Probability

[warwick]

I don't remember any probability from high school which was over nine years ago. This semester I'm taking Thermal Physics and Probability. So, I'm having to catch up with counting techniques and so forth and make sense of the logic behind the techniques.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110129_185742.jpg?t=1296349823

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110129_185800.jpg?t=1296349829

Did I do this problem correctly? It looks like I calculated the probability of the subsets of the powersets, i.e. the collection of sets with 0 elements all the way to 4 elements, respectively. The professor says to determine P(E) for every E from the powerset. Is that the probability for the subsets of the powerset or the probability of each set in the subsets of the powerset?

 

[pka]

First of all, it is very difficult to read those two images.
But I think the question asks you to find \(\displaystyle \mathcal{P}(X)\) for every
\(\displaystyle X\subseteq\{1,2,3,4\}\).
Clearly \(\displaystyle \mathcal{P}(\emptyset)=0\)
Now \(\displaystyle \mathcal{P}(X)=\sum\limits_{k \in X} {p_k }\)
Here is an example: \(\displaystyle \mathcal{P}(\{1,3\})=p_1+p_3\)
There are 16 subsets of \(\displaystyle \{1,2,3,4\}\).

 

[warwick]

First of all, it is very difficult to read those two images.
But I think the question asks you to find \(\displaystyle \mathcal{P}(X)\) for every
\(\displaystyle X\subseteq\{1,2,3,4\}\).
Clearly \(\displaystyle \mathcal{P}(\emptyset)=0\)
Now \(\displaystyle \mathcal{P}(X)=\sum\limits_{k \in X} {p_k }\)
Here is an example: \(\displaystyle \mathcal{P}(\{1,3\})=p_1+p_3\)
There are 16 subsets of \(\displaystyle \{1,2,3,4\}\).

Sorry about that. I no longer have a scanner.

If E[sub:2hlbs2sd]i[/sub:2hlbs2sd] and E[sub:2hlbs2sd]j[/sub:2hlbs2sd] are disjoint, then P(E[sub:2hlbs2sd]i[/sub:2hlbs2sd]?E[sub:2hlbs2sd]j[/sub:2hlbs2sd])= P(E[sub:2hlbs2sd]i[/sub:2hlbs2sd]) + P(E[sub:2hlbs2sd]j[/sub:2hlbs2sd])

p[sub:2hlbs2sd]1[/sub:2hlbs2sd] + p[sub:2hlbs2sd]2[/sub:2hlbs2sd] + p[sub:2hlbs2sd]3[/sub:2hlbs2sd] + p[sub:2hlbs2sd]4[/sub:2hlbs2sd] = 1

p[sub:2hlbs2sd]i[/sub:2hlbs2sd] + p[sub:2hlbs2sd]j[/sub:2hlbs2sd] + p[sub:2hlbs2sd]k[/sub:2hlbs2sd] + p[sub:2hlbs2sd]l[/sub:2hlbs2sd] = 1

If E[sub:2hlbs2sd]i[/sub:2hlbs2sd] ? E[sub:2hlbs2sd]j[/sub:2hlbs2sd] = E, then P(E) = 1 - p[sub:2hlbs2sd]k[/sub:2hlbs2sd] + p[sub:2hlbs2sd]l[/sub:2hlbs2sd]