composed of elements 4, 5, 6, 7, 8, 9, 10.
Such elements can be seen as 7, then the probability of event C can be calculated as 7 divided by 10, it is equal to 0.7.
Consider the following example: now there are not multi-colored balls in the urn, but 8 black balls.
Again a ball is taken out of the urn.
The first problem to be solved is what is considered a basic outcome of this experience.
The first thought is that a basic outcome is the color of the ball, i.e., the set of basic outcomes consisting of two outcomes: a white ball, a black ball.
But as there are more black balls it is clear that a black ball will appear more frequently than a white one.
So these two basic outcomes are not equally probable.
As a consequence, the classical probability cannot be used here.
But if we number the balls in such a way and take the space of basic outcomes, consisting of two white balls, that is, w1, w2, we assume that there are some points on all balls, black 1, black 2, and so on, black 8 then any ball will be taken out with equal probability.
Since we pull balls completely randomly. So, now we've got a lot of basic outcomes, where all basic outcomes are equally probable.
So, for event A, when a white ball is taken out, favorable basic outcomes are B1 and B2.
There are two elementary events, then probability of the event is calculated as two to ten.
Since the number of basic events in set "omega" 10.
Consider a slightly more complex example: now we still have 2 white and 8 black balls in the urn, but now 3 ballsare taken out simultaneously, that is, we had some an urn with 10 balls, and we randomly take 3 balls what can be considered as a basic outcome in this case?
Basic outcome will be a set of 3 balls at the same time it does not matter in what order the elements go in this set.
Totalle there are 10 sets of 3.
If we calculate, it will be 10 * 9/3 !.Itwillbe 120.
All basic outcomesare 10 to 3 or 10 * 9 * 8/3! , i.e. 120.
Let us will look at some statements in this experiment
Statement 1, among chosen balls there is no white ball, there will be an event when 3 balls are all black, but the number of basic events favorable to event A is the number of sets of black balls, as there are 8 black balls, there will be 8 of 3 or 8 * 7 * 6/3 !, i.e. 56.
Then we can calculate the probability of event A, as we are absolutely randomly choose three, we can assume that all threesare equally probable.
Therefore, we can calculate the probability of event A by the formula of classical probability - this is the number of favorable outcomes divided by the number of all outcomes.
Consider the second statement: a white ball was taken out first.
Here it is necessary to decide whether or not this statement is an event?
All basic outcomes are sets, unordered sets, so no ball can be the first.
And, then, B will not be an event in the space of basicoutcomes.
Next: the white ball taken out earlier than the black, for the same reason, it will not be an event in the given space.
D: there were more white balls taken out than black ones.
It will be an event in the case if among three taken out balls there are two or three white balls, the rest are black.
So statement D: there are more white balls taken out than black ones, this be an event in this experiment, with favorable outcomes are those basic ones where among the three taken out balls there will be two or three white balls.
But as wehave only two white balls, the sample cannot have three white balls, which means that favorable to event D samples are those where there are two white and one black ball.
Hence favorable to event D are those basic outcomes where there are two white and one black ball.
So the number of favorable outcomes of event D is two of two, we choose white, one of the eight black, it turns eight.
So the probability of event D is equal to 8/120
Well, the third statement: three white balls were taken out.
As there are only two white balls, none basic outcomes will be favorable to the event, but the statement will be an event, it will match an empty set.
So the probability of event E as aN impossible event is 0.
Consider another example: all the same, there are 2 white and 8 black balls in the urn, but balls are taken out one by one.
That is, first one ball, then one of the rest, and then one of the remaining eight balls.
In this case, the basic outcome is a sequential set of three balls.
An ordered set of three balls.
Total number of such sets of 10 * 9 * 8, that is 720, but in this case, all five statements will be events of their probability.
The number of outcomes favorable to event A, when among the selected balls there is no white ball, that is, all balls are black, will be equal to 8 * 7 * 6, then the probability of event A is equal to 8 * 7 * 6/10 * 9 * 8, we reduce and get 7/20 .
Event B: the white ball was taken out first, which means that for the first time we took out a white ball and then, in general, can takeany one.
The following options will be suitable: white, white, black; white, white, white are impossible, because there are only two white balls; white, black, white, and white, black, black.
Thatis, for B.
In this case, we get favorable outcomes: WWB 2 * 1 * 8; WBW 2 * 8 * 1, then 16 and then 16 in both cases, and in the latter case WBB 2 * 8 * 7, that is, 112 favorable outcomes.
Accordingly, event B has 32 + 112 = 144 favourablebasic outcomes, well, then the probability of event B is 144/720.
A white ball is taken out earlier thana black one - it means all the cases considered earlier suit: WWB, WBW, WBB, and statement C –a white ball was taken out before a black one, it means that a white ball was taken out first.
So event C and event B are one and the same event, their probabilities are equal.
Event D: there are more taken out white balls than black ones.
Just as in the previous case, those sets in which there are two white balls suit us, which means it will sets WWB, WBW, BWW
Sets of the first type are 1 * 2 * 8, the second type are 2 8 * 1 * and the third type are 8 * 2 * 1.
I.e.there will be 16 sets of each type, total 48 sets.
Then the probability of event B can be calculated as the 48/720.
And event E is still impossible, that is, its probability will be equal to 0.
The examples are over, you are ready to do practical tasks on hand-outs