Probability theory. Problem solving (2020)
There will also be tasks for an independent solution, to which you can see the answers.
The general formulation of the problem: the probabilities of some events are known, but the probabilities of other events that are associated with these events need to be calculated. In these tasks, there is a need for such actions on probabilities as addition and multiplication of probabilities.
For example, when hunting, two shots are fired. Event A- hitting a duck from the first shot, event B- hit from the second shot. Then the sum of events A and B- hit from the first or second shot or from two shots.
Tasks of a different type. Several events are given, for example, the coin is tossed three times. It is required to find the probability that either the coat of arms will be dropped all three times, or that the coat of arms will be drawn at least once. This is a problem of multiplying probabilities.
Adding the probabilities of inconsistent events
Addition of probabilities is used when you need to calculate the probability of a union or logical sum of random events.
Sum of events A and B denote A + B or A ∪ B... The sum of two events is an event that occurs if and only when at least one of the events occurs. It means that A + B- an event that occurs if and only when an event occurred during observation A or event B, or at the same time A and B.
If events A and B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one test is calculated using the addition of the probabilities.
The addition theorem for probabilities. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:
For example, while hunting, two shots are fired. Event A- hitting a duck from the first shot, event V- hit from the second shot, event ( A+ V) - hit from the first or second shot or from two shots. So if two events A and V- incompatible events, then A+ V- the onset of at least one of these events or two events.
Example 1. In the box there are 30 balls of the same size: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be taken without looking.
Solution. Let's assume that the event A- "the red ball is taken", and the event V- "a blue ball is taken." Then the event is “a colored (not white) ball is taken”. Find the probability of an event A:
and events V:
Developments A and V- mutually incompatible, since if one ball is taken, then you cannot take balls of different colors. Therefore, we use the addition of probabilities:
The theorem of addition of probabilities for several inconsistent events. If events make up the complete set of events, then the sum of their probabilities is 1:
The sum of the probabilities of opposite events is also equal to 1:
Opposite events form a complete set of events, and the probability of a complete set of events is 1.
The probabilities of opposite events are usually denoted in small letters p and q... In particular,
from which the following formulas for the probability of opposite events follow:
Example 2. The target in the shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the target in the first zone is 0.15, in the second zone - 0.23, in the third zone - 0.17. Find the probability that the shooter will hit the target and the probability that the shooter will miss the target.
Solution: Let's find the probability that the shooter will hit the target:
Let's find the probability that the shooter misses the target:
More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various problems on addition and multiplication of probabilities".
Addition of probabilities of mutually compatible events
Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a dice, the event A the fall of the number 4 is considered, and the event V- an even number dropped out. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of one of the mutually joint events.
Probability addition theorem for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events is as follows:
Since events A and V compatible, event A+ V occurs if one of three possible events occurs: or AB... According to the theorem of addition of incompatible events, we calculate as follows:
Event A will occur if one of two inconsistent events occurs: or AB... However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:
Likewise:
Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:
When using formula (8), it should be borne in mind that events A and V may be:
- mutually independent;
- mutually dependent.
Probability formula for mutually independent events:
Probability formula for mutually dependent events:
If events A and V are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for inconsistent events is as follows:
Example 3. In a car race, when driving the first car, there is a chance of winning, when driving in the second car. Find:
- the likelihood that both cars will win;
- the probability that at least one car will win;
1) The probability that the first car will win does not depend on the result of the second car, therefore the events A(the first car wins) and V(second car wins) - independent events. Let's find the probability that both cars will win:
2) Let's find the probability that one of the two cars will win:
More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various problems on addition and multiplication of probabilities".
Solve the probability addition problem yourself, and then see the solution
Example 4. Two coins are thrown. Event A- falling out of the coat of arms on the first coin. Event B- falling out of the coat of arms on the second coin. Find the probability of an event C = A + B .
Multiplication of probabilities
Probability multiplication is used when calculating the probability of the logical product of events.
Moreover, random events must be independent. Two events are called mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.
The multiplication theorem for probabilities for independent events. Probability of simultaneous occurrence of two independent events A and V is equal to the product of the probabilities of these events and is calculated by the formula:
Example 5. The coin is thrown three times in a row. Find the probability that the coat of arms will be dropped all three times.
Solution. The probability that on the first toss of the coin the coat of arms will appear, the second time, the third time. Let's find the probability that the coat of arms will be drawn all three times:
Solve probability multiplication problems yourself, and then see the solution
Example 6. Includes a box of nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, played and unplayed are not distinguished. What is the probability that after three games there will be no balls left in the box?
Example 7. 32 letters of the Russian alphabet are written on the cards of the split alphabet. Five cards are taken out at random one after the other and placed on the table in the order of appearance. Find the probability that the letters will form the word "end".
Example 8. From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards are of different suits.
Example 9. The same problem as in example 8, but after being taken out, each card is returned to the deck.
More difficult tasks in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events - on the page "Various problems on addition and multiplication of probabilities".
The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula.
The union (logical sum) of N events is called an event 
that is observed every time at least one of events 
... In particular, the union of events A and B is the event A+
B(some authors
), which is observed when comesor
A,or
Bor
both of these events at the same time(Fig. 7). The sign of intersection in the textual wording of events is the union "or".
Rice. 7. Combining events A + B
It should be borne in mind that the probability of the event P (A) corresponds to both the left part of the shaded part in Fig. 7 of the figure, and its central part, marked as 
... And the outcomes corresponding to event B are located both on the right side of the shaded figure and in the marked 
central part. Thus, when adding 
and 
area 
will actually enter this sum twice, and the exact expression for the area of the shaded figure is 
.
So, likelihood of unification two events A and B is equal to
For a large number of events, the general calculation expression becomes extremely cumbersome due to the need to take into account numerous variants of the mutual overlap of regions. However, if the events being combined are incompatible (see p. 33), then the overlapping of areas turns out to be impossible, and the favorable zone is determined directly by the sum of the areas corresponding to individual events.
Probability amalgamations arbitrary number inconsistent events 
defined by the expression
|
|
Corollary 1: A complete group of events consists of inconsistent events, one of which is necessarily realized in the experience. As a result, if events
…
,form a complete group, then for them
Thus,
WITHConsequence 3 Let us take into account that the opposite statement “will occur at least one of the events 
…
"Is the statement" none of the events 
…
not implemented. " That is, in other words, “in experience, events will be observed 
, and 
, and ..., and 
", Which is already the intersection of events opposite to the original set. Hence, taking into account (2 .0), to combine an arbitrary number of events, we obtain
|
|
Corollaries 2, 3 show that in cases where direct calculation of the probability of an event is problematic, it is useful to assess the complexity of researching the event opposite to it. After all, knowing the meaning 
, obtain from (2 .0) the required value 
no longer presents any work.
Examples of calculating the probabilities of complex events
Example 1 : Two students (Ivanov and Petrov) together Irushed to the defense of laboratory work, having learned the first 8 conof the 10 available questions to this work. Checking the readiness, nthe teacher asks everyone only onen a randomly selected question. Determine the likelihood of the following events:
A= “Ivanov will defend laboratory work”;
B= “Petrov will defend laboratory work”;
C= “Both defend laboratory work”;
D= “At least one of the students will defend the job”;
E= “Only one of the students will defend the job”;
F= "None of them will defend the job."
Solution. Note that the ability to defend a job like Ivanova, tas and Petrova separately is determined only by the number of questions mastered, the poetat... (Note: in this example, the values of the resulting fractions were deliberately not reduced to simplify the comparison of the calculation results.)
EventCcan be formulated differently as "the work will be protected by both Ivanov and Petrov", i.e. will happenand eventA, and eventB... Thus, the eventCis the intersection of eventsAandB, and in accordance with (2 .0)
where the factor “7/9” appears due to the fact that the occurrence of the eventAmeans that Ivanov got a “good” question, which means that Petrov now has only 7 “good” questions out of the remaining 9 questions.
EventDimplies that “the work will be protectedor Ivanov,or Petrov,or they are both together, ”that is, at least one of the events will occurAandB... So the eventDis a union of eventsAandB, and in accordance with (2 .0)
which is in line with expectations, since even for each student individually, the chances of success are quite high.
WITHoccurrence E means that “either Ivano will defend the jobc, and Petrov "pfalls ",or Ivanov will be caught unsuccessful injust ask, and Petrov will cope with the defense. " The two alternatives are mutually exclusive (inconsistent), so
Finally, the statementFwill be true only if “and Ivanov,and Petrov with protectionnot will cope. " So,
This completes the solution of the problem, but it is useful to note the following points:
1. Each of the obtained probabilities satisfies condition (1 .0),oh if for
and 
get conflictwith(1 .0) is in principle impossible, then for
try andusing (2 .0) instead of (2 .0) would lead to an obviously uncorrectedect value
... It is important to remember that such a value of probability is fundamentally impossible, and upon obtaining such a paradoxical result, immediately start looking for an error.
2. The found probabilities satisfy the relationsm
.
NSthen it is quite expected, since developmentsC, EandFform fullgroup, and eventsDandFare opposite to each other. Consideration of theseratios on the one hand can be usedvan to double-check the calculations, and in another situation it can serve as the basis for an alternative way to solve the problem.
NS note : Don't neglect the written fixationthe exact formulation of the event, otherwise, in the course of solving the problem, you can involuntarily switch to a different interpretation of the meaning of this event, which will entail errors in reasoning.
Example 2 : In a large batch of microcircuits that have not passed the final quality control, 30% of the products are defective.If you choose any two microcircuits from this batch at random, then what isthe probability that among them:
A= “Both valid”;
B= "Exactly 1 usable microcircuit";
C= “Both defective”.
Let us analyze the following version of reasoning (careful, contains a bug):
Since we are talking about a large batch of products, the removal of several microcircuits from it practically does not affect the ratio of the number of good and defective products, which means that choosing some microcircuits from this batch several times in a row, we can assume that in each of the cases there remain unchanged probabilities
=
P(defective item selected) = 0.3 and
=
P(good product selected) = 0.7.
For the event to occurAit is necessary thatand first,and for the second time a suitable product was chosen, and therefore (taking into account the independence from each other of the success of the choice of the first and second microcircuits) for the intersection of events we have
Similarly, for event C to occur, both products must be defective, and to obtain B, you need to select a suitable product once, and a defective product once.
Error symptom. NSalthough all the probabilities obtained aboveand look plausible, when analyzed together, it is easy tonote that .However, casesA, BandCform a completethe group of events for which to run .This contradiction indicates the presence of some kind of error in reasoning.
WITH a lot of mistakes. We introduce into consideration two auxiliaryflax events:
= "The first microcircuit is good, the second is defective";
= "The first microcircuit is defective, the second is good."
It is obvious that, however, it is precisely this version of the calculation that was used above to obtain the probability of an eventBalthough eventsBand
are not uhequivalent... Actually,
since formulationdevelopmentsBrequires that among the microcircuits exactlyone
but absolutelynot necessarily the first
was good (and the other was defective). Therefore, although
event 
is not a duplicate of the event 
, but should take into accountindependently. Given the inconsistency of events 
and 
,
the probability of their logical sum will be
After the indicated correction of the calculations, we have
which indirectly confirms the correctness of the found probabilities.
Note : Pay special attention to the difference in the wording of events such as “onlyfirst of the listed elements must ... "and" onlyone of the listed aleents should ... ”. The latter event is clearly broader and includesTin its composition the first as one of the (possibly numerousx) options. These alternatives (even if their probabilities coincide) should be considered independently of each other.
NS note : The word "percentage" comes from "per cent”, I.e."One hundred". The representation of frequencies and probabilities in percentages allows you to operate with larger values, which sometimes makes it easier to perceive the values "by ear". However, it is cumbersome and inefficient to use multiplication or division by “100%” in calculations for correct normalization. In this regard, not sWhen using values, be sure to mentionexpressed as a percentage, substitute them in the calculated expressions forthe same in the form of fractions of one (for example, 35% in the calculation is writteni as “0.35”) to minimize the risk of erroneous normalization of the results.
Example 3 : The resistor set contains one resistor4 kΩ nominal, three 8 kΩ resistors and six resistorsorors with a resistance of 15 kOhm. Three resistors chosen at random are connected in parallel with each other. Determine the probability of obtaining the final resistance not exceeding 4 kOhm.
Resh enie. Parallel connection resistance cuthistory can be calculated by the formula
.
This allows events such as
A= “Three 15 kΩ resistors selected” = “
”;
B= “Intwo resistors of 15 kOhm and one with resistancem 8 kOhm "="
”…
The complete group of events corresponding to the condition of the problem includes a number of more options, and it is precisely those thatwhich comply with the requirement to obtain a resistance of not more than 4 kOhm. However, although the “direct” solution path, involving the calculation (and subsequent sumsration) of the probabilities characterizing all these events, and is correct, it is impractical to act in this way.
Note that to obtain a final resistance of less than 4 kΩ dit is sufficient that at least one resistor with resistanceless than 15 kOhm. Thus, only in the caseAthe requirement of the task is not met, i.e. eventAis anopposite investigated. However,
.
Thus, .
NS
ri
tagging
: Calculating the probability of some eventA, do not forget to analyze the complexity of the definitionI the likelihood of an event opposite to it. If rassread
easy, then it is with this that you need to start resolvedits tasks, completing it by applying the relation (2 .0).
NS Example 4 : The box containsnwhite,mblack andkred balls. The balls are taken out of the box one at a timeand returned back after each extraction. Determine the probabilitydevelopmentsA= “White ballwill be extracted earlier than black” .
Resh enie. Consider the following set of events
= “The white ball was taken out on the first try”;
= “First took out the red ball and then the white one”;
= “Twice removed the red ball, and the third time - the white”…
So toAs the balls are returned, the sequence of sobout
can be formally infinitely long.
These events are inconsistent and in the aggregate make up the set of situations in which the event occurs.A... Thus,
It is easy to see that the terms included in the sum formgeometric progression
with initial element
and the denominator 
... But the sumsand the elements of an infinite geometric progression is
|
|
.
Thus, . LIt is curious that this probability (as follows from the obtainedth expression) does not depend on the number of red balls in the box.
How to calculate the probability of an event?
I understand that everyone wants to know in advance how the sporting event will end, who will win and who will lose. With this information, you can bet on sporting events without fear. But is it possible at all, and if so, how to calculate the probability of an event?
Probability is a relative value, therefore it cannot speak with accuracy about any event. This value allows you to analyze and assess the need to place a bet on a particular competition. Determining probabilities is a whole science that requires careful study and understanding.
Probability Coefficient in Probability Theory
In sports betting, there are several options for the outcome of the competition:
- victory of the first team;
- victory of the second team;
- draw;
- total.
Each outcome of the competition has its own probability and frequency with which this event will occur, provided that the initial characteristics are preserved. As mentioned earlier, it is impossible to accurately calculate the probability of any event - it may or may not coincide. Thus, your bet can win or lose.
There can be no exact 100% prediction of the results of the competition, since many factors influence the outcome of the match. Naturally, the bookmakers do not know in advance the outcome of the match and only assume the result, making a decision on their analysis system and offer certain odds for bets.
How to calculate the probability of an event?
Let's say the bookmaker's coefficient is 2. 1/2 - we get 50%. It turns out that the coefficient 2 is equal to the probability of 50%. By the same principle, you can get a break-even odds ratio - 1 / probability.
Many players think that after several repeated defeats, there will definitely be a win - this is a misconception. The probability of winning a bet does not depend on the number of losses. Even if you throw several heads in a row in a coin game, the probability of throwing heads remains the same - 50%.
Want to know what the mathematical odds of your bet will be successful? Then there are two good news for you. First: to calculate the cross-country ability, you do not need to carry out complex calculations and spend a lot of time. It is enough to use simple formulas, which will take a couple of minutes to work with. Second, after reading this article, you can easily calculate the probability of passing any of your trade.
To correctly determine the patency, you need to take three steps:
- Calculate the percentage of the probability of the outcome of the event in the opinion of the bookmaker;
- Calculate the probability from statistical data yourself;
- Find out the value of the bet, considering both probabilities.
Let's consider each of the steps in detail, using not only formulas, but also examples.
The first step is to find out with what probability the bookmaker himself estimates the chances of a particular outcome. After all, it is clear that bookmaker's odds are not set just like that. To do this, we use the following formula:
PB= (1 / K) * 100%,
where P B is the probability of the outcome according to the bookmaker's office;
K is the bookmaker's coefficient for the outcome.
Let's say there is a coefficient of 4 for the victory of London Arsenal in a duel against Bayern Munich. This means that the probability of his Victoria BC is regarded as (1/4) * 100% = 25%. Or Djokovic plays against Yuzhny. There is a multiplier of 1.2 for Novak to win, and his chances are (1 / 1.2) * 100% = 83%.
This is how the bookmaker itself estimates the chances of success for each player and team. After completing the first step, we move on to the second.
Calculation of the probability of an event by the player
The second point of our plan is our own assessment of the likelihood of an event. Since we cannot take into account mathematically such parameters as motivation, game tone, we will use a simplified model and will only use the statistics of previous meetings. To calculate the statistical probability of the outcome, we use the formula:
PAND= (UM / M) * 100%,
wherePAND- the probability of the event in the opinion of the player;
UM - the number of successful matches in which such an event took place;
M is the total number of matches.
To make it clearer, we will give examples. Andy Murray and Rafael Nadal have played 14 matches. In 6 of them the total was less than 21 in games, in 8 - the total was more. It is necessary to find out the probability that the next fight will be played by the total more: (8/14) * 100 = 57%. Valencia played 74 matches in Mestalla against Atlético, in which they won 29 victories. Valencia's chance of winning: (29/74) * 100% = 39%.
And we learn all this only thanks to the statistics of previous games! Naturally, such a probability cannot be calculated for a new team or player, so this betting strategy is suitable only for matches in which opponents have not met for the first time. Now we are able to determine the bookmaker's and our own probability of outcomes, and we have all the knowledge to proceed to the last step.
Determining the value of a bet
The value (value) of the bet and the passability have a direct connection: the higher the value, the higher the chance of passing. The value is calculated as follows:
V =PAND* K-100%,
where V is the value;
P AND - the probability of the outcome in the better's opinion;
K is the bookmaker's coefficient for the outcome.
Let's say we want to bet on Milan's victory in the match against Roma and calculated that the probability of victory for the “red-blacks” is 45%. The bookmaker offers us a coefficient of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V = 45% * 2.5-100% = 12.5%. Great, this is a valuable bet with good odds of passage.
Let's take another case. Maria Sharapova plays against Petra Kvitova. We want to make a deal for Maria to win, the probability of which, according to our calculations, is 60%. The offices offer a multiplier of 1.5 for this outcome. Determine the value: V = 60% * 1.5-100 = -10%. As you can see, this rate is of no value and should be refrained from.
The probability of passing the bet: conclusion
When calculating the passability of the bet, we used a simple model, which is based only on statistics. When calculating the probability, it is advisable to take into account many different factors that are individual in each sport. It happens that it is not statistical factors that have more influence. Without this, everything would be simple and predictable. By choosing your niche, you will eventually learn to take into account all these nuances and give a more accurate assessment of your own likelihood of events, including many other influences. The main thing is to love what you are doing, gradually move forward and step by step improve your skills. Good luck and success in the exciting world of betting!
TOPIC 1 ... The classic formula for calculating the probability.
Basic definitions and formulas:
An experiment whose outcome cannot be predicted is called random experiment(SE).
An event that in a given SE may or may not happen is called random event.
Elementary outcomes call events that meet the requirements:
1. for any implementation of SE, one and only one elementary outcome occurs;
2. every event is a certain combination, a certain set of elementary outcomes.
The set of all possible elementary outcomes fully describes the SE. Such a set is usually called space of elementary outcomes(PEI). The choice of the SEI for the description of this SE is ambiguous and depends on the problem being solved.
P (A) = n (A) / n,
where n is the total number of equally possible outcomes,
n (A) is the number of outcomes that make up event A, as they also say, favorable to event A.
The words “at random”, “at random”, “randomly” just guarantee the equal possibility of elementary outcomes.
Solution of typical examples
Example 1. From an urn containing 5 red, 3 black and 2 white balls, 3 balls are taken at random. Find the probabilities of events:
A- “all extracted balls are red”;
V- “all extracted balls are of the same color”;
WITH- “among those extracted there are exactly 2 black ones”.
Solution:
The elementary outcome of this FE is a triplet (disordered!) Balls. Therefore, the total number of outcomes is the number of combinations: n == 120 (10 = 5 + 3 + 2).
Event A consists only of those triplets that were drawn from five red balls, i.e. n (A) == 10.
Event V in addition to 10 red triplets, black triplets are also favored, the number of which is = 1. Therefore: n (B) = 10 + 1 = 11.
Event WITH those triplets of balls that contain 2 black and one non-black are favored. Each method of choosing two black balls can be combined with the choice of one non-black (out of seven). Therefore: n (C) = = 3 * 7 = 21.
So: P (A) = 10/120; P (B) = 11/120; P (C) = 21/120.
Example 2. Under the conditions of the previous problem, we will assume that the balls of each color have their own numbering, starting from 1. Find the probabilities of events:
D- “the maximum number extracted is 4”;
E- “the maximum number extracted is 3”.
Solution:
To calculate n (D), we can assume that the urn contains one ball with number 4, one ball with a higher number, and 8 balls (3k + 3h + 2b) with lower numbers. Event D those triplets of balls that necessarily contain a ball with number 4 and 2 balls with lower numbers are favored. Therefore: n (D) = 
P (D) = 28/120.
To calculate n (E), we consider: in the urn there are two balls with number 3, two with large numbers and six balls with lower numbers (2k + 2h + 2b). Event E consists of two types of triplets:
1. one ball with number 3 and two with lower numbers;
2.two balls with number 3 and one with a lower number.
Therefore: n (E) = 
P (E) = 36/120.
Example 3. Each of the M different particles is thrown at random into one of the N cells. Find the probabilities of events:
A- all particles hit the second cell;
V- all particles hit one cell;
WITH- each cell contains no more than one particle (M £ N);
D- all cells are occupied (M = N +1);
E- the second cell contains exactly To particles.
Solution:
For each particle, there are N ways to get into one or another cell. According to the basic principle of combinatorics for M particles, we have N * N * N *… * N (M-times). So, the total number of outcomes in this SE n = N M.
For each particle we have one opportunity to get into the second cell, therefore n (A) = 1 * 1 *… * 1 = 1 M = 1, and P (A) = 1 / N M.
To get into one cell (to all particles) means to get all into the first, or all into the second, or etc. everyone in the N-th. But each of these N options can be implemented in one way. Therefore, n (B) = 1 + 1 +… + 1 (N-times) = N and P (B) = N / N M.
Event C means that each particle has one less placement method than the previous particle, and the first can fall into any of N cells. That's why:
n (C) = N * (N -1) * ... * (N + M -1) and P (C) = 
In the particular case when M = N: P (C) =
Event D means that one of the cells contains two particles, and each of the (N -1) remaining cells contains one particle. To find n (D), we argue as follows: choose a cell in which there will be two particles, this can be done = N ways; then we select two particles for this cell, there are ways for this. After that we will distribute the remaining (N -1) particles one by one into the remaining (N -1) cells, for this we have (N -1)! ways.
So n (D) = 
.
The number n (E) can be calculated as follows: To particles for the second cell can be ways, the remaining (M - K) particles are randomly distributed over the (N -1) cell (N -1) M-K ways. That's why:
