Are there combinations of permutations there in GRE

Determine the probabilities using counting strategies

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1 R. Brinmann page Determining the probability pages with the help of counting strategies The previous tasks on the probability calculation could essentially be processed with clear result trees. But this method has its limits. This is shown by the example of throwing a dice several times. Orderly sample with retraction. Example: A die is thrown. According to the urn model, this means that a ball is drawn from an urn that contains balls with the numbers to when you put it back. A: With every throw or move you get a 4. a) What is the probability of getting a 4 with each of the throws or moves? b) How many elements does the result set contain (number of all possibilities)? / Level branches / level 2 = branches = 2 / / level 3 3 = branches 2 3/4 P (A) = ... times 5 level ... = branches times Created by R. Brinmann p9_w_rechnung_09.doc: 27 page from 2

2 R. Brinmann page a) Since the experiment is a Laplace experiment, where the same probability can be assumed for all results, the following applies to the probability of getting a 4 after each level: P (A) = ... = = times b) Since the number of branches in the tree diagram increases sixfold with each level, the following applies to the th level: ... = branches, that is exactly the number of all possibilities. times The number of possibilities can also be found via the probability of the individual results. The following applies to each individual result: P (ei) = But since the sum of the probabilities of all individual results must be, the number of all possible options can also be calculated from this: x = x = If one generalizes this law in such a way that one says: There is an urn n balls of the same kind with the numbers, 2, ..., n, where you sometimes move with back, then the number of possibilities is n. Mers: From n different elements of a set you get * n n ... n = nn, N ordered samples with retraction. times () Example: With a football bet (Toto) you have to predict whether the home team will win (tip:) or the visiting team (tip: 2) or whether both teams will play a draw (tip: 0) for games on a weekend. a) How many opportunities are there to fill out a tote sheet? b) What is the probability of a correct tip? Solution: a) Modeling with the urn model: An urn contains three balls with the numbers 0; and 2. It is drawn with putting back. n = 3 result set for a move: E = {0 ;; 2} = The random experiment is - stage. Number of possible results: x = n = 3 = 7747 b) Since there is only one possible tip with correct ones, the probability for this is: 7747 Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 Page 2 of 2

3 R. Brinmann Seite Exercise: A bicycle lock (combination lock) consists of four wheels that can be moved independently of one another and each contain digits (from to). The lock only opens with a very specific combination of numbers. How many positions (number combinations) does the bicycle lock have and what is the probability of opening the lock the first time it is set? Solution: Modeling with the urn model: An urn contains n = balls with the numbers to. It is drawn = 4 times with putting back. 4 The number of number combinations is: n = = 29 The probability of trying to find the right combination is 0, Exercise: From the 2 letters of the alphabet three letters are taken blindly one after the other with putting them back. What is the probability of drawing the same letter three times? Solution: Modeling with the urn model: An urn contains n = 2 balls with the letters A to Z. It is drawn = 3 times with putting back. 3 The number of letter combinations is: n = 2 = 757. B. to draw the letter A 3 times is times B, or C or any other letter has the same probability. There are a total of 2 favorable cases with the 2 letters of the Alpabet. This gives the probability of drawing three identical letters:, 0048 Ordered sample without moving back. Example: In an urn there are 4 balls with the colors red, yellow, green and blue. You draw a ball, register the number, put the ball aside and repeat the process. A total of 4 moves are possible, then the urn is empty. How many elements does the result set contain (number of all possibilities)? Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 Page 3 of 2

4 R. Brinmann page Level 4 branches Level 2 () 4 4 branches Level branches () () Level branches () () () As can be easily seen from the tree diagram, the number of branches decreases from level to level. The number of possibilities is 4 (4) (4 2) (4 3) = = 24 The law to be read from the tree diagram can be generalized. If one now looks at an urn with n balls numbered from to n and makes moves without replacing, the following applies to the number of possibilities: nnn 2 n 3 ... n + () () () () A product in which everyone Folgefator is called Faultät it is called 4 - Faultät and writes 4! For the number n we have n! = n n n 2 n n! read n - Faultät (n)! = (n) (n) (n 2) or in short form (n)! = (n) The expression n (n) (n 2) (n 3) ... (n +) can be transformed as follows: n (n) (n 2) (n 3) ... (n +) (n) n! = n n! () () () () () () Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 Page 4 of 2

5 R. Brinmann Seite Mere: From n different elements of a set one obtains by drawing (n) (n 2) (n 3) ... (n) + = ordered samples without putting them back. Definition: 0! = and! = n! (n)! Example: A computer program is protected by a password. This password consists of 4 different letters. a) How many passwords are possible? b) What is the probability that the code can be genacted in one attempt? Solution: a) All 2 letters of the alphabet are available exactly once. For the first letter of the word, all 2 letters of the alphabet are used, for the second only 25 letters, etc. It is an orderly sample with no setting back. From n = 2 letters = 4 letters are drawn. n! 2! Number of possibilities: = = = (n)! 22! b) Since there is only one correct code, the probability of success is calculated immediately: P (A) = 0, exercise: There are lots with the numbers to in a lottery wheel. A player draws three tickets one after the other. If he draws the numbers 2, 4 and in that order, he wins. Calculate the probability of winning. Solution: First the number of possibilities is calculated, of which there is only one that leads to a profit, namely the sequence of numbers 2, 4 ,. It is an orderly sample without moving back. From n = numbers = 3 numbers are drawn. n !! Number of possibilities: = = = 5 4 = 20 (n)! 3! 3 2 P (A) = 0, created by R. Brinmann p9_w_rechnung_09.doc: 27 page 5 of 2

6 R. Brinmann page Disordered sample without moving back. Example: When drawing the lottery numbers, numbers are drawn from a total of 49 numbers. This is a pull without putting back. n! 49! The number of possibilities as an ordered sample is: = (n)! 43! Since the order of the drawn numbers does not matter in the drawing, the number of possibilities is reduced by the part of how often the drawn numbers can be arranged. Are z. If, for example, the numbers 3, 2, 7, 22, 3 and 4 are drawn, they can also be arranged in the form 7, 22, 4, 3, 3 and 2. That has a meaning for the profit. In order to find out the number of possibilities in the lottery, we need to find out the number of possible swaps of the numbers. In other words, we need to figure out how many different ways these numbers can be arranged. The solution can easily be found through an urn experiment. In an urn there are n = balls with the numbers from to. If you now move in sequence (pulling without putting back) = times until the urn is empty, you have found all possible ways to arrange the numbers. n !!!!! (n)! = ()! = 0! = = If an urn with n elements is drawn (pulling without putting it back) until the urn is empty, then it is called an orderly census. In this case n =. Mere: For n different elements there are n! Full surveys. In other words: A set of n different elements can be reduced to n! arrange different ways. Let's go back to our lottery example. So far we have determined how many ways there are to draw numbers from 49 numbers. Since the order of the drawn numbers is not important for the evaluation, the number of possibilities must go through! to be shared. This means that the number of opportunities to have the lottery is correct: n! 49! = = =! n !! 43! () Mere: If one selects elements from a set with n different elements, then the number of unordered samples is n n! =! (n)! n The expression (read n over) is called the binomial coefficient. Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 page of 2

7 R. Brinmann Seite Example: 4 cards are drawn from a deck of 32 cards. What is the probability that these are 4 jacks? Solution: Disordered random sample without moving back. n = 32; = 4 number of possibilities: n n! 32! = = = = 3590! (n)! 4! 28! P (A) = 0, exercise: 8 cards are drawn from a deck of 32 cards. What is the probability that these are 8 diamonds? Solution: 8 cards are drawn from a deck of 32 cards.

What is the probability that these are 8 diamonds? n = 32; = 8 number of possibilities: n n! 32! = = = =! (n)! 8th! 24! P (A) = somewhat more sophisticated pocket calculators have function keys for the above formulas, with which the calculation process can be greatly simplified. For the TI 30 eco RS from Texas Instruments, for example, the following applies: Special General 8 x 2 2 y 8 = 25 n n y = ... 2! n! 2 2nd npr 4 = n 2nd npr = ... 4! n! 32 n 32 2nd ncr 4 = 3590 n 2nd ncr =! 4 x! = 24 n! n x! = ... (2) () x Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 Page 7 of 2

8 R. Brinmann Page Summary: Arrangement of elements: Number of possibilities :! Ordered sample with moving back: Number of possibilities: Draw n times () Ordered sample without moving back: n! Number of possibilities: draw n times! () () Disordered sample without setting back: n n! Number of possibilities: =! (n)! pull times n n Eas: 0! =! = and = = 0 Lotto out of 49 Probability for less than correct and additional number. Preliminary consideration: The definition of the probability of the occurrence of an event A is defined as follows: () () P A = number of all possibilities that belong to A number of all possibilities of the random attempt In the German Lotto, 49 numbers are used. The number of ways to do this is 49! = = = 39838! 43! The number of possibilities of winning numbers is to be judged exactly! = =! 0! The event whose probability is to be determined is: A: six correct ones in the lottery To A belongs exactly one possibility out of a total of possibilities. Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 page 8 of 2

9 R. Brinmann Seite This means that P (A) = 0, the probability of having exactly the right one for a tip. As a new event we define B: 4 correct ones in the lottery. This means that 4 of the winning numbers were used, 2 of the used numbers belong to the 49 = 43 non-winning numbers. The number of possibilities of winning numbers 4 is to be trimmed! 5 30 = = = = 5 4 4! 2! 2 2 The number of possibilities of the 43 non-winning numbers 2 is to be scored: 43 43! = = = =! 4! 2 2 This is the number of options for 4 correct ones in the lottery: 43 = = B includes a total of 3545 options out of a total of possible options. This means that P (B) = 0, the probability of having exactly 4 correct bets for a tip. The following scheme should once again illustrate the need to multiply 5 by 903: from 43 In one line, the used ggggnn ... ggggnn winning numbers (gggg) remain the same, the used non-winning numbers (nn) change ggggnn ... ggggnn 4 In one column, the applied non-winning numbers (nn) remain the same, the applied winning numbers (gggg) change. Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 page 9 of 2

10 R. Brinmann Seite We define C: 5 correct with an additional number as a new event. Number of possibilities for: 5 winning numbers added (5 from) = 5 additional numbers added (from) = 42 0 not used winning numbers (0 from 42) = PC () = = = 0, If the probability of a tip is exactly 5 correct with an additional number to have. Additional information on the lottery In Germany, the Deutsche Lotto- und Totobloc, an association of state lottery companies, operates the lottery game. You can also take part in the Super and Game 77 games. In addition to the numbers, an additional number and a super number are drawn. The bonus number is drawn from the remaining 43 balls as the seventh, after the first numbers. It increases the profit by one level for the lower profit classes. In contrast, the super number (only) for the Jacpot results from the numbers 0 to 9 from the last digit, the game 77 or super number already reduced on the game receipt. This is another lot, so to speak - but with the result that this chance is ten times lower. Prepared by R. Brinmann p9_w_rechnung_09.doc: 27 Page 0 of 2

11 R. Brinmann Page Profit Letting: Profit Letting Num. the correct probability for a tip class with super number = 0, class 2 = 0, class 3 5 with additional number = 0, class = 0, class 5 4 with additional number = 0, created by R. Brinmann p9_w_rechnung_09.doc: 27 page of 2