If you look at this retrospectively, you can use the hypergeometric distribution. Suppose we are given that it rained 7 of the 120 days and that the client wore the hat on 4 of those days. The probability that it rained on 3 of the four hat days would then be
(7C3 x 113C1 )/(120C4) = .00048146 (I think).
whew aCb refers to a things taken b at a time
Here "probability" refers to a repeated experiment in which one picks 4 of the 120 days at random to wear the hat, without knowledge of which of those are the rain days.
Al Feiveson
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Richard Goldstein
Sent: Wednesday, October 28, 2009 7:38 AM
To: statalist
Subject: st: probability question
it's been a long time since I thought about questions like this, but, as
a lead-in to a study, a client has asked the following question which he
thinks he understands and says is related to where he wants to go:
during a consecutive period of 120 days, if it rains on 7 days and my
client wears a hat on 4 days (these are independent of any knowledge of
the weather), what is the probability that it will rain on 3 of the days
on which he is wearing a hat?
my client swears that this is not a homework problem for him or his wife
or one of their kids!
Rich
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
