Counting techniques and probability

  • #1
Shackleford
1,666
2
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 [Broken]

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

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?
 
Last edited by a moderator:

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420


A probability measure on S satisfies [itex]P(\emptyset)=0[/itex] and [itex]P(S)=1[/itex] (this already contradicts your results), and it also satisfies [itex]P(A\cup B)=P(A)+P(B)[/itex] when A and B are disjoint. (You should look up the exact definition). You need to use this, together with the information you were given, which is that [itex]P(1)=p_1\geq 0[/itex] and so on. You should interpret P(n) as P({n}).

For example, if you want to calculate P({1,2}), you use the fact that {1,2} is the union of the disjoint sets {1} and {2} and that you know P({1}) and P({2}). Now, can you find a formula for P(E) in terms of the pn? (E is an arbitrary member of the powerset, so it can e.g. be {1,2}).

You counted the number of members of the powerset correctly, but it doesn't look like you need this result.
 
Last edited:
  • #3
Shackleford
1,666
2


A probability measure on S satisfies [itex]P(\emptyset)=0[/itex] and [itex]P(S)=1[/itex] (this already contradicts your results), and it also satisfies [itex]P(A\cup B)=P(A)+P(B)[/itex] when A and B are disjoint. (You should look up the exact definition). You need to use this, together with the information you were given, which is that [itex]P(1)=p_1\geq 0[/itex] and so on. You should interpret P(n) as P({n}).

For example, if you want to calculate P({1,2}), you use the fact that {1,2} is the union of the disjoint sets {1} and {2} and that you know P({1}) and P({2}). Now, can you find a formula for P(E) in terms of the pn? (E is an arbitrary member of the powerset, so it can e.g. be {1,2}).

You counted the number of members of the powerset correctly, but it doesn't look like you need this result.

I'm familiar with the definition. My probability is 1/4 for each P{1}, P{2}, P{3}, P{4}. They are disjoint and yield the sample space, and this adds to 1. That has to be just coincidence.

Well, my best guess is P(E) = 1 - P(Ec)
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420


My probability is 1/4 for each P{1}, P{2}, P{3}, P{4}. They are disjoint and yield the sample space, and this adds to 1. That has to be just coincidence.
Yes, it appears so. The probability P({1}) is =P(1)=p1. I don't see anything that would imply that it's specifically 1/4.

Well, my best guess is P(E) = 1 - P(Ec)
That follows from the P(A⋃B)=P(A)+P(B) rule, but it doesn't actually tell us anything. I told you how to find P({1,2}) (it's =p1+p2). That's the sort of answer you should be looking for, but for an arbitrary set E.
 
  • #5
Shackleford
1,666
2


Yes, it appears so. The probability P({1}) is =P(1)=p1. I don't see anything that would imply that it's specifically 1/4.


That follows from the P(A⋃B)=P(A)+P(B) rule, but it doesn't actually tell us anything. I told you the answer for P({1,2}) (it's =p1+p2). That's the sort of answer you should be looking for, but for an arbitrary set E.


That's what I got when I did the (6 1)T calculation.

If Ei and Ej are disjoint, then P(Ei⋃Ej)= P(Ei) + P(Ej)

p1 + p2 + p3 + p4 = 1

pi + pj + pk + pl = 1

If Ei ⋃ Ej = E, then P(E) = 1 - pk + pl
 
  • #6
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420


I don't know what you're doing now, but what you're supposed to do is to find a function [itex]P:\mathcal E\rightarrow [0,1][/itex] that satisfies the definition of a probability measure, and also the given conditions

P({1})=p1, P({2})=p2,...

In other words, you need to find P(E) for each of the 16 members of [itex]\mathcal E[/itex]. It's definitely not going to be easier to find P(Ec) first and compute P(E) from that.

p1 + p2 + p3 + p4 = 1

pi + pj + pk + pl = 1

If Ei ⋃ Ej = E, then P(E) = 1 - pk + pl
It looks like you just replaced each number with a symbol, and defined each Esomething to be a subset of S with only one member.
 
  • #7
Shackleford
1,666
2


I don't understand. First you say do it for an arbitrary E, which would be general and apply to all, then you say find P(E) for each and everyone.

That's the sort of answer you should be looking for, but for an arbitrary set E.

In other words, you need to find P(E) for each of the 16 members of [itex]\mathcal E[/itex].

Then my first step is to write out all 16 subsets, right?
 
  • #8
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,875
420


I meant one formula that holds for all E. No, you don't need to write out the subsets.

I'll just tell you the answer I had in mind. This is probably just the first of many exercises you'll do anyway.

[tex]P(E)=\sum_{n\in E}p_n[/tex]
 
  • #9
Shackleford
1,666
2


I meant one formula that holds for all E. No, you don't need to write out the subsets.

I'll just tell you the answer I had in mind. This is probably just the first of many exercises you'll do anyway.

[tex]P(E)=\sum_{n\in E}p_n[/tex]

I know that. If the sets are pairwise disjoint, the probability of their union is equal to the sum of their individual probabilities.

I understand that the equation in the problem gives you the union with the maximum number of disjoint sets.
 
Last edited:

Suggested for: Counting techniques and probability

Replies
4
Views
121
  • Last Post
Replies
19
Views
830
Replies
5
Views
230
Replies
11
Views
453
Replies
5
Views
391
Replies
2
Views
346
Replies
27
Views
462
Replies
5
Views
406
Replies
8
Views
112
Replies
6
Views
253
Top