Any math geeks?

[Man O' Vegas]

I'm working on a new stat model, just want to make sure in this part my math is right.

A= 780 situations
B= 1560 choices (2 sides per sittuation)
C= 199 times certain result occured for 1 choice in situation
D= % of occurence for C

My dilema is; would D be a result of C being divided by A or B?

Would appreciate any input.

 

[LetsMakeMoney]

:mj03: :s2: :look: :nooo: :shrug:

 

[bubbas1]

I would say divide by B.

Disregard..I reread it and now I have no clue. I should have known better than to try to think so early in the morning.

 

[eric stratton]

It's A.

 

[The Judge]

If I am reading your question correctly, D = Result occurred 25.5% of the time for one of the choices in each situation or 12.76% of the time for all choices.

 

[CANADA MAN]

You divide c into a which means your "certain result" happened 25.5% of the time.


:canada1

 

[PharoahUB]

I agree. A

 

[buddy]

We need a visit from Koty's dad for this thread. For those that don't know him.........Oh, man......

 

[THE HITMAN]

A fer shur

 

[KotysDad]

I agree with A also if I am reading the problem correctly. Man, tell me if this is what you're getting at.

A = 780 situation ( say 780 coins)
B = 1560 choices (head or tails for each coin)

The way I understand C is something similar to you observing say 199 heads (and the rest are tails). I have to assume you cant have both choices ever occur for any situation???

If thats the case, then D = C/A

 

[buddy]

Man O' Vegas,

You've been delivered.

 

[Man O' Vegas]

Thanks for the input fellas, makes my job a little easier. And now for the other 20 derivatives..... :scared