Probability - Dice question

[marimar12]

Hi, I need your help with a probability problem
Thank you!
The question is: Suppose you have two types of dice. One is a six-face dice and the other is eight face dice, both of which have an equal probability of each face to come up.
a) If the six-face dice is numbered by 1, 2, 3, 4, 5, and 6, but the eight-face dice is numbered by 1, 2, 3, 4, 5, 6, 6, and 6, what is the expected number if you roll each type of dice many times, respectively?

In this question, I did my calculations and I got an answer: E(X+Y)= E(X)+E(Y) = 3.5 + 4.125= 7.625. However, I am not very sure about it because I do not understand If I have to find the sum value or the expected value for each die.

b) Now if you randomly pick one dice from a black box with two six-face dices and one eight-face dice, what is the expected number you can roll?

 

[pka]

The question is: Suppose you have two types of dice. One is a six-face dice and the other is eight face dice, both of which have an equal probability of each face to come up.
a) If the six-face dice is numbered by 1, 2, 3, 4, 5, and 6, but the eight-face dice is numbered by 1, 2, 3, 4, 5, 6, 6, and 6, what is the expected number if you roll each type of dice many times, respectively? In this question, I did my calculations and I got an answer: E(X+Y)= E(X)+E(Y) = 3.5 + 4.125= 7.625. However, I am not very sure about it because I do not understand If I have to find the sum value or the expected value for each die.

b) Now if you randomly pick one dice from a black box with two six-face dices and one eight-face dice, what is the expected number you can roll?

I have never seen an eight sided die of this description. Although, I do agree with your calculations of the expected values in part a).
But part b) confuses me frankly. Let's say that \(Z\) is the value of a single roll of the blindly selected die.
What will you use for these two: \(\mathcal{P}(Z=5)~\&~\mathcal{P}(Z=6)~?\)

 

[Steven G]

Your answer to 1 is not correct at all.

You were asked what is the expected number if you roll each type of dice many times, respectively? At this point I am not talking probability or expected values. They are NOT asking for the expected value of the sum but rather they are asking you to find E(X) and E(Y). Now you did find them but never formally said what each equals. Just say E(X) = 3.5 and E(Y)= 4.125 and remove everything about E(X+Y).

part b) I guess that you have to assume that p(pick the 6-sided die) = p(pick the 8-sided die) = 1/2.

Then the expected value will be (.5)*(3.75) + (.5)*(4.125) = 3.8125.

If p(pick the 6-sided die) =p and p(pick the 8-sided die)= 1-p, then the expected value will be (p)*(3.75) + (1-p)*(4.125)

 

[pka]

Your answer to 1 is not correct at all.
You were asked what is the expected number if you roll each type of dice many times, respectively? At this point I am not talking probability or expected values. They are NOT asking for the expected value of the sum but rather they are asking you to find E(X) and E(Y). Now you did find them but never formally said what each equals. Just say E(X) = 3.5 and E(Y)= 4.125 and remove everything about E(X+Y)

Which of those two calculations are you saying is wrong? Is \(E(Y)\) wrong? if so why?
Surely you will agree that \(X~\&~Y\) are independent?
Jomo I would direct you to Jim Pitman's, @berkeley, Text on the exception as a linear function.

part b) I guess that you have to assume that p(pick the 6-sided die) = p(pick the 8-sided die) = 1/2.

Look Here

 

[Steven G]

I did not say that any calculations were wrong.

I said that the question asked the student to find E(X) and E(Y).

The student instead found, correctly, E(X+Y) using the fact that E(X+Y) = E(X) + E(Y).