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We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independently, one roll for each die. For i = 1, 2, let the random variable Xi represent the result of the i-th die, so that Xi is uniformly distributed over the set {1, 2, 3}. Define X = X2 − X1. 1. Calculate the numerical values of following probabilities:____________. (a) P(X = 0) = (b) P(X = 1) = (c) P(X = −2) = (d) P(X = 3) = Let Y = X2 . Calculate the following probabilities:_________. (a) P(Y = 0) =(b) P(Y = 1) = (c) P(Y = 2) =

Respuesta :

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

(c) P(Y = 2) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 : Result on 1st die.

- Random Variable X_2: Result on 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

 

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

                 ( X = -2 ):  { X_2 = 1 , X_1 = 3 }  

                P ( X = -2 ):  P ( X_2 = 1 ) * P ( X_1 = 3 )  

                                :  ( 1 / 3 ) * ( 1 / 3 )  

                                : ( 1 / 9 )    

                ( X = -1 ):  { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

                P ( X = -1 ):  P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

                                :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                : ( 2 / 9 )        

 ( X = 0 ):  { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } +  { X_2 = 3 , X_1 = 3 }

               P ( X = -1 ):3*P ( X_2 = 1 )*P ( X_1 = 1 )

                                :  3*( 1 / 3 ) * ( 1 / 3 )

                                : ( 3 / 9 ) = ( 1 / 3 )        

                  ( X = 1 ):  { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

               P ( X = 1 ): 2* P ( X_2 = 2 ) * P ( X_1 = 1 )

                                : 2* ( 1 / 3 ) * ( 1 / 3 )

                                : ( 2 / 9 )

                  ( X = 2 ):  { X_2 = 1 , X_1 = 3 }

                 P ( X = 2 ):  P ( X_2 = 3 ) * P ( X_1 = 1 )  

                                   :  ( 1 / 3 ) * ( 1 / 3 )  

                                   : ( 1 / 9 )                  

- The distribution Y = X_2,

                         P(Y=0) = 0

                         P(Y=1) =  1/3

                         P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

lhmarianateixeira

In this exercise we have to use the knowledge of probability to calculate the chance of an event to occur, so:

A) P(X = 0) = 1/3

B) P(X = 1) = 2/9

C) P(X = −2) = 1/9

D) P(X = 3) = 0

A) P(Y = 0) = 0

B) P(Y = 1) = 1/3

C) P(Y = 2) = 1/3

organizing the following information given in the text we have that:

  • Two 3-sided fair die.
  • Random Variable X_1 : Result on 1st die
  • Random Variable X_2: Result on 2nd die.
  • Random Variable X = X_2 - X_1.
  • Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

Then calculating the probability we find that:

A)  For P(X = 0) we have

[tex]( X = -2 ): { X_2 = 1 , X_1 = 3 } \\P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) \\ : ( 1 / 3 ) * ( 1 / 3 ) \\ : ( 1 / 9 )[/tex]

B)   For  P(X = 1) er have:

[tex]( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }\\ P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)\\ : ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )\\ : ( 2 / 9 )[/tex]  

C) For  P(X = −2)  we have:

[tex]( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }\\ P ( X = -1 ):3*P ( X_2 = 1 )*P ( X_1 = 1 )\\ : 3*( 1 / 3 ) * ( 1 / 3 )\\ : ( 3 / 9 ) = ( 1 / 3 )[/tex]    

D) For P(X = 3), we have:

[tex]( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }\\ P ( X = 1 ): 2* P ( X_2 = 2 ) * P ( X_1 = 1 )\\ : 2* ( 1 / 3 ) * ( 1 / 3 )\\ : ( 2 / 9 )\\ ( X = 2 ): { X_2 = 1 , X_1 = 3 }\\ P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 ) \\ : ( 1 / 3 ) * ( 1 / 3 ) \\ : ( 1 / 9 )[/tex]            

In this next case we have to take into account that the number will be the probability of falling 3, that is:

A) P(Y=0) = 0

B) P(Y=1) =  1/3

C) P(Y=2) = 1/ 3

See more about probability at brainly.com/question/795909

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