Aufgaben:Exercise 2.4: Number Lottery (6 from 49) - Revision history

Guenter at 15:48, 14 December 2021

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Guenter at 15:46, 14 December 2021

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Guenter at 15:39, 14 December 2021

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Guenter at 15:11, 14 December 2021

2021-12-14T15:11:42Z

Noah at 16:40, 12 December 2021

2021-12-12T16:40:29Z

Noah at 16:33, 12 December 2021

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Noah at 22:56, 10 December 2021

2021-12-10T22:56:26Z

Guenter at 16:43, 30 November 2021

2021-11-30T16:43:51Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.4: Zahlenlotto (6 aus 49) to Aufgaben:Exercise 2.4: Number Lottery (6 from 49)

2021-11-30T16:42:10Z

Guenter moved page Aufgabe 2.4: Zahlenlotto (6 aus 49) to Exercise 2.4: Number Lottery (6 from 49)

Javier: Text replacement - "”" to """

2021-05-28T14:26:38Z

Text replacement - "”" to """

@@ Line 17: / Line 17: @@
 :* 6 correct numbers without&nbsp; "super number" &nbsp; &rArr; &nbsp; $\text{"6R"}$,
 :* 5 correct numbers with&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R + Z"}$,
-:* 5 correct numbers without&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R - Z"}$,
+:* 5 correct numbers without&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R &ndash; Z"}$,
 :* 4 correct numbers &nbsp; &rArr; &nbsp; $\text{"4R"}$,
 :* 3 correct numbers &nbsp; &rArr; &nbsp; $\text{"3R"}$.
@@ Line 55: / Line 55: @@
 {What is the probability for&nbsp; $\text{"5 correct numbers without additional number"}$&nbsp;?
 |type="{}"}
-$\text{Pr("5R - Z")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
+$\text{Pr("5R &ndash; Z")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
@@ Line 117: / Line 117: @@
 *To subtract is still the case that also the additional number was drawn correctly.&nbsp; Therefore:
-:$$\rm \text{ Pr("5R - Z")}\approx 18.5\cdot 10^{-6}-0.429\cdot 10^{-6}\hspace{0.15cm} \underline{\approx 18.1\cdot 10^{-6}}.$$
+:$$\rm \text{ Pr("5R &ndash; Z")}\approx 18.5\cdot 10^{-6}-0.429\cdot 10^{-6}\hspace{0.15cm} \underline{\approx 18.1\cdot 10^{-6}}.$$

@@ Line 123: / Line 123: @@
 :$$\rm Pr(\it k\hspace{0.15cm} \rm correct\hspace{0.1cm}numbers)=\left({\it m \atop {\it k}}\right)\cdot\left({\it n-m \atop {m-\it k}}\right)\cdot\left({\it n \atop {m}}\right)^{-1}.$$
-*The last term in this equation gives the probability for a very specific "$m$ out of $n$" combination.
+*The last term in this equation gives the probability for a very specific&nbsp; "$m$ out of $n$"&nbsp; combination.
 *The first term describes the number of permutations that numbers drawn from&nbsp; $m$&nbsp; are exactly&nbsp; $k$&nbsp; correct. For&nbsp; $m = 6$&nbsp; and&nbsp; $k= 5$&nbsp; this gives the factor&nbsp; $6$.
@@ Line 130: / Line 130: @@
 *As a special case we get for the "6 out of 49" with&nbsp; $k= 4$:
-:$$\rm Pr(4\hspace{0.15cm}correct\hspace{0.1cm}numbers)=\left({6 \atop {4}}\right)\cdot\left({43 \atop {2}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=0.969\cdot10^{-3}}.$$
+:$$\rm \text{ Pr("4R")}=\left({6 \atop {4}}\right)\cdot\left({43 \atop {2}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=0.969\cdot10^{-3}}.$$
-'''(6)'''&nbsp; According to the general formula calculated in&nbsp; '''(5)'''&nbsp;, it further holds:
+'''(6)'''&nbsp; According to the general formula calculated in&nbsp; '''(5)''',&nbsp; it further holds:
-:$$\rm Pr(3\hspace{0.15cm}correct\hspace{0.1cm}numbers)=\left({6 \atop {3}}\right)\cdot\left({43 \atop {3}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=18.1\cdot10^{-3}}.$$
+:$$\rm \text{ Pr("3R")}=\left({6 \atop {3}}\right)\cdot\left({43 \atop {3}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=18.1\cdot10^{-3}}.$$
-'''(7)'''&nbsp; <u>Both statements</u> are correct:
+'''(7)'''&nbsp; <u>Both statements</u>&nbsp; are correct:
-*The probabilities are equal, of course.&nbsp; Because many people know this and want to demonstrate their knowledge&nbsp; (at least to themselves)&nbsp; many more people play this combination than a rather unspectacular one.
+*The probabilities are equal,&nbsp; of course.&nbsp; Because many people know this and want to demonstrate their knowledge&nbsp; (at least to themselves)&nbsp; many more people play this combination than a rather unspectacular one.
-*However, since the amount to be paid out in a winning class is determined in advance and this win then has to be divided by a larger number of winners, there is of course less left for the individual.
+*However,&nbsp; since the amount to be paid out in a winning class is determined in advance and this win then has to be divided by a larger number of winners,&nbsp; there is of course less left for the individual.
-*On the other hand, since many lottery players bet on their date of birth, the combination chosen here is not so favorable either.&nbsp; The numbers "3", "8" and "9" can indicate both the day and the month and, moreover, almost all players' year of birth starts with "19".
+*On the other hand,&nbsp; since many lottery players bet on their date of birth,&nbsp; the combination chosen here is not so favorable either.&nbsp; The numbers&nbsp; "3",&nbsp; "8" and&nbsp; "9"&nbsp; can indicate both the day and the month and,&nbsp; moreover,&nbsp; almost all players' year of birth starts with&nbsp; "19".
 {{ML-Fuß}}

@@ Line 17: / Line 17: @@
 :* 6 correct numbers without&nbsp; "super number" &nbsp; &rArr; &nbsp; $\text{"6R"}$,
 :* 5 correct numbers with&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R + Z"}$,
-:* 5 correct numbers without&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R"}$,
+:* 5 correct numbers without&nbsp; "additional number" &nbsp; &rArr; &nbsp; $\text{"5R - Z"}$,
 :* 4 correct numbers &nbsp; &rArr; &nbsp; $\text{"4R"}$,
 :* 3 correct numbers &nbsp; &rArr; &nbsp; $\text{"3R"}$.
@@ Line 55: / Line 55: @@
 {What is the probability for&nbsp; $\text{"5 correct numbers without additional number"}$&nbsp;?
 |type="{}"}
-$\text{Pr("5R")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
+$\text{Pr("5R - Z")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Assuming that the player has ticked the numbers „1“ to „6“ , the following applies:
+'''(1)'''&nbsp; Assuming that the player has ticked the numbers&nbsp; "1"&nbsp; to&nbsp; "6",&nbsp; the following applies:
-:$${\rm \Pr}(6\hspace{0.1cm}{\rm correct\hspace{0.1cm}numbers})={\rm \Pr}(123456\ \cup 123465\cup \ \text{...} \ \ \cup 654321).$$
+:$$\text{Pr("6R")}={\rm \Pr}(123456\ \cup 123465\cup \ \text{...} \ \ \cup 654321).$$
 *This takes into account that the order does not matter in the drawing.
-*In total, there are exactly&nbsp; $6! = 1 &middot; 2 &middot; 3 &middot; 4 &middot; 5 &middot; 6$&nbsp; gequally probable permutations for the set of numbers:
+*In total,&nbsp; there are exactly&nbsp; $6! = 1 &middot; 2 &middot; 3 &middot; 4 &middot; 5 &middot; 6$&nbsp; equally probable permutations for the set of numbers:
-:$$\rm \Pr(6\hspace{0.1cm}correct\hspace{0.1cm}numbers)=6!\cdot Pr(123456).$$
+:$$\text{Pr("6R")}=6!\cdot \rm Pr(123456).$$
-*The probability of the number „1“ being drawn as the first ball is&nbsp; $1/49$. The probability of „2“ being drawn as the second ball is correspondingly&nbsp; $1/48$&nbsp; (since there is now one less ball in the drum).&nbsp; Thus one receives as final result:
+*The probability of the number&nbsp; "1"&nbsp; being drawn as the first ball is&nbsp; $1/49$.&nbsp; The probability of&nbsp; "2"&nbsp; being drawn as the second ball is correspondingly&nbsp; $1/48$&nbsp; (since there is now one less ball in the drum).&nbsp; Thus one receives as final result:
 :$$\rm Pr(123456)=\frac{1}{49}\cdot\frac{1}{48}\cdot    \frac{1}{47} \cdot    \frac{1}{46}\cdot   \frac{1}{45}  \cdot    \frac{1}{44},$$
-:$$\rm Pr(6\hspace{0.1cm}correct\hspace{0.1cm}numbers)=\frac{1\cdot 2\cdot 3 \cdot 4 \cdot 5 \cdot 6}{49\cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44}.$$
+:$$\text{Pr("6R")}=\frac{1\cdot 2\cdot 3 \cdot 4 \cdot 5 \cdot 6}{49\cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44}.$$
-*This is exactly the inverse of „49 over 6“. It follows that:
+*This is exactly the inverse of&nbsp; "49 over 6".&nbsp; It follows that:
-:$$\rm Pr(6\hspace{0.15cm}correct\hspace{0.1cm}numbers)=\left({49 \atop {6}}\right)^{-1}=13\hspace{0.08cm}983\hspace{0.08cm}816^{-1}\hspace{0.15cm} \underline{\approx71.5\cdot 10^{-9}}.$$
+:$$\text{Pr("6R")}=\left({49 \atop {6}}\right)^{-1}=13\hspace{0.08cm}983\hspace{0.08cm}816^{-1}\hspace{0.15cm} \underline{\approx71.5\cdot 10^{-9}}.$$
@@ Line 96: / Line 96: @@
 '''(2)'''&nbsp; The probability that the final digit of a lottery ticket matches the super number is&nbsp; $10\%$.
 *But since the drawing of the super number is independent of the normal drawing, we now get for the probability we are looking for
-:$$\text{ Pr(6R + S)}	\hspace{0.15cm}\underline{\approx \ 7.15\cdot 10^{-9}}.$$
+:$$\text{ Pr("6R + S")}	\hspace{0.15cm}\underline{\approx \ 7.15\cdot 10^{-9}}.$$
 '''(3)'''&nbsp; In what follows,&nbsp; $Z$&nbsp; stands for "additional number". Then holds:
-:$$\rm Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers\hspace{0.1cm}plus\hspace{0.1cm}additional\hspace{0.1cm}number)={\rm Pr(12345} Z \ \cup \ {\rm 1234} Z\rm 6\cup \ \text{...} \ \cup \  Z {\rm 23456)}.$$
+:$$\text{ Pr("5R + Z")}={\rm Pr(12345} Z \ \cup \ {\rm 1234} Z\rm 6\cup \ \text{...} \ \cup \  Z {\rm 23456)}.$$
 *There are six permutations here. The probability for $\rm 12345Z$ is the same as for $123456$. It follows with the result of subtask&nbsp; '''(1)''':
-:$$\rm Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers\hspace{0.1cm}plus\hspace{0.1cm}additional\hspace{0.1cm}number)=6\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=0.429\cdot10^{-6}}.$$
+:$$\text{ Pr("5R + Z")}=6\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=0.429\cdot10^{-6}}.$$
-'''(4)'''&nbsp; If we denote by&nbsp; $X$&nbsp; a drawn number that is not one of those checked, we can write for the probability we are looking for:
+'''(4)'''&nbsp; If we denote by&nbsp; $X$&nbsp; a drawn number that is not one of those checked,&nbsp; we can write for the probability we are looking for:
-:$${\rm \Pr(5\hspace{0.1cm}correct\hspace{0.1cm}numbers)=Pr(12345} X \ \cup \ {\rm 1234} X6\ \cup \ \text{...} \ \cup \ X23456).$$
+:$${\rm \text{ Pr("5R")}=Pr(12345} X \ \cup \ {\rm 1234} X6\ \cup \ \text{...} \ \cup \ X23456).$$
 *There are six different possibilities for the location of&nbsp; $X$&nbsp;, all of which are equally probable. It follows that:
-:$$\rm Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers)=6\cdot Pr(12345\it X\rm ).$$
+:$$\rm \text{ Pr("5R")}=6\cdot Pr(12345\it X\rm ).$$
-*Here$X$&nbsp;can stand for any of the numbers&nbsp; $7$&nbsp; to&nbsp; $49$&nbsp; , which again are all equally probable.
+*Here&nbsp; $X$&nbsp;can stand for any of the numbers&nbsp; $7$&nbsp; to&nbsp; $49$,&nbsp; which again are all equally probable.
-*Thus, with&nbsp; $\rm Pr(123457) = Pr(123456)$, the result of subtask&nbsp; '''(1)''', we get:
+*Thus,&nbsp; with&nbsp; $\rm Pr(123457) = Pr(123456)$,&nbsp; the result of subtask&nbsp; '''(1)''',&nbsp; we get:
-:$$\rm Pr(5\hspace{0.1cm}correct\hspace{0.1cm}numbers)=6\cdot 43\cdot\left({49 \atop {6}}\right)^{-1}=1.85\cdot10^{-5}.$$
+:$$\rm \text{ Pr("5R")}=6\cdot 43\cdot\left({49 \atop {6}}\right)^{-1}=1.85\cdot10^{-5}.$$
-*To subtract is still the case that also the additional number was drawn correctly. Therefore:
+*To subtract is still the case that also the additional number was drawn correctly.&nbsp; Therefore:
-:$$\rm \Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers\hspace{0.1cm}without\hspace{0.1cm}additional\hspace{0.1cm}number)\approx 18.5\cdot 10^{-6}-0.429\cdot 10^{-6}\hspace{0.15cm} \underline{\approx 18.1\cdot 10^{-6}}.$$
+:$$\rm \text{ Pr("5R - Z")}\approx 18.5\cdot 10^{-6}-0.429\cdot 10^{-6}\hspace{0.15cm} \underline{\approx 18.1\cdot 10^{-6}}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID103__Sto_A_2_4.jpg|right|frame|Lottery ticket for&nbsp; $\text{6/49 or 6 out of 49}$]]
+[[File:P_ID103__Sto_A_2_4.jpg|right|frame|Lottery ticket for&nbsp; $\text{"6 out of 49"}$]]
-For the number lotto the following conditions should apply:
+For the&nbsp; "number lotto"&nbsp; the following conditions should apply:
-Six winning numbers are drawn from the&nbsp; $49$&nbsp; numbers (balls) in a drum &nbsp; ("6 out of 49"), then as the seventh ball the so-called additional number&nbsp; $(Z)$.
+*Six winning numbers are drawn from the&nbsp; $49$&nbsp; numbers&nbsp; (balls)&nbsp; in a drum &nbsp; ("6 out of 49"),
-In this task, the following winning classes are considered:
+Independently of this,&nbsp; a&nbsp; "super number"&nbsp; $S \in  \{0, \ 1,\hspace{0.1cm} \text{...} \hspace{0.1cm}, \ 9\}$&nbsp; is selected at random.&nbsp; If this number matches the final digit of the lottery ticket,&nbsp; the main prize is increased decisively.
@@ Line 31: / Line 29: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Binomial_Distribution|Binomial Distribution]].
+*This exercise was designed around 2002.&nbsp;  It is possible that the rules of the&nbsp; (German)&nbsp; lottery have been changed in the meantime.
@@ Line 41: / Line 38: @@
 <quiz display=simple>
-{What is the probability for&nbsp; $\text{6 correct numbers}$&nbsp;?
+{What is the probability for&nbsp; $\text{"6 correct numbers"}$&nbsp;?
 |type="{}"}
-$ \rm Pr(6\hspace{0.1cm}correct\hspace{0.1cm}numbers)	\ = \ $ { 71.5 3% } $\ \cdot 10^{-9}$
+$ \text{Pr("6R")}	\ = \ $ { 71.5 3% } $\ \cdot 10^{-9}$
-{What is the probability for&nbsp; $\text{6 correct numbers plus super number}$&nbsp;?
+{What is the probability for&nbsp; $\text{"6 correct numbers plus super number"}$&nbsp;?
 |type="{}"}
-$ \rm Pr(6R + S)	\ =  \ $ { 7.15 3% } $\ \cdot 10^{-9}$
+$ \text{Pr("6R + S")}	\ =  \ $ { 7.15 3% } $\ \cdot 10^{-9}$
-{What is the probability for&nbsp; $\text{5 correct numbers plus additional number}$&nbsp;?
+{What is the probability for&nbsp; $\text{"5 correct numbers plus additional number"}$&nbsp;?
 |type="{}"}
-$ \rm Pr(5R + Z)	\   =  \ $ { 0.429 3% } $\ \cdot 10^{-6}$
+$ \text{Pr("5R + Z")}	\   =  \ $ { 0.429 3% } $\ \cdot 10^{-6}$
-{What is the probability for&nbsp; $\text{5 correct numbers without additional number}$&nbsp;?
+{What is the probability for&nbsp; $\text{"5 correct numbers without additional number"}$&nbsp;?
 |type="{}"}
-$\rm Pr(5\hspace{0.1cm}correct\hspace{0.1cm}numbers)\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
+$\text{Pr("5R")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-6}$
-{What is the probability for&nbsp; $\text{4 correct numbers}$&nbsp;?&nbsp; What is the general result for&nbsp; $k$&nbsp; correct numbers in the&nbsp; $m&ndash;\text{out of}&ndash;n$&ndash;lotto&nbsp;?
+{What is the probability for&nbsp; $\text{"4 correct numbers"}$&nbsp;?&nbsp; What is the general result for&nbsp; $k$&nbsp; correct numbers in the&nbsp; "$m\hspace{0.15cm} \text{out of}\hspace{0.15cm} n$"&ndash;lotto&nbsp;?
 |type="{}"}
-$\rm Pr(4\hspace{0.1cm}correct\hspace{0.1cm}numbers)\ =  \ $ { 0.969 3% } $\ \cdot 10^{-3}$
+$\text{Pr("4R")}\ =  \ $ { 0.969 3% } $\ \cdot 10^{-3}$
-{What is the probability for&nbsp; $\text{3 correct numbers}$&nbsp;?
+{What is the probability for&nbsp; $\text{"3 correct numbers"}$&nbsp;?
 |type="{}"}
-$\rm Pr(3\hspace{0.1cm}correct\hspace{0.1cm}numbers)\ =  \ $ { 18.1 3% } $\ \cdot 10^{-3}$
+$\text{Pr("3R")}\ =  \ $ { 18.1 3% } $\ \cdot 10^{-3}$

@@ Line 116: / Line 116: @@
 :$$\rm Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers)=6\cdot Pr(12345\it X\rm ).$$
-*$X$&nbsp;  kann hierbei f&uuml;r jede der Zahlen&nbsp; $7$&nbsp; bis&nbsp; $49$&nbsp; stehen, die wiederum alle gleichwahrscheinlich sind.
+*Here$X$&nbsp;can stand for any of the numbers&nbsp; $7$&nbsp; to&nbsp; $49$&nbsp; , which again are all equally probable.
-*Mit&nbsp; $\rm Pr(123457) = Pr(123456)$, dem Ergebnis der Teilaufgabe&nbsp; '''(1)''', erh&auml;lt man somit:
+*Thus, with&nbsp; $\rm Pr(123457) = Pr(123456)$, the result of subtask&nbsp; '''(1)''', we get:
 :$$\rm Pr(5\hspace{0.1cm}correct\hspace{0.1cm}numbers)=6\cdot 43\cdot\left({49 \atop {6}}\right)^{-1}=1.85\cdot10^{-5}.$$
-*Zu subtrahieren ist noch der Fall, dass auch die Zusatzzahl richtig gezogen wurde. Deshalb gilt:
+*To subtract is still the case that also the additional number was drawn correctly. Therefore:
 :$$\rm \Pr(5\hspace{0.15cm}correct\hspace{0.1cm}numbers\hspace{0.1cm}without\hspace{0.1cm}additional\hspace{0.1cm}number)\approx 18.5\cdot 10^{-6}-0.429\cdot 10^{-6}\hspace{0.15cm} \underline{\approx 18.1\cdot 10^{-6}}.$$
-'''(5)'''&nbsp; Nach einigen &Uuml;berlegungen kommt man zum Ergebnis:
+'''(5)'''&nbsp; After some considerations, we arrive at the result:
 :$$\rm Pr(\it k\hspace{0.15cm} \rm correct\hspace{0.1cm}numbers)=\left({\it m \atop {\it k}}\right)\cdot\left({\it n-m \atop {m-\it k}}\right)\cdot\left({\it n \atop {m}}\right)^{-1}.$$
-*Der letzte Term in dieser Gleichung gibt die Wahrscheinlichkeit f&uuml;r eine ganz bestimmte "$m$ aus $n$"&ndash;Kombination an.
+*The last term in this equation gives the probability for a very specific "$m$ out of $n$" combination.
-*Der erste Term beschreibt die Anzahl der Permutationen, dass von&nbsp; $m$&nbsp; gezogenen Zahlen genau&nbsp; $k$&nbsp; richtig sind. F&uuml;r&nbsp; $m = 6$&nbsp; und&nbsp; $k= 5$&nbsp; ergibt das den Faktor&nbsp; $6$.
+*The first term describes the number of permutations that numbers drawn from&nbsp; $m$&nbsp; are exactly&nbsp; $k$&nbsp; correct. For&nbsp; $m = 6$&nbsp; and&nbsp; $k= 5$&nbsp; this gives the factor&nbsp; $6$.
-*Der mittlere Faktor gibt die Anzahl der M&ouml;glichkeiten an, dass alle restlichen gezogenen Zahlen&nbsp; $($also&nbsp; $m-k)$&nbsp; eine der ung&uuml;nstigen Zahlen&nbsp; $($hierf&uuml;r gibt es&nbsp; $n-m)$&nbsp; ist.&nbsp; F&uuml;r&nbsp; $m = 6$&nbsp; und&nbsp; $k= 5$&nbsp; ergibt dieser Term entsprechend  Punkt&nbsp; '''(4)'''&nbsp; den Wert&nbsp; $43$.
+*The middle factor gives the number of possibilities that all remaining drawn numbers&nbsp; $($thus&nbsp; $m-k)$&nbsp; is one of the unfavorable numbers&nbsp; $($for this there are&nbsp; $n-m)$&nbsp;.&nbsp; For&nbsp; $m = 6$&nbsp; and&nbsp; $k= 5$&nbsp; this term gives the value&nbsp; $43$ according to point&nbsp; '''(4)'''&nbsp;.
-*Als Sonderfall erhalten wir beim „6 aus 49“ mit&nbsp; $k= 4$:
+*As a special case we get for the "6 out of 49" with&nbsp; $k= 4$:
 :$$\rm Pr(4\hspace{0.15cm}correct\hspace{0.1cm}numbers)=\left({6 \atop {4}}\right)\cdot\left({43 \atop {2}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=0.969\cdot10^{-3}}.$$
-'''(6)'''&nbsp; Entsprechend der unter&nbsp; '''(5)'''&nbsp; berechneten allgemeinen Formel gilt weiter:
+'''(6)'''&nbsp; According to the general formula calculated in&nbsp; '''(5)'''&nbsp;, it further holds:
 :$$\rm Pr(3\hspace{0.15cm}correct\hspace{0.1cm}numbers)=\left({6 \atop {3}}\right)\cdot\left({43 \atop {3}}\right)\cdot\left({49 \atop {6}}\right)^{-1}\hspace{0.15cm} \underline{=18.1\cdot10^{-3}}.$$
-'''(7)'''&nbsp; <u>Beide Aussagen</u> sind richtig:
+'''(7)'''&nbsp; <u>Both statements</u> are correct:
-*Die Wahrscheinlichkeiten sind nat&uuml;rlich gleich.&nbsp; Weil das viele wissen und diese ihr Wissen auch&nbsp; (zumindest sich selbst)&nbsp; demonstrieren wollen, spielen sehr viel mehr Menschen diese Kombination als eine eher unspektakul&auml;re.
+*The probabilities are equal, of course.&nbsp; Because many people know this and want to demonstrate their knowledge&nbsp; (at least to themselves)&nbsp; many more people play this combination than a rather unspectacular one.
-*Da aber der in einer Gewinnklasse auszuzahlende Betrag vorher festgelegt wird und dieser Gewinn dann durch eine gr&ouml;&szlig;ere Anzahl von Gewinnern zu teilen ist, bleibt f&uuml;r den Einzelnen nat&uuml;rlich weniger.
+*However, since the amount to be paid out in a winning class is determined in advance and this win then has to be divided by a larger number of winners, there is of course less left for the individual.
-*Da andererseits viele Lottospieler auf ihr Geburtsdatum setzen, ist die hier gew&auml;hlte Komination auch nicht so g&uuml;nstig.&nbsp; Die Zahlen „3“, „8“ und „9“ k&ouml;nnen sowohl den Tag als auch das Monat angeben und zudem beginnt fast bei allen Spielern das Geburtsjahr mit "19".
+*On the other hand, since many lottery players bet on their date of birth, the combination chosen here is not so favorable either.&nbsp; The numbers "3", "8" and "9" can indicate both the day and the month and, moreover, almost all players' year of birth starts with "19".
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Sginaltheorie/Binomialverteilung
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Binomial_Distribution
 }}
-[[File:P_ID103__Sto_A_2_4.jpg|right|frame|Lottoschein für&nbsp; $\text{6 aus 49}$]]
+[[File:P_ID103__Sto_A_2_4.jpg|right|frame|Lottery ticket for&nbsp; $\text{6/49 or 6 out of 49}$]]
-Für das Zahlenlotto sollen folgende Voraussetzungen gelten:
+For the number lotto the following conditions should apply:
-Es werden aus den&nbsp; $49$&nbsp; Zahlen (Kugeln) in einer Trommel sechs Gewinnzahlen gezogen&nbsp; („6 aus 49 “), danach als siebente Kugel die sogenannte Zusatzzahl&nbsp; $(Z)$.
+Six winning numbers are drawn from the&nbsp; $49$&nbsp; numbers (balls) in a drum &nbsp; („6 out of 49 “), then as the seventh ball the so-called additional number&nbsp; $(Z)$.
 Unabh&auml;ngig davon wird noch eine Superzahl&nbsp; $S \in  \{0, \ 1,\hspace{0.1cm} \text{...} \hspace{0.1cm}, \ 9\}$&nbsp; per Zufall ausgew&auml;hlt.&nbsp; Stimmt diese mit der Endziffer des Lottoscheins &uuml;berein, so wird der Hauptgewinn entscheidend vergr&ouml;&szlig;ert.

← Older revision		Revision as of 16:43, 30 November 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^2.3 ~~Binomialverteilung~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^2.3 Binomial Distribution^]]

← Older revision	Revision as of 16:42, 30 November 2021
(No difference)

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-'''(3)'''&nbsp; Im Folgenden steht&nbsp; $Z$&nbsp; f&uuml;r "Zusatzzahl&rdquo;. Dann gilt:
+'''(3)'''&nbsp; Im Folgenden steht&nbsp; $Z$&nbsp; f&uuml;r "Zusatzzahl". Dann gilt:
 :$$\rm Pr(5\hspace{0.15cm}Richtige\hspace{0.1cm}mit\hspace{0.1cm}Zusatzzahl)={\rm Pr(12345} Z \ \cup \ {\rm 1234} Z\rm 6\cup \ \text{...} \ \cup \  Z {\rm 23456)}.$$
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 :$$\rm Pr(\it k\hspace{0.15cm} \rm Richtige)=\left({\it m \atop {\it k}}\right)\cdot\left({\it n-m \atop {m-\it k}}\right)\cdot\left({\it n \atop {m}}\right)^{-1}.$$
-*Der letzte Term in dieser Gleichung gibt die Wahrscheinlichkeit f&uuml;r eine ganz bestimmte "$m$ aus $n$&rdquo;&ndash;Kombination an.
+*Der letzte Term in dieser Gleichung gibt die Wahrscheinlichkeit f&uuml;r eine ganz bestimmte "$m$ aus $n$"&ndash;Kombination an.
 *Der erste Term beschreibt die Anzahl der Permutationen, dass von&nbsp; $m$&nbsp; gezogenen Zahlen genau&nbsp; $k$&nbsp; richtig sind. F&uuml;r&nbsp; $m = 6$&nbsp; und&nbsp; $k= 5$&nbsp; ergibt das den Faktor&nbsp; $6$.
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 *Die Wahrscheinlichkeiten sind nat&uuml;rlich gleich.&nbsp; Weil das viele wissen und diese ihr Wissen auch&nbsp; (zumindest sich selbst)&nbsp; demonstrieren wollen, spielen sehr viel mehr Menschen diese Kombination als eine eher unspektakul&auml;re.
 *Da aber der in einer Gewinnklasse auszuzahlende Betrag vorher festgelegt wird und dieser Gewinn dann durch eine gr&ouml;&szlig;ere Anzahl von Gewinnern zu teilen ist, bleibt f&uuml;r den Einzelnen nat&uuml;rlich weniger.
-*Da andererseits viele Lottospieler auf ihr Geburtsdatum setzen, ist die hier gew&auml;hlte Komination auch nicht so g&uuml;nstig.&nbsp; Die Zahlen „3“, „8“ und „9“ k&ouml;nnen sowohl den Tag als auch das Monat angeben und zudem beginnt fast bei allen Spielern das Geburtsjahr mit "19&rdquo;.
+*Da andererseits viele Lottospieler auf ihr Geburtsdatum setzen, ist die hier gew&auml;hlte Komination auch nicht so g&uuml;nstig.&nbsp; Die Zahlen „3“, „8“ und „9“ k&ouml;nnen sowohl den Tag als auch das Monat angeben und zudem beginnt fast bei allen Spielern das Geburtsjahr mit "19".
 {{ML-Fuß}}