Aufgaben:Exercise 1.5Z: Probabilities of Default - Revision history

Guenter at 12:26, 1 December 2021

2021-12-01T12:26:08Z

Guenter at 12:10, 1 December 2021

2021-12-01T12:10:09Z

Guenter at 16:29, 30 November 2021

2021-11-30T16:29:15Z

Guenter at 15:56, 30 November 2021

2021-11-30T15:56:03Z

Noah at 21:03, 28 November 2021

2021-11-28T21:03:14Z

Guenter at 16:08, 12 October 2021

2021-10-12T16:08:35Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.5Z: Ausfallwahrscheinlichkeiten to Aufgaben:Exercise 1.5Z: Probabilities of Default

2021-10-12T16:04:28Z

Guenter moved page Aufgabe 1.5Z: Ausfallwahrscheinlichkeiten to Exercise 1.5Z: Probabilities of Default

Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:44:25Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:51:34Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

Guenter at 16:41, 9 November 2019

2019-11-09T16:41:00Z

@@ Line 54: / Line 54: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Since the two subdevices default independently, set-theoretically holds:
+'''(1)'''&nbsp; Since the two sub-devices default independently,&nbsp; set-theoretically holds:
-:$$\rm Pr(\it G \rm \hspace{0.1cm}f\ddot{a}llt\hspace{0.1cm}aus) = Pr(\it T_{\rm 1}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus) \cdot Pr(\it T_{\rm 2}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus). $$
+:$${\rm Pr(}G \text{ drops out)} = {\rm Pr(}T_1 \text{ drops out)} \cdot {\rm Pr(}T_2 \text{ drops out)}. $$
-*Moreover, since subdevices&nbsp; $T_1$&nbsp;  and&nbsp; $T_2$&nbsp; are identical in construction, they default with the same probability&nbsp; $p_{\rm T}$&nbsp;. It follows that:
+*Moreover,&nbsp; since the sub-devices&nbsp; $T_1$&nbsp;  and&nbsp; $T_2$&nbsp; are identical in construction,&nbsp; they default with the same probability&nbsp; $p_{\rm T}.$&nbsp; It follows that:
-:$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \rm bzw. \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}}  \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$
+:$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}}  \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$
 '''(2)'''&nbsp; This result is easier to determine using the complementary event:
 :$$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm functions) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm functions \cap \it B_{\rm 2} \hspace{0.1cm} \rm functions \cap \it B_{\rm 3}\hspace{0.1cm} \rm functions).$$
-:$$\Rightarrow 1- p_{\rm T}= (1-p_{\rm A})^{3} \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}
+:$$\Rightarrow 1- p_{\rm T}= (1-p_{\rm B})^{3} \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}
 -p_{\rm T}=(0.9)^3= 0.729 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} p_{\rm T}\hspace{0.15cm}\underline {= 0.271 = 27.1\%}.$$
+'''(3)'''&nbsp; With&nbsp; $p_{\rm B} = 0.01$,&nbsp; we obtain&nbsp; $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$
-'''(3)'''&nbsp; With&nbsp; $p_{\rm A} = 0.01$&nbsp;, we obtain&nbsp; $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$
+*In general,&nbsp; the approximation is:&nbsp; $p_{\rm T} \approx n \cdot p_{\rm B}\; (= 3\%)$.
-'''(4)'''&nbsp; With the approximation of the last subtaks&nbsp; $\underline{n = 5}$ follows directly.
+'''(4)'''&nbsp; With the approximation of the last subtask &nbsp; &rArr; &nbsp; $\underline{n = 5}$&nbsp; follows directly.
-*For larger&nbsp; $p_{\rm A}$&nbsp;, one would have to proceed as follows:
+*For larger&nbsp; $p_{\rm B}$,&nbsp; one would have to proceed as follows:
 :$$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$
 {{ML-Fuß}}

@@ Line 37: / Line 37: @@
 $p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$
-{Let the default probability of all components be&nbsp; $\underline{p_{\rm B} = 0.1}$.&nbsp; Let each sub-device consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
+{Let the default probability of all components be&nbsp; $\underline{p_{\rm B} = 0.1}$.&nbsp; Let each sub-device consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a sub-device defaults.
 |type="{}"}
 $p_{\rm T} \ = \ $ { 27.1 3% } $ \ \%$
@@ Line 45: / Line 45: @@
 $p_{\rm T} \ = \ $ { 2.97 3% } $ \ \%$
-{Now apply&nbsp; $p_{\rm B} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the sub-device can contain?&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold.
+{Now apply&nbsp; $p_{\rm B} = 0.4\%$&nbsp; for the default probability of all components. &nbsp; What is the maximum number of components the sub-device can contain?&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold.
 |type="{}"}
 $n \ = \ $  { 5 3% }

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}
-[[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device with two identical parts&nbsp; $T_1$&nbsp; and&nbsp; $T_2$&nbsp;]]
+[[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device with two identical parts&nbsp; $T_1$&nbsp; and&nbsp; $T_2$]]
-A device part is composed of the components&nbsp; (German:&nbsp; "Bauteile" &nbsp; &rArr; &nbsp; "B") &nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; where the respective functionality can be assumed to be independent of all other components.
+A sub-device is composed of the components &nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; where the respective functionality can be assumed to be independent of all other components.
-*Assume that all components default with equal probability&nbsp; $p_{\rm A}$.
+*Assume that all components default with equal probability&nbsp; $p_{\rm B}$.
-*Part&nbsp; (German:&nbsp; "Teilgerät" &nbsp; &rArr; &nbsp; "T")&nbsp; $T_1$&nbsp; functions only if all&nbsp; $n$&nbsp; components are functional.
+*Sub-device&nbsp; $T_1$&nbsp; functions only if all&nbsp; $n$&nbsp; components are functional.
-To increase reliability,&nbsp; important assemblies are often duplicated.&nbsp; The device&nbsp; (German:&nbsp; "Gesamtgerät" &nbsp; &rArr; &nbsp; "G")&nbsp; $G$&nbsp; can be described in terms of set theory as follows:
+To increase reliability,&nbsp; important assemblies are often duplicated.&nbsp; The overall device&nbsp;  $G$&nbsp; can be described in terms of set theory as follows:
 :$$ G = T_1 \cup T_2.$$
-This means: &nbsp; Device&nbsp; $G$&nbsp; is already operational if at least one of the two identical subassemblies&nbsp; $(T_1$&nbsp; or&nbsp; $T_2)$&nbsp; is functional.
+This means: &nbsp; The overall device&nbsp; $G$&nbsp; is already operational if at least one of the two identical sub-devices&nbsp; $(T_1$&nbsp; or&nbsp; $T_2)$&nbsp; is functional.
@@ Line 24: / Line 24: @@
 *The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
 ::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".
@@ Line 29: / Line 33: @@
 <quiz display=simple>
-{The default probability&nbsp; $p_{\rm G}$&nbsp; of the total device must not be greater than&nbsp; $0.04\%$. <br>How large may then the default probabilities&nbsp; $p_{\rm T}$&nbsp; of the two identical device parts existing in parallel be at most?
+{The default probability&nbsp; $p_{\rm G}$&nbsp; of the overall device must not be greater than&nbsp; $0.04\%$. <br>How large may then the default probabilities&nbsp; $p_{\rm T}$&nbsp; of the two identical sub-devices existing in parallel be at most?
 |type="{}"}
 $p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$
-{Let the default probability of all components be&nbsp; $\underline{p_{\rm B} = 0.1}$.&nbsp; Let each subdevice consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
+{Let the default probability of all components be&nbsp; $\underline{p_{\rm B} = 0.1}$.&nbsp; Let each sub-device consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
 |type="{}"}
 $p_{\rm T} \ = \ $ { 27.1 3% } $ \ \%$
@@ Line 41: / Line 45: @@
 $p_{\rm T} \ = \ $ { 2.97 3% } $ \ \%$
-{Now apply&nbsp; $p_{\rm B} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the subdevice can contain?&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold.
+{Now apply&nbsp; $p_{\rm B} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the sub-device can contain?&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold.
 |type="{}"}
 $n \ = \ $  { 5 3% }

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}
-[[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device]]
+[[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device with two identical parts&nbsp; $T_1$&nbsp; and&nbsp; $T_2$&nbsp;]]
-A device part is composed of the components&nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; where the respective functionality can be assumed to be independent of all other components.
+A device part is composed of the components&nbsp; (German:&nbsp; "Bauteile" &nbsp; &rArr; &nbsp; "B") &nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; where the respective functionality can be assumed to be independent of all other components.
-*Assume that all components default with equal probability&nbsp; $p_{\rm A}$&nbsp;.
+*Assume that all components default with equal probability&nbsp; $p_{\rm A}$.
-*Part&nbsp; $T_1$&nbsp; functions only if all&nbsp; $n$&nbsp; components are functional.
+*Part&nbsp; (German:&nbsp; "Teilgerät" &nbsp; &rArr; &nbsp; "T")&nbsp; $T_1$&nbsp; functions only if all&nbsp; $n$&nbsp; components are functional.
-To increase reliability, important assemblies are often duplicated. The device&nbsp; $G$&nbsp; can thus be described in terms of set theory as follows:
+To increase reliability,&nbsp; important assemblies are often duplicated.&nbsp; The device&nbsp; (German:&nbsp; "Gesamtgerät" &nbsp; &rArr; &nbsp; "G")&nbsp; $G$&nbsp; can be described in terms of set theory as follows:
 :$$ G = T_1 \cup T_2.$$
 This means: &nbsp; Device&nbsp; $G$&nbsp; is already operational if at least one of the two identical subassemblies&nbsp; $(T_1$&nbsp; or&nbsp; $T_2)$&nbsp; is functional.
@@ Line 25: / Line 23: @@
 *The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
-:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".
+::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".
@@ Line 35: / Line 33: @@
 $p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$
-{Let the default probability of all components be&nbsp; $\underline{p_{\rm A} = 0.1}$.&nbsp; Let each subdevice consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
+{Let the default probability of all components be&nbsp; $\underline{p_{\rm B} = 0.1}$.&nbsp; Let each subdevice consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
 |type="{}"}
 $p_{\rm T} \ = \ $ { 27.1 3% } $ \ \%$
-{What value is obtained for&nbsp; $\underline{p_{\rm A} = 0.01}$?&nbsp; In what form can you approximate&nbsp; $p_{\rm T}$&nbsp; for small values of&nbsp; $p_{\rm A}$&nbsp;?
+{What value is obtained for&nbsp; $\underline{p_{\rm B} = 0.01}$?&nbsp; In what form can you approximate&nbsp; $p_{\rm T}$&nbsp; for small values of&nbsp; $p_{\rm B}$&nbsp;?
 |type="{}"}
 $p_{\rm T} \ = \ $ { 2.97 3% } $ \ \%$
-{Now apply&nbsp; $p_{\rm A} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the subdevice can contain&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold?
+{Now apply&nbsp; $p_{\rm B} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the subdevice can contain?&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold.
 |type="{}"}
 $n \ = \ $  { 5 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Statistische Abhängigkeit und Unabhängigkeit}}
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence}}
-[[File:P_ID87__Sto_Z_1_5.png|right|frame|Geräte&ndash;Funktionsschaltbild]]
+[[File:P_ID87__Sto_Z_1_5.png|right|frame|Functional circuit diagram of a device]]
-Ein Geräteteil ist aus den Bauteilen&nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; aufgebaut, wobei die jeweilige Funktionsfähigkeit unabhängig von allen anderen Bauteilen angenommen werden kann.
+A device part is composed of the components&nbsp; $B_1, \ B_2,\ \text{...} \ , B_n$&nbsp; where the respective functionality can be assumed to be independent of all other components.
-*Gehen Sie davon aus, dass alle Bauteile mit gleicher Wahrscheinlichkeit&nbsp; $p_{\rm A}$&nbsp; ausfallen.
+*Assume that all components default with equal probability&nbsp; $p_{\rm A}$&nbsp;.
-*Das Teil&nbsp; $T_1$&nbsp; funktioniert nur dann, wenn alle&nbsp; $n$&nbsp; Bauteile funktionsfähig sind.
+*Part&nbsp; $T_1$&nbsp; functions only if all&nbsp; $n$&nbsp; components are functional.
-Zur Erhöhung der Zuverlässigkeit werden wichtige Baugruppen häufig dupliziert. Das Gerät&nbsp; $G$&nbsp; kann somit mengentheoretisch wie folgt beschrieben werden:
+To increase reliability, important assemblies are often duplicated. The device&nbsp; $G$&nbsp; can thus be described in terms of set theory as follows:
 :$$ G = T_1 \cup T_2.$$
-Das heißt: &nbsp; Das Gerät&nbsp; $G$&nbsp; ist bereits dann einsatzbereit, wenn zumindest eines der beiden baugleichen Teilgeräte&nbsp; $(T_1$&nbsp; oder&nbsp; $T_2)$&nbsp; funktionsfähig ist.
+This means: &nbsp; Device&nbsp; $G$&nbsp; is already operational if at least one of the two identical subassemblies&nbsp; $(T_1$&nbsp; or&nbsp; $T_2)$&nbsp; is functional.
@@ Line 21: / Line 21: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abhängigkeit_und_Unabhängigkeit|Statistische Abhängigkeit und Unabhängigkeit]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence|Statistical dependence and independence]].
-*Eine Zusammenfassung der theoretischen Grundlagen mit Beispielen bringt das Lernvideo&nbsp;
+*The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
-::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]].
+:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical dependence and independence".
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Die Ausfallwahrscheinlichkeit&nbsp; $p_{\rm G}$&nbsp; des Gesamtgeräts darf nicht größer sein als&nbsp; $0.04\%$. <br>Wie groß dürfen dann die Ausfallwahrscheinlichkeiten&nbsp; $p_{\rm T}$&nbsp; der zwei parallel vorhandenen identischen Geräteteile höchstens sein?
+{The default probability&nbsp; $p_{\rm G}$&nbsp; of the total device must not be greater than&nbsp; $0.04\%$. <br>How large may then the default probabilities&nbsp; $p_{\rm T}$&nbsp; of the two identical device parts existing in parallel be at most?
 |type="{}"}
 $p_\text{T, max} \ = \ $ { 2 3% } $ \ \%$
-{Die Ausfallwahrscheinlichkeit aller Bauteile sei&nbsp; $\underline{p_{\rm A} = 0.1}$.&nbsp; Jedes Teilgerät bestehe aus&nbsp; $n = 3$&nbsp; Bauteilen. <br>Berechnen Sie die Wahrscheinlichkeit&nbsp; $p_{\rm T}$&nbsp; exakt, dass ein Teilgerät ausfällt.
+{Let the default probability of all components be&nbsp; $\underline{p_{\rm A} = 0.1}$.&nbsp; Let each subdevice consist of&nbsp; $n = 3$&nbsp; components. <br>Calculate the probability&nbsp; $p_{\rm T}$&nbsp; exactly that a subdevice defaults.
 |type="{}"}
 $p_{\rm T} \ = \ $ { 27.1 3% } $ \ \%$
-{Welcher Wert ergibt sich für&nbsp; $\underline{p_{\rm A} = 0.01}$?&nbsp; In welcher Form kann man&nbsp; $p_{\rm T}$&nbsp; für kleine Werte von&nbsp; $p_{\rm A}$&nbsp; annähern?
+{What value is obtained for&nbsp; $\underline{p_{\rm A} = 0.01}$?&nbsp; In what form can you approximate&nbsp; $p_{\rm T}$&nbsp; for small values of&nbsp; $p_{\rm A}$&nbsp;?
 |type="{}"}
 $p_{\rm T} \ = \ $ { 2.97 3% } $ \ \%$
-{Nun gelte für die Ausfallwahrscheinlichkeit aller Bauteile&nbsp; $p_{\rm A} = 0.4\%$.&nbsp; Wieviele Bauteile kann das Teilgerät höchstens enthalten, wenn&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; gelten soll?
+{Now apply&nbsp; $p_{\rm A} = 0.4\%$ for the default probability of all components. &nbsp; What is the maximum number of components the subdevice can contain&nbsp; $p_{\rm T} ≤ 2\%$&nbsp; is to hold?
 |type="{}"}
 $n \ = \ $  { 5 3% }
@@ Line 50: / Line 50: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Da die beiden Teilger&auml;te unabh&auml;ngig voneinander ausfallen, gilt mengentheoretisch:
+'''(1)'''&nbsp; Since the two subdevices default independently, set-theoretically holds:
 :$$\rm Pr(\it G \rm \hspace{0.1cm}f\ddot{a}llt\hspace{0.1cm}aus) = Pr(\it T_{\rm 1}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus) \cdot Pr(\it T_{\rm 2}\rm \hspace{0.1cm} f\ddot{a}llt \hspace{0.1cm}aus). $$
-*Da die Teilgeräte&nbsp; $T_1$&nbsp;  und&nbsp; $T_2$&nbsp; zudem baugleich sind, fallen sie mit der gleichen Wahrscheinlichkeit&nbsp; $p_{\rm T}$&nbsp; aus. Daraus folgt:
+*Moreover, since subdevices&nbsp; $T_1$&nbsp;  and&nbsp; $T_2$&nbsp; are identical in construction, they default with the same probability&nbsp; $p_{\rm T}$&nbsp;. It follows that:
 :$$p_{\rm G} = \it p_{\rm T}^{\rm 2} \hspace{0.5cm} \rm bzw. \hspace{0.5cm} \rm \it p_{\rm T,\hspace{0.1cm}max}= \sqrt{\it p_{\rm G}}  \le \rm\sqrt{0.0004} \hspace{0.15cm}\underline {= 2\%}.$$
-'''(2)'''&nbsp; Dieses Ergebnis ist einfacher &uuml;ber das Komplement&auml;rereignis zu bestimmen:
+'''(2)'''&nbsp; This result is easier to determine using the complementary event:
-:$$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm funktioniert) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm funktioniert \cap \it B_{\rm 2} \hspace{0.1cm} \rm funktioniert \cap \it B_{\rm 3}\hspace{0.1cm} \rm funktioniert).$$
+:$$\rm Pr(\it T_{\rm 1}\hspace{0.1cm}\rm functions) = \rm Pr(\it B_{\rm 1} \hspace{0.1cm}\rm functions \cap \it B_{\rm 2} \hspace{0.1cm} \rm functions \cap \it B_{\rm 3}\hspace{0.1cm} \rm functions).$$
 :$$\Rightarrow 1- p_{\rm T}= (1-p_{\rm A})^{3} \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}
 -p_{\rm T}=(0.9)^3= 0.729 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} p_{\rm T}\hspace{0.15cm}\underline {= 0.271 = 27.1\%}.$$
@@ Line 66: / Line 66: @@
-'''(3)'''&nbsp; Mit&nbsp; $p_{\rm A} = 0.01$&nbsp; erh&auml;lt man&nbsp; $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$
+'''(3)'''&nbsp; With&nbsp; $p_{\rm A} = 0.01$&nbsp;, we obtain&nbsp; $p_{\rm T}\hspace{0.15cm}\underline {= 2.97\%}.$
-*Allgemein gilt die N&auml;herung:&nbsp; $p_{\rm T} \approx n \cdot p_{\rm A}\; (= 3\%)$.
+*In general, the approximation is:&nbsp; $p_{\rm T} \approx n \cdot p_{\rm A}\; (= 3\%)$.
-'''(4)'''&nbsp; Mit der N&auml;herung der letzten Teilaufgabe folgt direkt&nbsp; $\underline{n = 5}$.
+'''(4)'''&nbsp; With the approximation of the last subtaks&nbsp; $\underline{n = 5}$ follows directly.
-*Bei gr&ouml;&szlig;erem&nbsp; $p_{\rm A}$&nbsp; m&uuml;sste man wie folgt vorgehen:
+*For larger&nbsp; $p_{\rm A}$&nbsp;, one would have to proceed as follows:
 :$$0.996^{\it n}\ge 0.98 \hspace{0.5cm} \rm\Rightarrow \hspace{0.5cm} \it n\le\rm\frac{log(0.98)}{log(0.996)} = 5.0406\hspace{0.15cm}\underline { \approx 5}.$$
 {{ML-Fuß}}

← Older revision		Revision as of 16:08, 12 October 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 ~~Statistische (Un-)Abhängigkeit~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistical Dependence/Independence^]]

← Older revision		Revision as of 12:44, 23 March 2021
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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^1.3 Statistische (Un-)Abhängigkeit^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^1.3 Statistische (Un-)Abhängigkeit^]]