binomial probability distribution

[PG1995]

Hi

Could you please help me with this query?

Regards
PG

 

[MrAl]

Hi,

To start this off and to begin with, is that formula the correct one for this problem? It's been a while for me on this topic but that looks like the formula to find the probability of seeing an occurrence of r successes out of n trials, where the r successes have to be one right after the other such as:
11100010 which shows a successful find with r=3 and n=8, however it is not successful for r=4 and n=8 because the fourth '1' is separated by 0's (failures).

Check it out and correct this if wrong. Since we know that the probability of seeing 50 successes out of 100 is close to 0.5, and that number gets closer and closer to 0.5 as we let n get higher, this formula appears to make it look harder and harder to get 50 successes and that it probably because it wants to see them all in a row. For example, 0001111000 would be a success for r=4 but 0001110100 would not.

So the probability of this formula not being the right probability formula is 99 percent

 

[PG1995]

Thank you, MrAl.

Sorry but I don't understand whatever you are saying about '0' and '1'. The fomula is correct one because that binomial formula is the one which is used in given set of problems. I was able to solve all other problems correctly. Do you see any explanation that how the given solution reached that value for 'p'? Kindly help me. Thanks.

Regards
PG

 

[MrAl]

Hello again,

Ok well i'll read it over again. To me it sounded like we would want to use a formula which did not place emphasis on whether or not any target 'successes' came all in a row or one now, one later, another one still later. But if they want you to use that formula then i guess you have no choice

LATER:
Your only choice then is to do the calculations and see if they match up. Check your math carefully perhaps doing it twice.
Also, you can simplify the formula a tiny bit to eliminate q simply by replacing q with 1-p. No big deal but it helps a little when comparing with other similar formulas.

 

[PG1995]

Hi

It seems there is no way out! Is there? Kindly help me. Thanks.

Regards
PG

 

[MrAl]

Hi PG,


Sorry i didnt help more yet, i've been busy lately. Had a lot of running around to do again today too. Now im beat but if nobody else responds by tomorrow i'll read it again when i am wide awake and see what i can find out.

 

[PG1995]

No problem, MrAl.

Help me whenever you have free time and feel comfortable. Thanks.

Regards
PG

 

[misterT]

I think the author used tables to solve 5-24a (p=0.40). What kind of tables you have available in the book? If you have cumulative binomial probability distribution, then all you need to do is to search for the value 0.0905. The other numbers in the problem give you a rough "neighborhood" where to look.. And you know that n=15 and r<=3.

I found the right spot in this table. Last page, n=15, x=3.
https://www.electro-tech-online.com/custompdfs/2013/05/BinomialDistribution_cumul.pdf

 

[PG1995]

Thanks a lot, misterT.

I believe you have it right because otherwise there is no easy way to solve it as far as I can tell. Yes, the attached table in your post gives the correct value for 'p'. I have also checked the table given in the book but unfortunately I don't even know how to read the values of 'p'. For instance, how do I find p=0.4 from table? As you can see values for 'p' are given as: 0.01, 0.02, 0.03, 0.04, etc. Please help me. Thank you.

Could you please also help me with this query? Thanks.

Regards
PG

 

[misterT]

Thanks a lot, misterT.

I believe you have it right because otherwise there is no easy way to solve it as far as I can tell. Yes, the attached table in your post gives the correct value for 'p'. I have also checked the table given in the book but unfortunately I don't even know how to read the values of 'p'. For instance, how do I find p=0.4 from table? As you can see values for 'p' are given as: 0.01, 0.02, 0.03, 0.04, etc. Please help me. Thank you.

Could you please also help me with this query? Thanks.

The "four times" does not mean that the price will be 4 times larger than before. It means that the prices go up that many number of times. It does not matter how much the prices go up each time they increase them.
.. hard to explain that one with my english skills

And yes, the tables you posted are kind of strange. Are you missing some pages? I could not figure out how to read them .. At least how would I get p=0.4 from what you posted. Maybe there is explanation or example somewhere in the book.

EDIT: The tables you posted are not for cumulative binomial probability distribution. They are simply Binomial probability tables.. there is a difference.

 

[MrAl]

Hi

It seems there is no way out! Is there? Kindly help me. Thanks.

Regards
PG

I think it would help a lot if we could see the entire chapter that leads up to these questions. Im pretty sure we could provide a more definite answer in that way. is there any way we can see the entire chapter that led up to this part?

 

[misterT]

no way out! Is there? Kindly help me. Thanks.

I plugged that equation in WolframAlpha and it gives the right answer: https://www.wolframalpha.com/input/...*(p^2)*(1-p)^13+++455*(p^3)*(1-p)^12+=+0.0905
(Of course there are 15 possible solutions, but one of them is p=0.40).
It would be a burden to solve that by hand. So, I really think that the problem is supposed to be solved using tables or computer.

Good thing is that all the work you did is correct. The equation you ended up with gives the right solution. I would not like to solve it by hand.. I know there are plenty of tricks you can do, but I have forgotten most of them. If I would try to solve that by hand first I would take logarithm from both sides. That would change the exponentials to multiplications..

 

[MrAl]

I plugged that equation in WolframAlpha and it gives the right answer: https://www.wolframalpha.com/input/...*(p^2)*(1-p)^13+++455*(p^3)*(1-p)^12+=+0.0905
(Of course there are 15 possible solutions, but one of them is p=0.40).
It would be a burden to solve that by hand. So, I really think that the problem is supposed to be solved using tables or computer.

Good thing is that all the work you did is correct. The equation you ended up with gives the right solution. I would not like to solve it by hand.. I know there are plenty of tricks you can do, but I have forgotten most of them. If I would try to solve that by hand first I would take logarithm from both sides. That would change the exponentials to multiplications..

Hi,

Plot the function

 

[misterT]

Hi,

Plot the function

https://www.wolframalpha.com/input/...*(p^2)*(1-p)^13+++455*(p^3)*(1-p)^12+=+0.0905

 

[MrAl]

Hi,

I meant "plot the function" in that plotting the function helps to find the real solutions. Nice to see it there too though

 

[misterT]

I meant "plot the function" in that plotting the function helps to find the real solutions. Nice to see it there too though

It is not easy to plot 15th order function.

 

[MrAl]

It is not easy to plot 15th order function.

Hi,

Yes you are certainly right about that

But i think maybe more important is that the table you provided and the solution you found was for a different function than PG had shown in his original post. His post showed the binomial distribution while your table indicates the cumulative binomial distribution.

As we know from his post the BD is:
[LATEX]
\frac{n!\,{\left( 1-p\right) }^{n-r}\,{p}^{r}}{\left( n-r\right) !\,r!}
[/LATEX]

while the CBD is:
[LATEX]
n!\,\sum_{r=0}^{x}\frac{{\left( 1-p\right) }^{n-r}\,{p}^{r}}{\left( n-r\right) !\,r!}
[/LATEX]

or alternately written the CBD is:
[LATEX]
n!\,\sum_{x=0}^{r}\frac{{\left( 1-p\right) }^{n-x}\,{p}^{x}}{\left( n-x\right) !\,x!}
[/LATEX]

 

[misterT]

Yes, but the solution requires the use of cumulative probability.. I don't understand your point there. Are you saying that I made a mistake or PG made a mistake?
The table I provided (cumulative table) is the one that should be used.. PG posted the wrong table.

 

[MrAl]

Yes, but the solution requires the use of cumulative probability.. I don't understand your point there. Are you saying that I made a mistake or PG made a mistake?
The table I provided (cumulative table) is the one that should be used.. PG posted the wrong table.

Hello there MisterT,

Well originally i had brought PG's posted formula into question (back several posts) so i thought that the table you provided (and so the new formula) was more correct. I suspected all long that the original formula posted was not the right one to use. But it's been so long since i had to do these problems i like to have some way to verify everything.

 

[misterT]

Yeah.. I did not go through all that math. I just think that tables are the way to go and it kind of escalated from there. Good thing that you go through the details.