The matter can be considered as in terms of standard methods and approaches of combination theory, and with application of probability theory. It allows to broaden a little horizons and also to look at an objective from the non-standard point of view.
Instruction
1. It is known that the probability of simple events is determined by classical formula P (A) =m/n in which number of events (outcomes) of course and ravnovozmozhno. At this n - the total number of outcomes, and m – number of favorable outcomes (statement of the problem). Now, it is necessary to consider three most widespread formulas of combination theory: shifts, combinations and placements.
2. PerestanovkiPredstavte to themselves that on a table five cards on which invisible party figures are written lie: 1, 2, 3, 4 and 5. Arbitrarily, on one, they are taken out, turn over and keep within in turn. What probability that the taken combination will be number 12345? The quantity of favorable outcomes of m is obvious – m=1. While all n=5 options! =120 where "!" - the sign of a factorial will be whole 120, and the required probability of this event P = 1/120, respectively. In this example the total number of outcomes was looked for as number of various shifts of five elements on five positions. Therefore and in any case of n of elements this number is called number of shifts and designate Pn (Pn=n!)
3. SochetaniyaSleduet to review the following example. In a basket there is a number of spheres of two flowers, equal n. In such problem statement, number of combinations from n of elements on m call a set of the ways differing from each other in quantity of spheres of different color in each combination. At this n – the total number of spheres (elements), m – number of elements in the taken combination. Combinations are various if they differ at least in one element. Designation of number of combinations and a formula for calculation are given in figure 1.
4. Presumably, it is necessary to calculate prize probability in a sport lotto of 6 of 49 where "is guessed" 4 of 6. It is obvious that at the same time the formula for a combination is used. Total number of outcomes With (from 49 on 6)=49!/43! 6! The favorable number of outcomes can be found for the following reasons. There are 6 49 numbers, "good" from total. Concerning a task there are enough 4 coincidence. From 6 "good" it is possible to choose 4 With (from 6 on 4) ways. At the same time from remained 43 "bad" get out 2 for addition of the chosen combination to six elements C (of 43 po2) ways. It sounds as follows.
5. The number of favorable situations gathers as With (from 6 on 4) and With (from 37 on 2) (a situation of logical multiplication). M= C (from 6 on 4) s means (from 43 on 2). Thus, the probability even of the "scantiest" prize P =m/n= With (from 6 on 4) s (from 43 on 2) / With (from 49 on 6)= (6!/2! 4!) (43!/2! 41!) / (49!/6! 43!) =15*21*43/66*92*47*49=9*43/92*47*154=0.000347.
6. RazmeshcheniyaEsli in a task about combinations to consider a sequence of elements in the chosen combination from m of elements, will appear a task about placements. The question on the basis of which is accepted by the decision on application of a formula of number of combinations has to additionally (in comparison with combinations) to contain data on need of accounting of an order of arrangement of elements for the chosen combinations. If m of elements is chosen, then calculating number of placements it is necessary to increase number of combinations by number of shifts of Pm=m!. Designation of number of placements and a formula for its calculation are given in fig. 2.