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- bobbym
- 2012-11-10 02:28:39
Hi;
Is sometimes true and sometimes false.
Is sometimes true and sometimes false.
Please post the original problem and maybe something can be done.
- Karimazer1
- 2012-11-10 02:25:43
Thank you!
Do the other ones work too? This is bad news for me, means I messed up somewhere. I just tried to input e.g. 0.9 =p and 0.9 = q and it did not work, you are correct. darn!!
- bobbym
- 2012-11-10 02:22:07
Hi;
Is sometimes true and sometimes false.
- Karimazer1
- 2012-11-10 02:20:30
Apologies, I see what you mean, normally that is the case but here it is not because it is an extention of that form in this example.
It is not the case, here p + (1-p) = 1, and q + (1-q) = 1.
(1-q) and (1-p) don't feature here in that form because I had to use them to derive the given inequalities. So I guess here, p+(1-p)+q+(1-q)=2 ? But that cancel's out to 2=2 so i guess is not useful...
I look forward to your reply! 
- bobbym
- 2012-11-10 02:15:58
Hi;
When you use the variables p and q and mention the word probability it is standard for the relationship between p and q to be p+q=1. That is what confused me. Is this the case?
Also p+q>1 +(pq)/2 is sometimes true and sometimes false.
- Karimazer1
- 2012-11-10 02:13:31
Thank you Bobby!
But why does p+q = 1 ?
If p = 0.2 (so 1-p=0.8) and q=0.3 (so 1-q = 0.7) then p+q = 0.5 whilst still being viable given restrictions on p and q... ?
- Karimazer1
- 2012-11-10 02:01:58
I am so sorry bobby and everyone, nothing comes after the negative sign!! nothing at all, i just put that to add the note for the brackets!
- bobbym
- 2012-11-10 01:59:50
Hi;
p+q > 1+ (pq)/2 -
What comes after the minus sign in all three problems?
- Karimazer1
- 2012-11-10 01:56:46
Ahhhhhh. I am so bad at this.
Ok, completely new assumptions. Discard the above. I didn't include r.
So, 0<p<1 so that (1-p) + p = 1. (Probability)
Also, 0<q<1 so that (1-q) + q = 1.
Now;
1. p+q > 1+ (pq)/2 - (I have cancelled this down to this, why can this NOT work? I can't understand the intuition / maths)
2. p+q > 6/15 + 2pq - (I have cancelled this down to this, why can this NOT work? I can't understand the intuition / maths)
3. 1 > p + q + pq - (I have cancelled this down to this, why can this NOT work? I can't understand the intuition / maths)
- Karimazer1
- 2012-11-10 01:49:22
THANKS. I found a mistake because of your comment, thank you, i now have new inequalities after looking back, i will attempt. hopefully will not need anyhelp, i hope! let me see now i will work on it!!
- Karimazer1
- 2012-11-10 01:45:52
All three must hold given the assumptions. I want to contradict why these can not work. So for the first one for example, even though p2 and p3 are not there, they are both still positive values less than one and greater than 0. Does this help?
Thank you!!
- anonimnystefy
- 2012-11-10 01:30:46
Those do not always hold. It would be good if you posted the inequality you are strugling with.
- Karimazer1
- 2012-11-10 01:26:55
Want to prove three different things; this should all work for my whole answer to be corrrect but in the last stages I am struggling with the proof/confirmation.
Ok, 0<pi<1, i=1,2,3,4 . BUT the SUM =1, so Sum p1+p2+p3+p4 =1. But none of them can be 0 or 1, all inbetween whilst summing to one.
Now, given this.... I want to understand WHY these inequalities CANNOT hold:
1. 9(p1) > 6(p1) + 6(p4)
2. 9(p2) + 9(p3) > 6(p1) + 6(p4)
3. 9(p4) > 6(p1) + 6(p4)
Treat all of these seperately! I don't know if this is correct or not, basically I think all is correct until this point, but this is the checking part...i.e. if these DO NOT hold, I am correct. Thank you I really cannot wait until I see the response!!!!