Difference between revisions of "Aufgaben:Exercise 3.1: Probabilities when Rolling Dice"

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'''(1)'''  Assuming fair dice, the probability that
 
'''(1)'''  Assuming fair dice, the probability that
a "6" is rolled with the red die:
+
a  "6"  is rolled with the red die:
 
:$$\underline{{\rm Pr}(R=6) = 1/6} = 0.1667 \hspace{0.05cm},$$
 
:$$\underline{{\rm Pr}(R=6) = 1/6} = 0.1667 \hspace{0.05cm},$$
* a "1" or a "2" is rolled with the blue die:
+
* that a  "1"  or a  "2"  is rolled with the blue die:
 
:$$\underline{{\rm Pr}(B\le 2) = 1/3} = 0.3333 \hspace{0.05cm},$$
 
:$$\underline{{\rm Pr}(B\le 2) = 1/3} = 0.3333 \hspace{0.05cm},$$
* both dice show the same number of points:
+
* that both dice show the same number of points:
 
:$$\underline{{\rm Pr}(R=B) = 6/36} = 0.1667 \hspace{0.05cm}.$$
 
:$$\underline{{\rm Pr}(R=B) = 6/36} = 0.1667 \hspace{0.05cm}.$$
  
 
The latter is based on the 2D representation on the specification sheet and on the "Classical Definition of Probability" corresponding to  $K/M$:  
 
The latter is based on the 2D representation on the specification sheet and on the "Classical Definition of Probability" corresponding to  $K/M$:  
 
*$K = 6$  of the total  $M = 36$  equally probable elementary events  $R \cap B$  can be assigned to the event  $R=B$  derived from this.
 
*$K = 6$  of the total  $M = 36$  equally probable elementary events  $R \cap B$  can be assigned to the event  $R=B$  derived from this.
*These lie on the diagonal.  Dice players speak in this case of a "double".
+
*These lie on the diagonal.  Dice players speak in this case of a  "double".
  
  
  
 
'''(2)'''  The solution is again based on the Classical Definition of Probability:
 
'''(2)'''  The solution is again based on the Classical Definition of Probability:
* In  $K = 2$  of the  $M = 36$  elementary fields there is a "3"   ⇒   ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
+
* In  $K = 2$  of the  $M = 36$  elementary fields there is a  "3"   ⇒   ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
* In  $K = 6$  of the  $M = 36$  elementary fields there is a "7"  ⇒   ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
+
* In  $K = 6$  of the  $M = 36$  elementary fields there is a  "7"  ⇒   ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
* In  $K = 18$  of the  $M = 36$  fields there is an odd number   ⇒   ${\rm Pr}(S\text{is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$
+
* In  $K = 18$  of the  $M = 36$  fields there is an odd number   ⇒   ${\rm Pr}(S\text{ is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$
  
  
 
*This last result could be obtained in another way:
 
*This last result could be obtained in another way:
 
:$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) =
 
:$${\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  ] +
 
(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}. $$
 
(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})\big  ]\hspace{0.05cm}. $$
 
*With  ${\rm Pr}(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) = {\rm Pr} (R\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) =  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} even})=  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})  = 1/2$  it also follows:
 
*With  ${\rm Pr}(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) = {\rm Pr} (R\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) =  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} even})=  {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})  = 1/2$  it also follows:
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'''(3)'''  The probability for the event that at least one of the two dice shows a "6" is:
+
'''(3)'''  The probability for the event that at least one of the two dice shows a  "6" is:
 
:$${\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = K/M = 11/36 \hspace{0.15cm} \underline{= 0.3056}
 
:$${\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = K/M = 11/36 \hspace{0.15cm} \underline{= 0.3056}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
*The second probability stands alone for the "double sixes":
+
*The second probability stands alone for the  "double sixes":
 
:$${\rm Pr}\big [(R= 6) \cap (B= 6) \big ] = K/M = 1/36 \hspace{0.15cm} \underline{= 0.0278}
 
:$${\rm Pr}\big [(R= 6) \cap (B= 6) \big ] = K/M = 1/36 \hspace{0.15cm} \underline{= 0.0278}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
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*The probability  $p_2$  can be expressed with  $p_1$  as follows:
 
*The probability  $p_2$  can be expressed with  $p_1$  as follows:
 
:$$p_2 = (1 - p_1) \cdot p_1 = \frac{25}{36} \cdot \frac{11}{36} \hspace{0.15cm} \underline{= 0.2122} \hspace{0.05cm}. $$
 
:$$p_2 = (1 - p_1) \cdot p_1 = \frac{25}{36} \cdot \frac{11}{36} \hspace{0.15cm} \underline{= 0.2122} \hspace{0.05cm}. $$
: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   ⇒   probability  $1-p_1$, but there is at least one "6" in the second roll   ⇒   probability  $p_1$.  
+
: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   ⇒   probability  $1-p_1$, but there is at least one  "6"  in the second roll   ⇒   probability  $p_1$.  
  
 
*Correspondingly, for the probability "first 6 in the third throw":
 
*Correspondingly, for the probability "first 6 in the third throw":
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\hspace{0.05cm}. $$
 
\hspace{0.05cm}. $$
 
*Accordingly, we obtain for the probability of the complementary event:
 
*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 ]
 
= 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}
 
\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::
 
*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$$
+
{\rm Pr}(L \text{ is odd | first „6”)}  = 1$$
:$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})   \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1  \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})  = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} =  \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$
+
:$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L \text{ is even | first „6”)}  \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1  \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even})  = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} =  \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$
 
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Revision as of 14:40, 16 August 2021

Sum  $S= R + B$  of two dice

We consider the random experiment  »rolling one or two dice«.  Both dice are fair (the six possible outcomes are equally probable) and distinguishable by their colors:

  • The random quantity  $R = \{1, \ 2,\ 3,\ 4,\ 5,\ 6 \}$  denotes the number of eyes of the red cube.
  • The random quantity  $B = \{1,\ 2,\ 3,\ 4,\ 5,\ 6 \}$  denotes the number of eyes of the blue cube.
  • The random quantity  $S =R + B$  stands for the sum of both dice.


In this exercise different probabilities are to be computed with reference to the random variables  $R$,  $B$  and  $S$, 
whereby the scheme indicated above can be helpful.  This includes the sum  $S$  as a function of  $R$  and  $B$.





Hints:



Questions

1

Give the following probabilities:

$\text{Pr}(R = 6)\ = \ $

$\text{Pr}(B ≤ 2)\ = \ $

$\text{Pr}(R = B)\ = \ $

2

What are the following probabilities?

$\text{Pr}(S = 3)\ = \ $

$\text{Pr}(S = 7)\ = \ $

$\text{Pr(odd sum)}\ = \ $

3

State the following probabilities:

$\text{Pr}\big [(R = 6)\ \cup \ (B =6)\big]\ = \ $

$\text{Pr}\big[(R = 6)\ \cap \ (B =6)\big]\ = \ $

4

What is the probability that the first  "6"  will be on the  $L$–th double roll?

$L = 1\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \ $

$L = 2\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \ $

$L = 3\text{:}\hspace{0.5cm}\text{Pr(first „6”)} \ = \ $

5

What is the probability of the event  »You need an even number of double rolls, to get the first "6"«  ?
Using the nomenclature given in subtask  (4):

$\text{Pr(}L\text{ is even | first „6”)}\ = \ $


Solution

(1)  Assuming fair dice, the probability that a  "6"  is rolled with the red die:

$$\underline{{\rm Pr}(R=6) = 1/6} = 0.1667 \hspace{0.05cm},$$
  • that a  "1"  or a  "2"  is rolled with the blue die:
$$\underline{{\rm Pr}(B\le 2) = 1/3} = 0.3333 \hspace{0.05cm},$$
  • 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  $K/M$:

  • $K = 6$  of the total  $M = 36$  equally probable elementary events  $R \cap B$  can be assigned to the event  $R=B$  derived from this.
  • These lie on the diagonal.  Dice players speak in this case of a  "double".


(2)  The solution is again based on the Classical Definition of Probability:

  • In  $K = 2$  of the  $M = 36$  elementary fields there is a  "3"   ⇒   ${\rm Pr}(S = 3) = 2/36\hspace{0.15cm}\underline{ = 0.0556}.$
  • In  $K = 6$  of the  $M = 36$  elementary fields there is a  "7"  ⇒   ${\rm Pr}(S = 7) = 6/36\hspace{0.15cm}\underline{ = 0.1667}.$
  • In  $K = 18$  of the  $M = 36$  fields there is an odd number   ⇒   ${\rm Pr}(S\text{ is odd}) = 18/36\hspace{0.15cm}\underline{ = 0.5}.$


  • This last result could be obtained in another way:
$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) = {\rm Pr}\big [(R\hspace{0.12cm}{\rm is \hspace{0.15cm} odd}) \cap (B\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \big ] + {\rm Pr}\big [(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) \cap (B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd})\big ]\hspace{0.05cm}. $$
  • With  ${\rm Pr}(R\hspace{0.12cm}{\rm is\hspace{0.12cm} even}) = {\rm Pr} (R\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) = {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} even})= {\rm Pr}(B\hspace{0.12cm}{\rm is\hspace{0.12cm} odd}) = 1/2$  it also follows:
$${\rm Pr}(S\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}) = 1/2 \cdot 1/2 + 1/2 \cdot 1/2 = 1/2 \hspace{0.05cm}.$$


(3)  The probability for the event that at least one of the two dice shows a  "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":
$${\rm Pr}\big [(R= 6) \cap (B= 6) \big ] = K/M = 1/36 \hspace{0.15cm} \underline{= 0.0278} \hspace{0.05cm}.$$


(4)  The result for  $L = 1$  has already been determined in subtask  (3) :

$$p_1 = {\rm Pr}\big [(R= 6) \cup (B= 6) \big ] = {11}/{36} \hspace{0.15cm} \underline{= 0.3056} \hspace{0.05cm}.$$
  • The probability  $p_2$  can be expressed with  $p_1$  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   ⇒   probability  $1-p_1$, but there is at least one  "6"  in the second roll   ⇒   probability  $p_1$.
  • Correspondingly, for the probability "first 6 in the third throw":
$$p_3 = (1 - p_1)^2 \cdot p_1 = \frac{25}{36} \cdot \frac{25}{36} \cdot\frac{11}{36} \hspace{0.15cm} \underline{= 0.1474} \hspace{0.05cm}.$$


(5)  By extending the sample solution to subtask  (4) , we obtain:

$$\text{Pr(}L\text{ is even | first „6”)}\ = p_2 \hspace{-0.05cm}+ \hspace{-0.05cm}p_4 \hspace{-0.05cm}+ \hspace{-0.05cm} p_6 \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...} = (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^3 \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm}(1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^5 \cdot p_1 \hspace{-0.05cm}+ \hspace{-0.05cm} \text{...} = (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1) \cdot p_1 \cdot \left [ 1 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (1 \hspace{-0.05cm}- \hspace{-0.05cm} p_1)^4 +\text{...}\hspace{0.05cm} \right ] \hspace{0.05cm}. $$
  • Accordingly, we obtain for the probability of the complementary event:
$${\rm Pr}(L\text{ is odd | first „6”)}) = p_1 + p_3 + p_5 + \text{...} = p_1 \cdot \left [ 1 + (1 - p_1)^2 + (1 - p_1)^4 + \text{...} \hspace{0.15cm} \right ] \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{{\rm Pr}(L\text{ is odd | first „6”)}) } {{\rm Pr}(L\text{ is even | first „6”)})} = \frac{1}{1 - p_1} \hspace{0.05cm}. $$
  • Further, must hold::
$${\rm Pr}(L \text{ is even | first „6”)} + {\rm Pr}(L \text{ is odd | first „6”)} = 1$$
$$\Rightarrow \hspace{0.3cm} {\rm Pr}(L \text{ is even | first „6”)} \cdot \left [ 1 + \frac{1}{1 - p_1} \right ] = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(L\hspace{0.15cm}{\rm is\hspace{0.15cm} even}) = \frac{1 - p_1}{2 - p_1} = \frac{25/36}{61/36} = \frac{25}{61} \hspace{0.15cm} \underline{= 0.4098} \hspace{0.05cm}.$$