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There are two weather stations, station A and station B which are independent of each other. On average, the weather forecast accuracy of station A is $80\%$ and that of station B is $90\%$. Station A predicts that tomorrow will be sunny, whereas station B predicts rain. What is the probability that it rains tomorrow? We are not asking for the exact probability; we are just asking whether it is more likely to rain or not.

OK I suppose we must examine the following $4$ cases:

a) A and B make the same forecast and both are right

b) A and B make the same forecast and both are wrong

c) A and B make different forecasts and A is right

d) A and B make different forecasts and B is right

and of course we are in one of the cases c) or d), since we know they make different forecasts .

Samuel
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  • Sorry guys, just realized that I had missed a word!! Weather forecast ACCURACY!! – Samuel Apr 15 '16 at 20:08
  • The question cannot be answered because we do not know the a priori probability of rain. The answer will be very different in a place where rain is common compared to a place where rain is rare. This question (and its answers) can help you understand more: https://math.stackexchange.com/questions/2055535/cat-dead-or-alive – Thanassis Jan 13 '20 at 12:41

2 Answers2

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It is not clear what you mean by independent concerning A and B. It is clear that the forecast of A and B should not be independent as they are both linked to the actual weather of tomorrow. What I understand is that the event (A fails) is not linked to the event (B fails). Furthermore I interpret your percentage as the probability of forecasting the weather for instance P(Forecast=sunny | weather=sunny) which is different from P(weather=sunny|Forecast=sunny)

You need to know the marginal probability P(W) of the weather (sunny/rain) tomorrow. let $f_a$ and $f_b$ stand for the forecast of A and B You want to know So $P(W=rain|f_a,f_b)$.

You have that :

$P(W|f_a,f_b) \sim P(f_a,f_b|W)P(W)$ (Bayes theorem)

As the error probability of A and B are independent :

$P(f_a,f_b | W=rain)=P(f_a|W=rain)P(f_b|W=rain)=0.20*0.90 = 0.18$ $P(f_a,f_b | W=sunny)=P(f_a|W=sunny)P(f_b|W=sunny)=0.80*0.10 = 0.08$

So $P(W=rain|f_a,f_b)=\frac{0.18 P(W=rain)}{0.18 P(W=rain)+0.08 P(W=sunny)}$

Arnaud Mégret
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  • thx . I have corrected – Arnaud Mégret Apr 14 '16 at 17:14
  • Independent of each other, in the sense that each station's forecast does not depend on the other station's. – Samuel Apr 14 '16 at 18:55
  • Do you rather mean that a station forecast is independent of the other GIVEN THE REAL WEATHER ? Otherwise, complete independence looks strange as one station says that it will rain tomorrow, then you increase the chance that it will really rain tomorrow and, by that mean, the chance that the second station will forecast rain too. So, the two forecasts should not be independent. – Arnaud Mégret Apr 15 '16 at 09:00
  • The chance that it will rain tomorrow does not depend on any forecast. Station A gives a forecast without knowing anything about Station B forecast and vice versa. In this sense they are independent. – Samuel Apr 15 '16 at 10:50
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Denote by $r$ the a-priori probability of rain. Then the probability $P[A:S, B:R]$ of the forecasts A: sunny and B: rain amounts to $r\cdot0.2\cdot 0.9+(1-r)\cdot0.8\cdot 0.1$, and the probability that it actually rains under these premises is the first part of this sum. Therefore we obtain $$P[R\>| A:S, B:R]={0.18 r\over 0.1 r+0.08}\ .$$

Christian Blatter
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  • What is the a-priori probability of rain?? – Samuel Apr 14 '16 at 18:46
  • The a-priori probability of rain is the assumed probability of rain before any forecasts were anounced. – browngreen Apr 14 '16 at 22:02
  • So, if r is unknown, how do we calculate P? Or at least, how can we have an indication of whether it will rain tomorrow? – Samuel Apr 18 '16 at 18:23
  • @Samuel The value of $r$ is not given in the question, so there is no way to calculate $P$. Sometimes questions are poorly worded or do not make sense; it's better to leave the question alone. – Toby Mak Mar 13 '18 at 00:11