
#1
Feb1708, 12:59 PM

P: 190

Say I have the probability of something happening in period of a day. How am I going to tackle finding the probability of it happening in a period 7 days? Statistical independence of events or not.




#2
Feb1708, 04:00 PM

Sci Advisor
P: 5,941

If they are independent it is easy. Let p = prob(event in one day), q=1p. Then the prob of not happening in seven days is q^7, so prob happening at least once in 7 days is 1q^7.




#3
Feb1808, 01:57 AM

P: 190

does that derive from a union of independent events?




#4
Feb1808, 03:21 PM

Sci Advisor
P: 5,941

Probability over a period of time 



#5
Feb1908, 02:41 AM

P: 190

Ok, I think I understand the logic now. ty.




#6
May308, 08:23 AM

P: 190

and (10.9)^3 = 0.01 (only)? (assuming an event of 0.1 (and it not happening of 0.9)) shouldn't they be equivalent? (their addition leading to 1) EDIT: Oh, it may be I guess that the closer to 1 a probability is, the more likely to occur[in subsequent runs] in an exponential fashion, whereas the closer to 0, the less likely in a reversely exponential fashion. 



#7
May308, 03:05 PM

Sci Advisor
P: 5,941

I am not sure what you are trying to do. It looks like you are calculating probabilities for these two events: happening every day or never happening at all. Obviously they don't add up to 1.
Note (10.9)^3=0.001 



#8
May408, 11:57 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

If you have an event that has a probability of 0.1 of "happening" in a given trial, and so a probability of 0.9 of "not happening", then (1 0.1)^3 is the probability it does NOT happen in three consecutive trials. (1 0.9)^3= 0.1^3 is the probability it DOES happen in all three trials. They don't add to 1 because there are many other things that could happen: happen in the first trial but not in either of second or third trials, happen on the first and third trial but not on the second trial, etc. "happen on all trials" and "happen on no trials" do not exhaust all the possible outcomes. 



#9
May908, 01:35 PM

P: 190

I understood what I didn't understand a while after the last post of mine.
Simply: 1(1"probability_of_what_we_try_to_see_IF_IT_WILL_HAPPEN in_n_subsequent runs")^n i.e. if I put 0.01 there I try to see if that will happen in n runs. but if I put its complimentary 0.99, I'll be solving ANOTHER PROBLEM, trying to see if its complimentary will happen in n runs. 


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