Show that a sample space is valid by verifying properties

[a255c]

Homework Statement

http://puu.sh/nYQqE/2b0eaf2720.png

Homework Equations

http://puu.sh/nYSjQ/e48cad3a8b.png

The Attempt at a Solution

http://puu.sh/nYYjW/174ad8267c.png

My main issue is with part b) and part d). I think that part b) is mostly right, but part d) is definitely wrong and incomplete, and I have no idea how to do it.

 

[andrewkirk]

part d) is definitely wrong and incomplete, and I have no idea how to do it.

(d) is not too bad. The suggestions I would make are:
- in iii you need to state that A and B are disjoint, otherwise the equalities will not hold. Also the first $Pr$ should be $Pr'$. You need to be careful with your symbols in a problem like this, as misplaced or wrong symbols lead to confusion.
- in ii, the statement $\sum_{i\in S}\frac{n_i}{36}$ should be $\sum_{i\in S'}\frac{n_i}{36}$ (care with symbols again). You'll probably find it easier if you split it into two sums $\sum_{i=2}^6$ and $\sum_{i=7}^{12}$. Then use the above definitions of $n_i$ in terms of $i$.
- the proof that $P(\emptyset)=0$ is not correct because dice have nothing to do with it. What is $\sum_{i\in\emptyset}\frac{n_i}{36}$ (how many terms are there in the sum?)?

 

[a255c]

(d) is not too bad. The suggestions I would make are:
- in iii you need to state that A and B are disjoint, otherwise the equalities will not hold. Also the first $Pr$ should be $Pr'$. You need to be careful with your symbols in a problem like this, as misplaced or wrong symbols lead to confusion.
- in ii, the statement $\sum_{i\in S}\frac{n_i}{36}$ should be $\sum_{i\in S'}\frac{n_i}{36}$ (care with symbols again). You'll probably find it easier if you split it into two sums $\sum{i=2}^6$ and $\sum{i=7}^12$. Then use the above definitions of $n_i$ in terms of $i$.
- the proof that $P(\emptyset)=0$ is not correct because dice have nothing to do with it. What is $\sum_{i\in\emptyset}\frac{n_i}{36}$ (how many terms are there in the sum?)?

I have used your suggestions and have fixed my answers to this:
http://puu.sh/nZZZl/d243480f50.png

but I did not know what you mean about splitting the sums for d)ii, and I'm still unsure how to do d)i

 

[andrewkirk]

What are you trying to prove in (d)(i)? I think all the requirements of a probability space are contained in (d)ii and iii.

For splitting sums, replace $\sum_{j\in S'}$ by $\sum_{j=2}^6+\sum_{j=7}^{12}$ (I prefer to use $j$ rather than $i$ as an index variable because MathJax insists on inappropriately autocorrecting $i$ to $I$).

 

[a255c]

What are you trying to prove in (d)(i)? I think all the requirements of a probability space are contained in (d)ii and iii.

For splitting sums, replace $\sum_{j\in S'}$ by $\sum_{j=2}^6+\sum_{j=7}^{12}$ (I prefer to use $j$ rather than $i$ as an index variable because MathJax insists on inappropriately autocorrecting $i$ to $I$).

I'm trying to prove property 1 for d)i.

I don't really understand how you get j = 2 and j = 7 though for the sums or where you get the two sums from to begin with.

 

[andrewkirk]

where you get the two sums from to begin with.

You started with $\sum_{j\in S'}$
What is $S'$?

I'm trying to prove property 1 for d)i.

Assuming you prove $Pr(S')=1$ in ii, you just have to prove that $A\subseteq S'\Rightarrow Pr(A)\leq Pr(S')$. Try splitting $S'$ into $A$ and $S'-A$.

 

[a255c]

You started with $\sum_{j\in S'}$
What is $S'$?

Assuming you prove $Pr(S')=1$ in ii, you just have to prove that $A\subseteq S'\Rightarrow Pr(A)\leq Pr(S')$. Try splitting $S'$ into $A$ and $S'-A$.

S' is 36, but I still don't understand how you got 2 and 7 specifically..

If I split S' into A and S'-A, then I say something like \sum{i \in A} + \sum{i\in S'-A} = \sum{i \in S}?
I'm not sure how splitting could help me for part d)i.

 

[andrewkirk]

S' is 36

No it isn't. Look at the OP where $S'$ is defined.

 

[a255c]

No it isn't. Look at the OP where $S'$ is defined.

Oh, i see what you are saying. So then I evaluate for n_j/36 for the two sums?

I did 2(1+2+3+4+5+6+7)/36 but this does not equal 1

I have modified my answers to this
http://puu.sh/o111N/1376d907c3.png

 

[andrewkirk]

I did 2(1+2+3+4+5+6+7)/36 but this does not equal 1

That sum has fourteen terms (2 x 7). How many terms are there supposed to be in the sum?

 

[a255c]

That sum has fourteen terms (2 x 7). How many terms are there supposed to be in the sum?

10...

so it's 2(1 + 2 + 3 + 4 + 5)/36? That's 30/36 though