Every day it rains with probability $ \frac12 $, otherwise we have sunny weather. Weather forecast makes mistake with probability $ \frac13 $ (if Weather forecast says it will be sunny, with a probability $ \frac23 $ so be it.

Professor always takes an umbrella if the weather forecast announces rain. If the weather is sunny announced Professor take an umbrella with probability $ \frac13 $

(1) Calculate the probability that will announce the forecast rain.

My approach is this: There are two possible announcements, or rain, or sun. Therefore, $ \Omega = \{s, r \} $ As both events have equal chance of occurrence, I can use probability Classic. So the answer is $ \frac12 $ I do not know how well, but what could be wrong?

(2) Assuming it rains, calculate the probability that the professor does not have an umbrella. I'll think about space events - because I have to find it. I do not really see how to choose, and I try: It depends on what they said in the forecast. If predicting rain, then a professor took his umbrella. However, if the sun is a professor of predicting the probability of $\frac13 $ took the umbrella.

I mean, we have $\frac23 $ that the forecast was correct and the professor has an umbrella. If the forecast is wrong, it $\frac13 $ probability. Then a professor at $ \frac13 $ took an umbrella. From my deduction that $ \frac13 \cdot \frac13 + \frac23 \cdot1 $ But is it good? If you do not understand what is wrong?

It is very important for me to understand it.


You have the following conditional probabilities: $$\begin{array}{*{20}{l}} {p\left( R \right) = \tfrac{1}{2}}&{{\text{probability of rain}}} \\ {p\left( {r|R} \right) = \tfrac{2}{3}}&{{\text{probability that rain is forecast if rain is coming}}} \\ {p\left( S \right) = \tfrac{1}{2}}&{{\text{probability of sun}}} \\ {p\left( {s|S} \right) = \tfrac{2}{3}}&{{\text{probability that sun is forecast if sun is coming}}} \end{array}$$

From these we can derive the probability that sun is forecast when rain is coming: $$p\left( {s|R} \right) = 1 - p\left( {r|R} \right) = \tfrac{1}{3}$$

  1. From the rules of probability we have $$\begin{array}{*{20}{l}} {p\left( s \right)}& = &{p\left( {s{\text{ and }}S} \right) + p\left( {s{\text{ and }}R} \right)} \\ {}& = &{p\left( {s|S} \right)p\left( S \right) + p\left( {s|R} \right)p\left( R \right)} \\ {}& = &{\tfrac{2}{3} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot \tfrac{1}{2}} \\ {}& = &{\tfrac{1}{2}} \end{array}$$

  2. The professor never leaves his umbrella home when the forecast is rainy, and leaves it at home with probability $\frac{2}{3}$ if the forecast is sunny. Hence the conditional probability $$p\left( {{\text{professor doesn't take umbrella }}|R} \right)$$ is equal to simply $0\cdot p\left( {r|R} \right) + \frac{2}{3}\cdot p\left( {s|R} \right)$. Plugging in the number we calculated above, this yields a probability of $\frac{2}{9}$.

  • $\begingroup$ From what you know that $ p\left( {r|R} \right) = \frac{2}{3} \text{probability that rain is forecast if rain is coming} ??$ From text we know only that $p\left( R|r\right) = \frac{2}{3} $ $\endgroup$ – xaweyNEW Oct 3 '14 at 20:06
  • $\begingroup$ @xaweyNEW: The statement "weather forecast makes mistake with probability $\frac{1}{3}$" implies that whatever the weather turns out to be, the probability that the bureau forecasts the opposite is $\frac{1}{3}$. In other words $p(s|R) = p(r|S) = \frac{1}{3}$. Even if we interpret the statement as $p(R|r) = \frac{2}{3}$, by Bayes' rule we have $p(r|R) = p(R|r)\frac{p(r)}{p(R)}$. If the probability of forecasted error is equal for both sunny and rainy conditions, it is easy to show that $p(r) = p(R)$, hence $p(r|R) = p(R|r)$ and our choice of interpretation doesn't matter. $\endgroup$ – COTO Oct 4 '14 at 3:20
  • $\begingroup$ 2nd part, why it is 1/3 it should be 2/9 - coz 1/3 times the forecast is wrong, i.e. P(s/R) = 1/3 but professor doesn't take his umbrella only 2/3 times hence P(u'/R) = 2/3*1/3 = 2/9?? $\endgroup$ – codeomnitrix Nov 4 '14 at 1:39
  • $\begingroup$ @codeomnitrix: You're right. I overlooked the fact that the professor sometimes takes his umbrella even if rain isn't forecast. I'll fix the answer. $\endgroup$ – COTO Nov 4 '14 at 10:53

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