|
|Store|
|
|
|
|
|
|
|
|
CTK Exchange
George Arthur
guest
|
Mar-14-03, 10:37 AM (EST) |
|
|
"Simple(?) rain probability."
| |
If there is a 20% chance of rain today and a 20% chance of rain tomorrow, what is the probability that it will rain on at least one of those days?
Explain logic. |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
Graham C
Member since Feb-5-03
|
Mar-20-03, 03:06 PM (EST) |
    |
1. "RE: Simple(?) rain probability."
In response to message #0
| |
>If there is a 20% chance of rain today and a 20% chance of
>rain tomorrow, what is the probability that it will rain on
>at least one of those days?
>Explain logic. In case I'm helping someone who shouldn't be asking for it, here's a hint. Think about the probabilities that it won't rain.
|
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
 |
Rohan Dawate
guest
|
Apr-30-03, 01:30 PM (EST) |
|
|
13. "RE: Simple(?) rain probability."
In response to message #1
| |
Since it has been given that there is likely an occurence of 20% rain on 2 days , the probability of rain could also be got in the following manner. P(rain on 1st day) + P(rain on 2nd day) - P(rain on either days) = 0.20 + 0.20 - (0.20 * 0.20) = 0.36 i.e. P(A union B) = P(A) + P(B) - P(A intersection B) |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
Ben
guest
|
Mar-21-03, 11:39 AM (EST) |
|
|
2. "RE: Simple(?) rain probability."
In response to message #0
| |
I think Graham is being a little cryptic there. What he is saying is based on the logic that that given some event A the the propability that is happens or doesn't happen is connected by the logic <Probability that A Happens> = 1 - <Probability that A Doesn't Happen> This is true regardless of what a is so that <Probability that it rains on at least 1 day> = 1 - <Probability that it doesn't rain on either day> Alternitively you can consider all possible combinations of results and then add those combinations that match you criteria. Using P(A) to mean the probability that A happens then if you have 2 unrelated events A and B you have up to 4 possible outcomings of the events A and B
A and Not B
Not A and B
Not A and Not B Each of those outcomings has an associated probablity given by P(A and B) = P(A) x P(B)
P(A and Not B) = P(A) x P(Not B)
P(Not A and B) = P(Not A) x P(B)
P(Not A and Not B) = P(Not A) x P(Not B) If you want to know the probablity that a set of those possible outcomings occurs then you add them together. That is if I want the probability that A happens but Not B or that B happens but Not A but that the don't both happen (exclusive or (xor)) I can get that result by P(A xor B) = P(A and Not B) + P(Not A and B)
|
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
|
|
|
|
|
|
SteveSchaefer
Member since Apr-2-02
|
Mar-31-03, 10:32 AM (EST) |
|
|
6. "RE: Simple(?) rain probability."
In response to message #5
| |
OK, but a simple answer will not do. This is the sort of question that should be obvious, or you should work through it until it is obvious. There are two possibilities regarding rain on the first day. Either it rains (p = 0.20) or it doesn't (1 - p = 0.80). These two events can be designated as P(A) and P(not A). On the second day there are also two possibilities. The probability that it rains is P(B) = p and the probability that it doesn't rain is P(not B) = 1 - p. Now we have to consider the combined probabilities. It may rain on both days: P(A and B) = P(A)*P(B) = p*p. It may rain on the first day, but not on the second day: P(A and not B) = P(A)*P(not B) = p(1 - p). It may not rain on the first day, but rain on the second day: P(B not A) = P(not A)*P(B) = (1 - p)p. Finally, it may rain on neither day: P(not B and not A) = P(not A)*P(not B) = (1 - p)(1 - p). If we want to know whether it will rain on at least one day, then we want the sum of the first three probabilit's. Alternatively, since all four probabilities add up to * = 1*1 = 1, we want everything except the last possibility: P(A or B) = 1 - P(not B and not A) = 1 - 0.8*0.8 = 0.36 = 36%.
|
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
|
JACKAL
guest
|
Apr-30-03, 01:30 PM (EST) |
|
|
12. "RE: Simple(?) rain probability."
In response to message #6
| |
I completley agree with your logic and your method only further proves your intelligence in this matter. Thankyou for putting an end to this problem.
I also wish to comment one other thing; The Matrix Reloaded is coming out on May 16th and I cannot wait to veiw this in the cinema with all its splendor!!! |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
|
Bobby
guest
|
Apr-24-03, 09:00 PM (EST) |
|
|
8. "RE: Simple(?) rain probability."
In response to message #5
| |
I think I can provide a simple enough answer...
Chances of rain on Monday: 2 out of 10 chances (20%)
Chances of rain on Tuesday: 2 out of 10 chances (20%)
Thus, over the two days, there will be a total of 20 chances,
of which, there will be 4 chances of rain.
4/20 = 2/10 = 20%-Bobby
bobbysim@earthlink.net |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
|
RicBrad
Member since Nov-16-01
|
Apr-25-03, 04:21 AM (EST) |
|
|
9. "RE: Simple(?) rain probability."
In response to message #8
| |
>Thus, over the two days, there will be a total of 20 chances,
>of which, there will be 4 chances of rain.
>4/20 = 2/10 = 20% Unfortunately things are not quite this simple. What if we had 3 days? By your argument the chance of rain would be 6/30 = 20%. What about n days? Summing the "chances", prob. = 2*n / 10*n = 20%. So if this argument worked, it would be equally likely to not rain tomorrow as it would be to never rain again, for all of eternity. I hope you can agree that this is not the case. Read some of the above posts to see how to work out the true probabaility. Rich |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
Paul Austin
guest
|
Apr-25-03, 03:40 PM (EST) |
|
|
10. "RE: Simple(?) rain probability."
In response to message #0
| |
Your saying that there is going to be a 20% chance on each that it will rain so the probability that it will rain on one of those days is 20%. |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
lokesh gupta
guest
|
Nov-20-04, 10:38 AM (EST) |
|
|
14. "RE: Simple(?) rain probability."
In response to message #0
| |
if there is a 20% probability of raining on each day then the probability of raining on one of the two days will be more than 20%. P(today) + P(tommorrow) = 20+20= 40% |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
Me
guest
|
Nov-22-04, 05:47 PM (EST) |
|
|
16. "RE: Simple(?) rain probability."
In response to message #15
| |
Yes, so that would mean it would defineately would rain. That would be wrong. Origanator and the other person are both right, but they just used different formulae. Any other formulae? |
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
|
You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.
|Front page|
|Contents|
Copyright © 1996-2018 Alexander Bogomolny
71544993
|
|
Printer-friendly copy
Email this topic to a friend
