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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 --- Message posted from http://www.ExcelForum.com/ |
#2
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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. --- Message posted from http://www.ExcelForum.com/ |
#3
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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. --- Message posted from http://www.ExcelForum.com/ |
#4
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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! --- Message posted from http://www.ExcelForum.com/ |
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