Stats Question
 

Not sure if this an excel or VBA question.

I need a function that assesess the following:

On any day an event has a 3% probabilty of ocurring. I want to
assess, over the period of 1-year, what the probability is that the
event occurent. That is, after 365 trials, what is the prob that the
event occured?

I know its binomial but can't think of what funciton to use


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Stats Question
 

Just answered my own question:

prob of occuring in any given day= 3%
# of trials = 365 (1 year)
number of events occured = 0

BINOMDIST(0,365,0.03,TRUE) = .0015%

There is a .0015% chance of this event not occuring over a 365 day
period.


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Stats Question
 

If there is a 3% chance of event each day, there is 97% chance of it not
occurring in a day. So the chance of it not occurring in a year is
0.97^365=0.0000015 (or 0.0015%). No need for BIONOMIST here.

What is the probability in a room of 25 people that any two have a birthday
on the same day?
--
Bernard Liengme
www.stfx.ca/people/bliengme
remove CAPS in e-mail address


"ExcelMonkey " wrote in message
...
Just answered my own question:

prob of occuring in any given day= 3%
# of trials = 365 (1 year)
number of events occured = 0

BINOMDIST(0,365,0.03,TRUE) = .0015%

There is a .0015% chance of this event not occuring over a 365 day
period.


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Message posted from http://www.ExcelForum.com/

Stats Question
 

The odds that one person has a birthday in the year is 365/365 = 100%.
The odds that a second person does not have a birthday on the same day
as the first (in a party of two) is 364/365.

365/365*364/365=99.72% prob that they don't have same birthday.
1-99.72% = .18% that they do have same birthday

P(n,m) = 1-(n*[n-1][n-2]...[n-m+1])/m^n
Probability that N people have same birthday over m periods

Thus for a party of 25:
1-((365*364*363*362*......341)/(365^25))=56.87%

In a room of 25 people, there is a 57% prob that two of the people have
birthdays that fall on the same day.

Shocking!


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