Formula for finding conditional probability. Conditional probability. Independence of events. Unconditional probability of an elementary event
In fact, formulas (1) and (2) are a short record of conditional probability based on a contingency table of characteristics. Let's return to the example discussed (Fig. 1). Suppose we learn that a family is planning to buy a wide-screen television. What is the probability that this family will actually buy such a TV?
Rice. 1. Widescreen TV Buying Behavior
In this case, we need to calculate the conditional probability P (purchase completed | purchase planned). Since we know that the family is planning to buy, the sample space does not consist of all 1000 families, but only those planning to buy a wide-screen TV. Of the 250 such families, 200 actually bought this TV. Therefore, the probability that a family will actually buy a wide-screen TV if they have planned to do so can be calculated using the following formula:
P (purchase completed | purchase planned) = number of families who planned and bought a wide-screen TV / number of families planning to buy a wide-screen TV = 200 / 250 = 0.8
Formula (2) gives the same result:
where is the event A is that the family is planning to purchase a widescreen TV, and the event IN- that she will actually buy it. Substituting real data into the formula, we get:
Decision tree
In Fig. 1 families are divided into four categories: those who planned to buy a wide-screen TV and those who did not, as well as those who bought such a TV and those who did not. A similar classification can be performed using a decision tree (Fig. 2). The tree shown in Fig. 2 has two branches corresponding to families who planned to purchase a widescreen TV and families who did not. Each of these branches splits into two additional branches corresponding to households that did and did not purchase a widescreen TV. The probabilities written at the ends of the two main branches are the unconditional probabilities of events A And A'. The probabilities written at the ends of the four additional branches are the conditional probabilities of each combination of events A And IN. Conditional probabilities are calculated by dividing the joint probability of events by the corresponding unconditional probability of each of them.
Rice. 2. Decision tree
For example, to calculate the probability that a family will buy a wide-screen television if it has planned to do so, one must determine the probability of the event purchase planned and completed, and then divide it by the probability of the event purchase planned. Moving along the decision tree shown in Fig. 2, we get the following (similar to the previous) answer:
Statistical independence
In the example of buying a wide-screen TV, the probability that a randomly selected family purchased a wide-screen TV given that they planned to do so is 200/250 = 0.8. Recall that the unconditional probability that a randomly selected family purchased a wide-screen TV is 300/1000 = 0.3. This leads to a very important conclusion. Prior information that the family was planning a purchase influences the likelihood of the purchase itself. In other words, these two events depend on each other. In contrast to this example, there are statistically independent events whose probabilities do not depend on each other. Statistical independence is expressed by the identity: P(A|B) = P(A), Where P(A|B)- probability of event A provided that the event occurred IN, P(A)- unconditional probability of event A.
Please note that events A And IN P(A|B) = P(A). If in a contingency table of characteristics having a size of 2×2, this condition is satisfied for at least one combination of events A And IN, it will be valid for any other combination. In our example events purchase planned And purchase completed are not statistically independent because information about one event affects the probability of another.
Let's look at an example that shows how to test the statistical independence of two events. Let's ask 300 families who bought a widescreen TV if they were satisfied with their purchase (Fig. 3). Determine whether the degree of satisfaction with the purchase and the type of TV are related.
Rice. 3. Data characterizing the degree of satisfaction of buyers of widescreen TVs
Judging by these data,
In the same time,
P (customer satisfied) = 240 / 300 = 0.80
Therefore, the probability that the customer is satisfied with the purchase and that the family purchased an HDTV are equal, and these events are statistically independent because they are not related to each other.
Probability multiplication rule
The formula for calculating conditional probability allows you to determine the probability of a joint event A and B. Having resolved formula (1)
relative to joint probability P(A and B), we obtain a general rule for multiplying probabilities. Probability of event A and B equal to the probability of the event A provided that the event occurs IN IN:
(3) P(A and B) = P(A|B) * P(B)
Let's take as an example 80 families who bought a widescreen HDTV television (Fig. 3). The table shows that 64 families are satisfied with the purchase and 16 are not. Let us assume that two families are randomly selected from among them. Determine the probability that both customers will be satisfied. Using formula (3), we obtain:
P(A and B) = P(A|B) * P(B)
where is the event A is that the second family is satisfied with their purchase, and the event IN- that the first family is satisfied with their purchase. The probability that the first family is satisfied with their purchase is 64/80. However, the likelihood that the second family is also satisfied with their purchase depends on the first family's response. If the first family does not return to the sample after the survey (selection without return), the number of respondents is reduced to 79. If the first family is satisfied with their purchase, the probability that the second family will also be satisfied is 63/79, since there are only 63 left in the sample families satisfied with their purchase. Thus, substituting specific data into formula (3), we get the following answer:
P(A and B) = (63/79)(64/80) = 0.638.
Therefore, the probability that both families are satisfied with their purchases is 63.8%.
Suppose that after the survey the first family returns to the sample. Determine the probability that both families will be satisfied with their purchase. In this case, the probability that both families are satisfied with their purchase is the same, equal to 64/80. Therefore, P(A and B) = (64/80)(64/80) = 0.64. Thus, the probability that both families are satisfied with their purchases is 64.0%. This example shows that the choice of the second family does not depend on the choice of the first. Thus, replacing the conditional probability in formula (3) P(A|B) probability P(A), we obtain a formula for multiplying the probabilities of independent events.
The rule for multiplying the probabilities of independent events. If events A And IN are statistically independent, the probability of an event A and B equal to the probability of the event A, multiplied by the probability of the event IN.
(4) P(A and B) = P(A)P(B)
If this rule is true for events A And IN, which means they are statistically independent. Thus, there are two ways to determine the statistical independence of two events:
- Events A And IN are statistically independent of each other if and only if P(A|B) = P(A).
- Events A And B are statistically independent of each other if and only if P(A and B) = P(A)P(B).
If in a 2x2 contingency table, one of these conditions is met for at least one combination of events A And B, it will be valid for any other combination.
Unconditional probability of an elementary event
(5) P(A) = P(A|B 1)P(B 1) + P(A|B 2)P(B 2) + … + P(A|B k)P(B k)
where events B 1, B 2, ... B k are mutually exclusive and exhaustive.
Let us illustrate the application of this formula using the example of Fig. 1. Using formula (5), we obtain:
P(A) = P(A|B 1)P(B 1) + P(A|B 2)P(B 2)
Where P(A)- the likelihood that the purchase was planned, P(B 1)- the probability that the purchase is made, P(B 2)- the probability that the purchase is not completed.
BAYES' THEOREM
Conditional probability events takes into account information that some other event has occurred. This approach can be used both to refine the probability taking into account newly received information, and to calculate the probability that the observed effect is a consequence of a specific cause. The procedure for refining these probabilities is called Bayes' theorem. It was first developed by Thomas Bayes in the 18th century.
Let's assume that the company mentioned above is researching the market for a new TV model. In the past, 40% of the TVs created by the company were successful, while 60% of the models were not recognized. Before announcing the release of a new model, marketing specialists carefully research the market and record demand. In the past, 80% of successful models were predicted to be successful, while 30% of successful predictions turned out to be wrong. The marketing department gave a favorable forecast for the new model. What is the likelihood that a new TV model will be in demand?
Bayes' theorem can be derived from the definitions of conditional probability (1) and (2). To calculate the probability P(B|A), take formula (2):
and substitute instead of P(A and B) the value from formula (3):
P(A and B) = P(A|B) * P(B)
Substituting formula (5) instead of P(A), we obtain Bayes’ theorem:
where events B 1, B 2, ... B k are mutually exclusive and exhaustive.
Let us introduce the following notation: event S - TV is in demand, event S’ - TV is not in demand, event F - favorable prognosis, event F’ - poor prognosis. Let’s assume that P(S) = 0.4, P(S’) = 0.6, P(F|S) = 0.8, P(F|S’) = 0.3. Applying Bayes' theorem we get:
The probability of demand for a new TV model, given a favorable forecast, is 0.64. Thus, the probability of lack of demand given a favorable forecast is 1–0.64=0.36. The calculation process is shown in Fig. 4.
Rice. 4. (a) Calculations using the Bayes formula to estimate the probability of demand for televisions; (b) Decision tree when studying demand for a new TV model
Let's look at an example of using Bayes' theorem for medical diagnostics. The probability that a person suffers from a particular disease is 0.03. A medical test can check if this is true. If a person is truly sick, the probability of an accurate diagnosis (saying that the person is sick when he really is sick) is 0.9. If a person is healthy, the probability of a false positive diagnosis (saying that a person is sick when he is healthy) is 0.02. Let's say that the medical test gives a positive result. What is the probability that a person is actually sick? What is the likelihood of an accurate diagnosis?
Let us introduce the following notation: event D - the person is sick, event D’ - the person is healthy, event T - diagnosis is positive, event T’ - diagnosis negative. From the conditions of the problem it follows that P(D) = 0.03, P(D’) = 0.97, P(T|D) = 0.90, P(T|D’) = 0.02. Applying formula (6), we obtain:
The probability that with a positive diagnosis a person is really sick is 0.582 (see also Fig. 5). Please note that the denominator of the Bayes formula is equal to the probability of a positive diagnosis, i.e. 0.0464.
Lecture 4
The principle of practical impossibility of unlikely events
If a random event has a very low probability, then we can practically assume that this event will not occur in a single trial. It all depends on the specific task. If the probability of a parachute not opening is 0.01, then such a parachute cannot be used. If the train is late with a probability of 0.01, then you can be sure that it will arrive on time.
A sufficiently small probability at which an event in a given problem can be considered practically impossible is called level of significance. In practice, significance levels of 0.01 to 0.05 are usually accepted.
If a random event has a probability very close to one, then we can practically assume that this event will occur in a single trial.
Conditional probability
The product of two events A And B call the event AB, consisting in the joint appearance (combination) of these events. For example, if A- the part is suitable, IN- the part is painted, then AB- the part is usable and painted.
The product of several events call an event consisting of the joint occurrence of all these events yy. For example, if A, B, C- the appearance of the “coat of arms” in the first, second and third tosses of the coin, respectively, then ABC- loss of the “coat of arms” in all three tests.
In the introduction, a random event is defined as an event that, when a set of conditions is fulfilled, S may or may not happen.
If, when calculating the probability of an event, no other restrictions are imposed except for the conditions S, then such a probability is calledunconditional; if other additional conditions are imposed, then the probability of the event is called conditional.
For example, the probability of an event is often calculated B with the additional condition that the event occurred A. Unconditional probability, strictly speaking, is conditional, since the fulfillment of conditions S is assumed.
Conditional probability P A (B) or call the probability of event B, calculated under the assumption that event A has already occurred
Conditional probability is calculated according to the formula
This formula can be proven based on the classical definition of probability.
Example 3. There are 3 white and 3 black balls in the urn. One ball at a time is taken out of the urn twice without replacing them. Find the probability of a white ball appearing during the second trial (event IN), if on the first trial a black ball was drawn (event A).
Solution. After the first test, there are 5 balls left in the urn, 3 of which are white. The required conditional probability R A ( IN) = 3/5.
The same result can be obtained using the formula
R A ( IN) =P (AB)/P(A) ( P (A) > 0).
Indeed, the probability of a white ball appearing on the first trial
P(A) = 3/6 =1/2.
Let's find the probability P(AB) that in the first test a black ball will appear, and in the second - a white one according to formula (3.1). The total number of outcomes - the joint appearance of two balls, no matter what color, is equal to the number of placements = 6 5 = 30. Of this number of outcomes, the event AB is favored by 3 3 = 9 outcomes. Hence, P (AB) =9/30 = 3/10.
Conditional probability P A ( IN) =P(AB)/R(A) = (3/10)/(1/2) = 3/5. The same result was obtained.
We also learned how to solve standard problems with independent events, and now a much more interesting continuation will follow, which will not only allow us to master new material, but also, perhaps, provide practical everyday help.
Let us briefly repeat what independence of events is: events are INDEPENDENT if the probability of any of them does not depend from the occurrence or non-occurrence of another event. The simplest example is tossing two coins. The probability of heads or tails on one coin does not depend in any way on the result of tossing another coin.
The concept of dependency of events is also familiar to you and it’s time to take a closer look at them.
First consider the traditional set consisting of two events: the event is dependent , if, in addition to random factors, its probability depends on the occurrence or non-occurrence of the event. Probability of an event, calculated under the assumption that the event has already happened, called conditional probability the occurrence of the event and is denoted by . In this case, the events are called dependent events (although, strictly speaking, only one of them is dependent).
Cards in hand:
Problem 1
From a deck of 36 cards, 2 cards are drawn sequentially. Find the probability that the second card will be a heart if before:
a) a worm was extracted;
b) a card of a different suit was drawn.
Solution: consider the event: – the second card will be a heart. It is absolutely clear that the probability of this event depends on whether a worm or not a worm was drawn earlier.
a) If a heart was drawn first (event), then 35 cards remain in the deck, among which there are now 8 cards of the heart suit. By classical definition:
given that, that before this a worm was also extracted.
b) If a card of a different suit was drawn first (event), then all 9 hearts remained in the deck. By classical definition:
– the probability that the second card will be a heart given that that a card of a different suit was drawn before.
Everything is logical - if the probability of drawing hearts from a full deck is 
, then when the next card is drawn, a similar probability will change: in the first case, it will decrease 
(because there are fewer hearts), and in the second it will increase: 
(since all the hearts remained in the deck).
Answer:
Of course, there may be more dependent events. While the problem is still warm, let’s add one more thing: – a heart will be drawn with the third card. Suppose that the event occurred and then the event; then there are 34 cards left in the deck, including 7 hearts. By classical definition:
– probability of an event occurring given that that two hearts had been drawn before.
For independent training:
Problem 2
The envelope contains 10 lottery tickets, including 3 winning ones. Tickets are sequentially removed from the envelope. Find the probabilities that:
a) the 2nd ticket drawn will be a winner if the 1st was a winner;
b) the 3rd one will be winning if the previous two tickets were winning;
c) The 4th will be winning if the previous tickets were winning.
A short solution with comments at the end of the lesson.
Now let’s pay attention to one fundamental thing important point: in the considered examples it was necessary to find only conditional probabilities, while previous events were considered to have taken place reliably. But in reality they are also random! Thus, in a “heated” task, drawing hearts from a full deck is a random event, the probability of which is equal to .
In practice, it is much more often necessary to find the probability co-occurrence dependent events. How, for example, to find the probability of an event consisting of a full deck will worm extracted And then another heart? The answer to this question is given by
theorem for multiplying probabilities of dependent events: the probability of the joint occurrence of two dependent events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the assumption that the first event has already occurred:
In our case:
– the probability that 2 hearts will be drawn in a row from a full deck.
Likewise:
– the probability that a card of a different suit will be drawn first And then a heart.
The probability of the event turned out to be noticeably greater than the probability of the event, which, in general, was obvious without any calculations.
And, of course, you don’t need to have any special hopes that from an envelope with ten lottery tickets (Task 2) you will draw 3 winning tickets in a row:
, however, this is still a generous chance.
Yes, that’s absolutely right - the theorem for multiplying the probabilities of dependent events naturally extends to a larger number of them.
Let's consolidate the material with several typical examples:
Problem 3
There are 4 white and 7 black balls in the urn. Two balls are drawn from the urn at random, one after the other, without replacing them. Find the probability that:
a) both balls will be white;
b) both balls will be black;
c) first the white ball will be drawn, and then the black one.
Note the qualifier “without bringing them back.” This comment further emphasizes the fact that events are dependent. Indeed, what if the extracted balls are returned back? In the case of return sampling, the probabilities of drawing a black and white ball will not change, and in such a problem one should already be guided theorem of multiplication of probabilities of INDEPENDENT events.
Solution: total in the urn: 4 + 7 = 11 balls. Go:
a) Consider the events - the first ball will be white, - the second ball will be white and find the probability of the event that the 1st ball will be white And 2nd white.
According to the classical definition of probability: . Suppose that the white ball is removed, then there will be 10 balls left in the urn, including 3 white ones, therefore:
– the probability of drawing a white ball in the 2nd trial, provided that a white ball was drawn before.
– the probability that both balls will be white.
b) Find the probability of the event that the 1st ball will be black And 2nd black
According to the classical definition: – the probability that a black ball will be drawn in the 1st trial. Let a black ball be drawn, then there will be 10 balls left in the urn, including 6 black ones, therefore: 
– the probability that in the 2nd trial a black ball will be drawn, provided that a black ball was drawn before.
According to the theorem of multiplication of probabilities of dependent events: 
– the probability that both balls will be black.
c) Find the probability of the event (the white ball will be drawn first And then black)
After removing the white ball (with probability ) there will be 10 balls left in the urn, including 3 white and 7 black, thus: – the probability that in the 2nd trial a black ball will be drawn, provided that a white ball was drawn before.
According to the theorem of multiplication of probabilities of dependent events: 
– the desired probability.
Answer: 
This problem can be easily checked using the theorem for adding the probabilities of events forming a complete group. To do this, we find the probability of the 4th missing event: 
– that the black ball will be drawn first And then white.
Events 
form a complete group, so the sum of their probabilities must equal one:
, which was what needed to be checked.
And I immediately suggest checking how well you have mastered the material presented:
Problem 4
What is the probability that two aces will be drawn in a row from a deck of 36 cards?
Problem 5
There are 6 black, 5 red and 4 white balls in the urn. Three balls are drawn sequentially. Find the probability that
a) the third ball will turn out to be white if a black and a red ball were previously drawn;
b) the first ball will be black, the second – red and the third – white.
Solutions and answers at the end of the lesson.
It must be said that many of the problems under consideration can be solved in another way, but to avoid confusion, perhaps I will keep silent about it altogether.
Probably everyone has noticed that dependent events arise in cases where a certain chain of actions is carried out. However, the sequence of actions in itself does not guarantee the dependence of events. Let, for example, a student answer the questions of some test at random - although these events happen one after another, but ignorance of the answer to one question does not depend in any way on ignorance of other answers =) Although, of course, there are patterns here =) Then completely a simple example with repeated tossing of a coin - this exciting process It’s even called this: repeated independent tests.
I tried as best I could to delay this moment and select various examples, but if in problems on multiplication theorem for independent events shooters are in charge, then here there is a real invasion of urns with balls =) Therefore, there is no escape - again an urn:
Problem 6
From an urn containing 6 white and 4 black balls, three balls are drawn at random, one after the other. Find the probability that:
a) all three balls will be black;
b) there will be at least two black balls.
Solution:total: 6 + 4 = 10 balls in the urn.
There will be a lot of events in this task, and in this regard it is more advisable to use a mixed design style, denoting only the main events in capital letters. I hope you already understand the principle by which conditional probabilities are calculated.
a) Consider the event: – all three balls will be black.
According to the theorem of multiplication of probabilities of dependent events: 
b) The second point is more interesting, consider the event: – there will be at least two black balls. This event consists of 2 incompatible outcomes: either all balls will be black (event) or 2 balls will be black and 1 white - let’s denote the last event with the letter .
The event includes 3 incompatible outcomes:
in the 1st test, white was extracted And in the 2nd And in the 3rd tests - black balls
or
And in the 2nd – BS And in the 3rd – ChS
or
in the 1st test the BS was extracted And in the 2nd – ChS And in the 3rd – BS.
Those interested can familiarize themselves with more difficult examples from collection by Chudesenko, in which several balls are transferred. For special fans, I offer tasks of increased combinational complexity - with two consecutive movements of balls from the 1st to the 2nd urn, from the 2nd to the 3rd and the final extraction of the ball from the last urn - see the latest problems in the file Additional problems on probability addition and multiplication theorems. By the way, there are many other interesting tasks there.
And at the end of this article, we will analyze the most interesting problem with which I lured you in the very first lesson =) We won’t even analyze it, but will conduct a small practical study. The calculations in general will be too cumbersome, so let’s consider a specific example:
Petya takes an exam in probability theory, and he knows 20 tickets well and 10 poorly. Suppose that on the first day part of the group, for example, 16 people, including our hero, takes the exam. In general, the situation is painfully familiar: students, one after another, enter the classroom and pull out tickets.
It is obvious that the sequential retrieval of tickets represents a chain of dependent events, and an urgent question: In what case is Petya more likely to get a “good” ticket - if he goes “in the front row”, or if he goes “in the middle”, or if he is among the last to draw a ticket? When is the best time to come?
First, let's consider an “experimentally pure” situation in which Petya keeps his chances constant - he does not receive information about what questions his classmates have already received, he does not learn anything in the corridor, waiting for his turn, etc.
Let's consider the event: - Petya will be the very first to enter the audience and pull out a “good” ticket. According to the classical definition of probability: 
.
How will the probability of getting a successful ticket change if excellent student Nastya is passed ahead? In this case, two inconsistent hypotheses are possible:
– Nastya will draw a “good” (for Petya) ticket;
– Nastya will draw a “bad” ticket, i.e. will increase Petya's chances.
The event (Petya will come in second and get a “good” ticket) becomes dependent.
1) Suppose that Nastya is with probability 
“stole” one lucky ticket from Petya. Then there will be 29 tickets left on the table, among which 19 are “good”. According to the classical definition of probability:
2) Now suppose that Nastya with probability 
“rescued” Petya from the 1st “bad” ticket. Then there will be 29 tickets left on the table, among which there are still 20 “good” ones. According to the classical definition:
Using the theorems of adding the probabilities of incompatible events and multiplying the probabilities of dependent events, we calculate the probability that Petya will draw a “good” ticket, being second in line:
The probability... remains the same! Okay, let's consider the event: - Petya will go third, letting Nastya and Lena go ahead, and pull out a “good” ticket.
There will be more hypotheses here: ladies can “rob” a gentleman of 2 successful tickets, or vice versa – save him from 2 unsuccessful ones, or extract 1 “good” and 1 “bad” ticket. If we carry out similar reasoning and use the same theorems, then... we will get the same probability value!
Thus, from a purely mathematical point of view, no matter when to go, the initial probabilities will remain unchanged. BUT. This is only an average theoretical estimate, so, for example, if Petya goes last, this does not mean at all that he will have 10 “good” and 5 “bad” tickets to choose from in accordance with his initial chances. This ratio may vary for the better or for the worse, but it is still unlikely that among the tickets there will be “one freebie”, or vice versa – “sheer horror”. Although “unique” cases are not excluded, there are still not 3 million lottery tickets with almost zero probability of a big win. Therefore, “incredible luck” or “evil fate” would be too exaggerated statements. Even if Petya knows only 3 tickets out of 30, then his chances are 10%, which is noticeably higher than zero. And from personal experience I’ll tell you the opposite case: during the exam analytical geometry I knew 24 questions out of 28 well, so I came across two “bad” questions on the ticket; Calculate the probability of this event yourself :)
Mathematics and “pure experiment” are good, but what strategy and tactics are still more profitable to follow? in real conditions? Of course, subjective factors should be taken into account, for example, the teacher’s “discount” for the “brave” or his fatigue at the end of the exam. Often these factors can even be decisive, but in the final discussions I will try not to discount additional probabilistic aspects:
If you are well prepared for the exam, then it is probably better to go “in the forefront”. While there is a full set of tickets, the postulate “ unlikely events do not occur"works for you to a much greater extent. Agree that it is much more pleasant to have the ratio “30 tickets, including 2 bad” than “15 tickets, including 2 bad.” And the fact that two unsuccessful tickets on a separate exam (and not according to the average theoretical estimate!) they will remain on the table - it is quite possible.
Now let’s consider the “Petya situation” - when the student is prepared for the exam quite well, but on the other hand, he “swims” well too. In other words, “he knows more than he doesn’t know.” In this case, it is advisable to let 5-6 people go ahead and wait for the right moment outside the audience. Act according to the situation. Pretty soon information will begin to come in about what kind of tickets classmates pulled out. (dependent events again!) , and you don’t have to waste any more energy on “played” questions - learn and repeat other tickets, thereby increasing the initial probability of your success. If the “first batch” of examinees “saved” you from 3-4 difficult (for you personally) tickets at once, then it is more profitable to get to the exam as quickly as possible - right now the chances have increased significantly. Try not to miss the moment - just a few people let ahead, and the advantage will most likely melt away. If, on the contrary, there are few “bad” tickets, wait. After a few people, this “anomaly” is again, with a high probability, if it does not disappear, then it will smooth out for the better. In the “usual” and most common case, there is also a benefit: the “24 tickets/8 bad” ratio will be better than the “30 tickets/10 bad” ratio. Why? There are no longer ten difficult tickets, but eight! We are studying the material with redoubled energy!
If you are prepared no matter or poorly, then of course it is better to go in the “last rows” (although original solutions are also possible, especially if there is nothing to lose). There is a small, but still non-zero probability that you will remain relatively simple questions+ additional cramming + spurs, which will be given by fellow students who have shot =) And, yes - in a very critical situation there is still the next day, when the second part of the group takes the exam;-)
We have already said that the basis for determining the probability of an event is a certain set of conditions. If no restrictions, other than conditions, are imposed when calculating the probability, then such probabilities are called unconditional.
However, in a number of cases it is necessary to find the probabilities of events under the additional condition that some event B has occurred, which has a non-zero probability, i.e. We will call these probabilities conditional and denote them by the symbol; this means the probability of event A given that event B occurs.
Example 1. Two are thrown dice. What is the probability that the sum of the points dropped on them is 8 (event A), if it is known that this sum is an even number (event B)?
We write down all possible cases that may arise when throwing two dice in table 1.7.1, each cell of which contains a record of a possible event: in the first place in parentheses the number of points that fell on the first dice is indicated, in the second place - the number of points rolled on the second dice.
The total number of possible cases is 36, favoring event A is 5. Thus, the unconditional probability.
If event B occurs, then one of the 18 (not 36) possibilities has occurred and, therefore, the conditional probability is equal.
Example 2. Two cards are drawn sequentially from a deck of cards. Find: a) the unconditional probability that the second card will be an ace (it is not known which card was drawn first), and b) the conditional probability that the second card will be an ace if an ace was drawn initially.
Let us denote by A the event consisting in the appearance of an ace in second place, and by B the event consisting in the appearance of an ace in first place. It is clear that there is equality.
Due to the incompatibility of the events AB and AB we have:
When drawing two cards from a deck of 36 cards, 36 * 35 (considering the order!) cases can occur. Of these, 4*3 cases favor the AB event, and 32*4 cases favor the event. Thus,
If the first card is an ace, then there are 35 cards left in the deck and among them there are only three aces. Hence, .
The general solution to the problem of finding the conditional probability for the classical definition of probability is not difficult. In fact, let m events out of n uniquely possible, incompatible and equally probable events favor event A. If event B has occurred, then this means that one of the events favoring B has occurred. Under this condition, event A is favored by r and only r events Aj favoring AB. Thus,
In the same way one can deduce that
It is clear that
that is, the probability of two events occurring is equal to the product of the probability of one of these events by the conditional probability of the other, provided that the first occurred.
The multiplication theorem is also applicable in the case when one of the events A or B is an impossible event, since in this case the equalities and take place together with.
Conditional probability has all the properties of probability. This can be easily verified by checking that it satisfies all the properties formulated in the previous paragraphs. Indeed, the first property is satisfied in an obvious way, since for each event A a non-negative function is defined. If, then
Verification of the third property is also not difficult and we leave its implementation to the reader.
Note that the probability space for conditional probabilities is given by the following triple.
Definition 1. Event A is said to be independent of event B if equality occurs, i.e., if the occurrence of event B does not change the probability of the occurrence of event A.
If event A is independent of B, then the equality holds
From here we find: that is, event B is also independent of A. Thus, the property of independence of events is mutual.
If events A and B are independent, then events A and are also independent. Indeed, since
From here we draw an important conclusion: if events A and B are independent, then every two events are also independent.
The concept of independence of events plays a significant role in probability theory and its applications. In particular, most of the results presented in this manual were obtained under the assumption of independence of certain events under consideration.
So, for example, it is clear that the loss of a coat of arms on one coin does not change the probability of the appearance of a coat of arms (tails) on another coin, unless these coins are connected to each other during throwing (for example, they are not rigidly fastened). In the same way, the birth of a boy to one mother does not change the probability of the birth of a boy (girl) to another mother. These are independent events.
For independent events, the multiplication theorem takes a particularly simple form, namely, if events A and B are independent, then
We will now generalize the concept of independence of two events to a collection of several events.
Definition 2. Events are called collectively independent if for any event from their number and arbitrary from their number the events and are mutually independent. Due to the previous, this definition is equivalent: for any
Note that for independence in the aggregate of several events, pairwise independence is not enough. This can be seen in the following simple example.
Example S.N. Bernstein. Let's imagine that the faces of the tetrahedron are colored: the 1st - red (A), the 2nd - green (B), the third - blue (C) and the 4th - all these three colors (ABC). It is easy to see that the probability of a face on which a tetrahedron will fall when thrown and its color being red is 1/2: there are four faces and two of them are red in color.
events A, B, C, thus, are pairwise independent.
However, if we know that events B and C took place, then event A certainly took place, i.e. .
Thus, events A, B, C are collectively dependent. Thus, in the general case, by definition
(In the case, the conditional probability remains undefined.) This allows us to automatically transfer all the definitions and results of this section to the general concept of probability.
Let A And IN are the two events considered in this test. In this case, the occurrence of one of the events may influence the possibility of the occurrence of another. For example, the occurrence of an event A can influence the event IN or vice versa. To take into account such dependence of some events on others, the concept of conditional probability is introduced.
Definition. If the probability of an event IN is found under the condition that the event A happened, then the resulting probability of the event IN called conditional probability events IN. To denote such a conditional probability, the following symbols are used: R A ( IN) or R(IN / A).
Note 2. In contrast to conditional probability, “unconditional” probability is also considered when any conditions for the occurrence of some event IN are missing.
Example. There are 5 balls in the urn, including 3 red and 2 blue. One ball at a time is removed from it, with and without return. Find the conditional probability of drawing a red ball for the second time, provided that the first time is drawn: a) a red ball; b) blue ball.
Let the event A– drawing the red ball for the first time, and the event IN– drawing the red ball a second time. It's obvious that R(A) = 3 / 5; then in the case when the ball taken out for the first time returns to the urn, R(IN)=3/5. In the case when the removed ball is not returned, the probability of drawing a red ball is R(IN) depends on which ball was drawn the first time - red (event A) or blue (event). Then in the first case R A ( IN) = 2 / 4, and in the second ( IN) = 3 / 4.
The theorem for multiplying the probabilities of events, one of which occurs subject to the occurrence of the other
The probability of two events occurring is equal to the product of the probability of one of them and the conditional probability of the other, found under the assumption that the first event occurred:
R(A ∙ B) = R(A) ∙ R A ( IN) . (1.7)
Proof. Indeed, let n– the total number of equally possible and incompatible (elementary) test outcomes. Let it go n 1 – number of outcomes favorable to the event A, which comes first, and m– the number of outcomes in which the event occurs IN assuming that the event A it has arrived. Thus, m is the number of outcomes favorable to the event IN. Then we get:
Those. the probability of multiple events occurring is equal to the product of the probability of one of these events and the conditional probabilities of the others, and the conditional probability of each subsequent event is calculated under the assumption that all previous events have occurred.
Example. There are 4 masters of sports in a team of 10 athletes. By drawing lots, 3 athletes are selected from the team. What is the probability that all selected athletes are masters of sports?
Solution. Let us reduce the problem to the “urn” model, i.e. Let us assume that in an urn containing 10 balls there are 4 red balls and 6 white balls. 3 balls are drawn at random from this urn (selection S= 3). Let the event A consists of extracting 3 balls. The problem can be solved in two ways: according to the classical scheme and according to formula (1.9).
The first method, based on the combinatorics formula:
The second method (according to formula (1.9)). 3 balls are drawn sequentially from the urn without replacement. Let A 1 – the first ball drawn is red, A 2 – the second drawn ball is red, A 3 – the third drawn ball is red. Let also the event A means that all 3 drawn balls are red. Then: A = A 1 ∙ (A 2 / A 1) ∙ A 3 / (A 1 ∙ A 2), i.e.
Example. Let from the set of cards a, a, p, b, o, t cards are removed sequentially one at a time. What is the probability of receiving the word “ Job” when sequentially folding them into one line from left to right?
Let IN– the event in which the declared word is obtained. Then, using formula (1.9), we obtain:
R(IN) = 1/6 ∙ 2/5 ∙ 1/4 ∙ 1/3 ∙ 1/2 ∙ 1/1 = 1/360.
The probability multiplication theorem takes on its simplest form when the product is formed by events independent of each other.
Definition. Event IN called independent from the event A, if its probability does not change depending on whether the event occurred A or not. Two events are called independent (dependent) if the occurrence of one of them does not change (changes) the probability of the occurrence of the other. Thus, for independent events p(B/A) = R(IN) or = R(IN), and for dependent events R(IN/A)
