Largest and smallest possible values of a probability question

  • Thread starter Thread starter PhysicsMathGuy
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
The discussion centers on determining the largest and smallest possible values of P({2}) given certain probabilities in a random selection of positive integers. The known probabilities include P({1,2,3,4,5}) = 0.3, P({4,5,6}) = 0.4, and P({1}) = 0.1. Participants clarify that P({2}) is constrained between 0 and 0.2, with the reasoning that P({2,3,4,5}) can be derived from the provided probabilities. The conclusion emphasizes that without restrictions on P({3}) or P({6}), the bounds for P({2}) are likely accurate. Overall, the analysis highlights the complexity of the probability distribution involved.
PhysicsMathGuy
Messages
5
Reaction score
0
Hello,
So I've been trying to solve this problem for a while and I can't get my head around it.
"Suppose we choose a positive integer at random, according to some unknown distribution. Suppose we know P({1,2,3,4,5}) = 0.3, P({4,5,6}) = 0.4 and P({1}) = 0.1"

All I've managed to get so far is that P({2,3,4,5}) = P({1,2,3,4,5}) - P({1}) = 0.2
I'm sure that's not a good upper bound because there's probably a way the use the fact that the union of {1,2,3,4,5} and {4,5,6} is {4,5}.

Thanks in advance for your help!
 
Physics news on Phys.org
Just saw that this section is not for this. Sorry, my bad!
 
P({4,5}) ≤ P({2,3,4,5}) =0.2, P({6}) = P({4,5,6}) - P({4,5}) ≥ 0.2. What are you trying to find?
 
Thanks a lot for your reply! I'm trying to find the largest and smallest possible values of P({2}).
 
Off hand P({2}) is between 0 and 0.2. I suspect this is the best you can do since there is no restriction on P({3}) or P({6}).
 
Thanks a lot! That's what I thought, but I wasn't sure and thought I was missing something. Thanks a lot!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top