Probability and Probability Distributions question

[DragonMir]

Two 4-sided dice are tossed 10 times and the sum of the “down” faces is noted. What is the expected number of times a sum of three should occur? Provide evidence of your work by typing out your full solution.

So I did this but i think I am wrong
The sample space of an event where "down" face is noted on a 4-sided die is {(1,2,3),(1,2,4),(1,3,4),(2,3,4)}





From the sample space, we can observe that there is no event where a sum of three when "down" faces is noted on a 4-sided die.





Thus, the expected number of times a sum of three should occur when the "down" faces is noted on two 4-sided die is 0.

 

[Dr.Peterson]

Two 4-sided dice are tossed 10 times and the sum of the “down” faces is noted. What is the expected number of times a sum of three should occur? Provide evidence of your work by typing out your full solution.

So I did this but i think I am wrong
The sample space of an event where "down" face is noted on a 4-sided die is {(1,2,3),(1,2,4),(1,3,4),(2,3,4)}

From the sample space, we can observe that there is no event where a sum of three when "down" faces is noted on a 4-sided die.

Thus, the expected number of times a sum of three should occur when the "down" faces is noted on two 4-sided die is 0.

I think you're misinterpreting how they are reading the dice.

They are only determining the single face that is down on a give die, not the three other faces. So the sample space for one roll of one die is (presumably -- they didn't specify the markings!) {1, 2, 3, 4}.

The sum is the sum of the "down" faces on the two dice, not the sum of visible faces on one die; so the sample space for one roll is {(1,1), (1,2), ..., (4,4)}. The random variable is the sum of the two numbers.

 

[DragonMir]

I think you're misinterpreting how they are reading the dice.

They are only determining the single face that is down on a give die, not the three other faces. So the sample space for one roll of one die is (presumably -- they didn't specify the markings!) {1, 2, 3, 4}.

The sum is the sum of the "down" faces on the two dice, not the sum of visible faces on one die; so the sample space for one roll is {(1,1), (1,2), ..., (4,4)}. The random variable is the sum of the two numbers.

And now i have no clue how to solve it

 

[Steven G]

If each die has the numbers 1, 2, 3 and 4 on it, then what might be the sum if you tossed two dice. The answer is 2,3,4,5,6,7 or 8 coming from, as Dr Peterson pointed out, {(1,1), (1,2), ..., (4,4)}.
List ALL the possible outcomes with out ..... What is the expected sum of the toss of two dice? How about 10 dice?

Two 4-sided dice are tossed 10 times and the sum of the “down” faces is noted.They say down faces, not because these are multiple down faces on each toss--actually, there is only one down face, but because you are tossing two dice and will get two down faces

 

[DragonMir]

D1 and D2 then the outcome may be represented as an ordered pair (d1,d2) where d1 is the relevant face on D1, and d2 that on D2.

There are 16 possible outcomes: (1,1), (1,2), (1,3),(1,4), (2,1), (2,2), (2,3), (2,4), ..., (4,3), (4,4)
The favorable outcomes are those where d1+d2=3, count them

The expected number of favorable outcomes in 10 trials is 10p.

 

[DragonMir]

so we need D1 and D2 to be 3, therefore, we have 2,1 and 1,2 which means p = number of fav outcomes /num of possible outcomes if we toss the dice 10 times that gives us fav outcome is 20 so 20/16 = 1.25? is this correct