Aufgaben:Exercise 3.1: Probabilities when Rolling Dice - Revision history

Noah at 08:08, 24 September 2021

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Guenter at 08:59, 17 August 2021

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Guenter at 13:40, 16 August 2021

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Noah at 21:12, 9 August 2021

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Guenter at 13:31, 16 July 2021

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Guenter at 13:28, 16 July 2021

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← Older revision		Revision as of 08:08, 24 September 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Einige Vorbemerkungen zu zweidimensionalen Zufallsgrößen~~	+	{{quiz-Header\|Buchseite=Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables
	}}		}}

@@ Line 123: / Line 123: @@
 \hspace{0.05cm}. $$
 *Accordingly, we obtain for the probability of the complementary event:
-:$${\rm Pr}(L\text{ is odd | first „6”)})
+:$${\rm Pr}(L\text{ is odd | first „6”)}
 = p_1 + p_3 + p_5 + \text{...} = p_1 \cdot \left [ 1 + (1 - p_1)^2 + (1 - p_1)^4 + \text{...} \hspace{0.15cm} \right ]
 \hspace{0.05cm}\hspace{0.3cm}

@@ Line 66: / Line 66: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; Assuming fair dice, the probability that
-a "6" is rolled with the red die:
+a&nbsp; "6"&nbsp; is rolled with the red die:
 :$$\underline{{\rm Pr}(R=6) = 1/6} = 0.1667 \hspace{0.05cm},$$
-* a "1" or a "2" is rolled with the blue die:
+* that a&nbsp; "1"&nbsp; or a&nbsp; "2"&nbsp; is rolled with the blue die:
 :$$\underline{{\rm Pr}(B\le 2) = 1/3} = 0.3333 \hspace{0.05cm},$$
-* both dice show the same number of points:
+* that both dice show the same number of points:
 :$$\underline{{\rm Pr}(R=B) = 6/36} = 0.1667 \hspace{0.05cm}.$$
 The latter is based on the 2D representation on the specification sheet and on the "Classical Definition of Probability" corresponding to&nbsp; $K/M$:
 *$K = 6$&nbsp; of the total&nbsp; $M = 36$&nbsp; equally probable elementary events&nbsp; $R \cap B$&nbsp; can be assigned to the event&nbsp; $R=B$&nbsp; derived from this.
-*These lie on the diagonal.&nbsp; Dice players speak in this case of a "double".
+*These lie on the diagonal.&nbsp; Dice players speak in this case of a&nbsp; "double".
 '''(2)'''&nbsp; The solution is again based on the Classical Definition of Probability:
-* In&nbsp; $K = 2$&nbsp; of the&nbsp; $M = 36$&nbsp; elementary fields there is a "3" &nbsp; &#8658; &nbsp; ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
+* In&nbsp; $K = 2$&nbsp; of the&nbsp; $M = 36$&nbsp; elementary fields there is a&nbsp; "3" &nbsp; &#8658; &nbsp; ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
-* In&nbsp; $K = 6$&nbsp; of the&nbsp; $M = 36$&nbsp; elementary fields there is a "7"&nbsp; &#8658; &nbsp; ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
+* In&nbsp; $K = 6$&nbsp; of the&nbsp; $M = 36$&nbsp; elementary fields there is a&nbsp; "7"&nbsp; &#8658; &nbsp; ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
-* In&nbsp; $K = 18$&nbsp; of the&nbsp; $M = 36$&nbsp; fields there is an odd number &nbsp; &#8658; &nbsp; ${\rm Pr}(S\text{is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$
+* In&nbsp; $K = 18$&nbsp; of the&nbsp; $M = 36$&nbsp; fields there is an odd number &nbsp; &#8658; &nbsp; ${\rm Pr}(S\text{ is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$
 *This last result could be obtained in another way:
 :$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) =
-  {\rm Pr}\big [(R\hspace{0.12cm}{\rm ist\hspace{0.12cm} ungerade}) \cap
+  {\rm Pr}\big [(R\hspace{0.12cm}{\rm is \hspace{0.15cm} odd}) \cap
 (B\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \big  ] +
-{\rm Pr}\big [(R\hspace{0.12cm}{\rm ist\hspace{0.12cm} gerade}) \cap
+{\rm Pr}\big [(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \cap
 (B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})\big  ]\hspace{0.05cm}. $$
 *With&nbsp; ${\rm Pr}(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) = {\rm Pr} (R\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) =  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} even})=  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})   = 1/2$&nbsp; it also follows:
@@ Line 96: / Line 96: @@
-'''(3)'''&nbsp; The probability for the event that at least one of the two dice shows a "6" is:
+'''(3)'''&nbsp; The probability for the event that at least one of the two dice shows a&nbsp; "6" is:
 :$${\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = K/M = 11/36 \hspace{0.15cm} \underline{= 0.3056}
 \hspace{0.05cm}.$$
-*The second probability stands alone for the "double sixes":
+*The second probability stands alone for the&nbsp; "double sixes":
 :$${\rm Pr}\big [(R= 6) \cap (B= 6) \big ] = K/M = 1/36 \hspace{0.15cm} \underline{= 0.0278}
 \hspace{0.05cm}.$$
@@ Line 109: / Line 109: @@
 *The probability&nbsp; $p_2$&nbsp; can be expressed with&nbsp; $p_1$&nbsp; as follows:
 :$$p_2 = (1 - p_1) \cdot p_1 = \frac{25}{36} \cdot \frac{11}{36} \hspace{0.15cm} \underline{= 0.2122} \hspace{0.05cm}. $$
-:In words, the probability that a "6" is rolled for the first time in the second roll is equal to the probability that no "6" was rolled in the first roll &nbsp; &#8658; &nbsp; probability&nbsp; $1-p_1$, but there is at least one "6" in the second roll &nbsp; &#8658; &nbsp; probability&nbsp; $p_1$.
+:In words, the probability that a&nbsp; "6"&nbsp; is rolled for the first time in the second roll is equal to the probability that no&nbsp; "6"&nbsp; was rolled in the first roll &nbsp; &#8658; &nbsp; probability&nbsp; $1-p_1$, but there is at least one&nbsp; "6"&nbsp; in the second roll &nbsp; &#8658; &nbsp; probability&nbsp; $p_1$.
 *Correspondingly, for the probability "first 6 in the third throw":
@@ Line 123: / Line 123: @@
 \hspace{0.05cm}. $$
 *Accordingly, we obtain for the probability of the complementary event:
-:$${\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} odd})
+:$${\rm Pr}(L\text{ is odd | first „6”)})
 = p_1 + p_3 + p_5 + \text{...} = p_1 \cdot \left [ 1 + (1 - p_1)^2 + (1 - p_1)^4 + \text{...} \hspace{0.15cm} \right ]
 \hspace{0.05cm}\hspace{0.3cm}
-\Rightarrow \hspace{0.3cm} \frac{{\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} odd}) } {{\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})} = \frac{1}{1 - p_1} \hspace{0.05cm}. $$
+\Rightarrow \hspace{0.3cm} \frac{{\rm Pr}(L\text{ is odd | first „6”)}) } {{\rm Pr}(L\text{ is even | first „6”)})} = \frac{1}{1 - p_1} \hspace{0.05cm}. $$
 *Further, must hold::
-:$${\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})   +
+:$${\rm Pr}(L \text{ is even | first „6”)}   +
-{\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} odd})  = 1$$
+{\rm Pr}(L \text{ is odd | first „6”)}  = 1$$
-:$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})   \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1  \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})   = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} =  \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$
+:$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L \text{ is even | first „6”)}   \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1  \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})   = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} =  \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID2749__Inf_A_3_1.png|right|frame|Sum&nbsp; $S= R + B$&nbsp; of two dice]]
-We consider the random experiment "rolling one or two dice".&nbsp; Both dice are fair (the six possible outcomes are equally probable) and distinguishable by their colors:
+We consider the random experiment&nbsp; &raquo;rolling one or two dice&laquo;.&nbsp; Both dice are fair (the six possible outcomes are equally probable) and distinguishable by their colors:
-* The random quantity&nbsp; $R = \{1, \ 2,\ 3,\ 4,\ 5,\ 6 \}$&nbsp; denotes the number of eyes on the red cube.
+* The random quantity&nbsp; $R = \{1, \ 2,\ 3,\ 4,\ 5,\ 6 \}$&nbsp; denotes the number of eyes of the red cube.
 * The random quantity&nbsp; $B = \{1,\ 2,\ 3,\ 4,\ 5,\ 6 \}$&nbsp; denotes the number of eyes of the blue cube.
 * The random quantity&nbsp; $S =R + B$&nbsp; stands for the sum of both dice.
@@ Line 22: / Line 22: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on 2D random variables]].
-*In particular, the subject matter of the chapter&nbsp;   [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen|Probability Calculation]]&nbsp; in the book "Theory of Stochastic Signals" is repeated here.
+*In particular, the subject matter of the chapter&nbsp;   [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen|Probability Calculation]]&nbsp; in the book "Theory of Stochastic Signals"&nbsp; is repeated here.
@@ Line 50: / Line 50: @@
-{What is the probability that the first "6" will be on the&nbsp; $L$&ndash;th double roll?
+{What is the probability that the first&nbsp; "6"&nbsp; will be on the&nbsp; $L$&ndash;th double roll?
 |type="{}"}
 $L = 1\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \ $ { 0.3056 3% }
@@ Line 57: / Line 57: @@
-{What is the probability &nbsp; "To get the first "6" &nbsp; , you need an even number of double rolls? <br>Using the nomenclature given in subtask&nbsp; '''(4)''':
+{What is the probability of the event&nbsp; &raquo;You need an even number of double rolls, to get the first "6"&laquo;&nbsp; ? <br>Using the nomenclature given in subtask&nbsp; '''(4)''':
 |type="{}"}
-$\text{Pr(}L\text{ is even)}\ = \ $ { 0.4098 3% }
+$\text{Pr(}L\text{ is even | first „6”)}\ = \ $ { 0.4098 3% }
 </quiz>
@@ Line 117: / Line 117: @@
 '''(5)'''&nbsp; By extending the sample solution to subtask&nbsp; '''(4)'''&nbsp;, we obtain:
-:$$\text{Pr(even L)}= p_2 \hspace{-0.05cm}+ \hspace{-0.05cm}p_4  \hspace{-0.05cm}+ \hspace{-0.05cm} p_6  \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...} =
+:$$\text{Pr(}L\text{ is even | first „6”)}\ =
 (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1  \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^3 \cdot p_1  \hspace{-0.05cm}+ \hspace{-0.05cm}(1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^5 \cdot p_1  \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...}
 = (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1 \cdot \left [ 1  \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^2  \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^4 +\text{...}\hspace{0.05cm} \right ]

@@ Line 22: / Line 22: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Einige_Vorbemerkungen_zu_zweidimensionalen_Zufallsgrößen|Some preliminary remarks on 2D random variables]].
-*In particular, the subject matter of the chapter&nbsp;   [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen|Wahrscheinlichkeitsrechnung]] in the book "Stochastic Signal Theory" is repeated here.
+*In particular, the subject matter of the chapter&nbsp;   [[Theory_of_Stochastic_Signals/Einige_grundlegende_Definitionen|Probability Calculation]]&nbsp; in the book "Theory of Stochastic Signals" is repeated here.

← Older revision		Revision as of 13:31, 16 July 2021
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−	[[Category:Information Theory: Exercises\|^3.1 General Information on ~~Two-dimensional~~ Random Variables^]]	+	[[Category:Information Theory: Exercises\|^3.1 General Information on 2D Random Variables^]]

← Older revision		Revision as of 13:28, 16 July 2021
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−	[[Category:Information Theory: Exercises\|^3.1 ~~Allgemeines zu 2D~~-~~Zufallsgrößen~~^]]	+	[[Category:Information Theory: Exercises\|^3.1 General Information on Two-dimensional Random Variables^]]