Re: Probability of precipitation

It's 0.01" (measurable precipitation), not 0.10" (a tenth of an inch).

Also, you can't simply multiple the chances as you have done below. Each
time segment is a unique and independent set. The chance of precipitation
listed in a forecast is typically the highest probability that is found in
any subset of the day. Smart forecasts will even provide detail such as
"Chance of rain 30% this morning, increasing to 70% this afternoon".

"Jack Crane" <jdcrane7@xxxxxxxxx> wrote in message
news:Xns973BBB584E50jdcrane7yahoocom@xxxxxxxxxxxxxx
> What do the probabilities of precipitation mean in weather forecasts? I
> THINK that, say, 60% means the probability of 0.1 inch of precipitation
> is 0.6.
>
> If so, this leads to some fairly obvious contradictions. At
> http://www.wunderground.com/cgi-bin/findweather/getForecast?query=seattle
> right now the probability of precipitation on New Year's Day is 70%. But
> clicking on the "Details" link for that day shows the day broken up into
> 4 6-hour periods, for all of which the probability is also 70%. See
> http://www.wunderground.com/cgi-bin/findweather/getForecast?query=98101
> &hourly=1&yday=1&weekday=Monday
>
> Another example is Friday, Dec. 30. Probability for the day is 90%; the
> details page for Friday shows it broken into 8 3-hour periods, for which
> the probabilities are 70%, 70%, 90%, 90%, 90%, 90%, 80% and 80%,
> respectively.
>
> Now for the first example, the probability of less than 0.1 inch of rain
> during the day would be 30% (1 - 70%). The same for each of the 4 6-hour
> periods. And .3*.3*.3*.3 is 0.0089, the probablity of less than 0.1 inch
> in all of the 4 periods together. Therefore the probablilty that at least
> one of the 4 periods having more than 0.1 inch is 1 - 0.0089 = 0.991. And
> yet the probability of precipitation for the whole day is supposed to be
> only 70%.
>
> For the second example the analagous probabilities are 0.9999 vs. 0.9.
>
> Something seems to be wrong here. Am I correct?


.

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