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.
Best Answer
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: $$p\left( {s|R} \right) = 1 - p\left( {r|R} \right) = \tfrac{1}{3}$$
From the rules of probability we have $$\begin{array}{*{20}{l}} {p\left( s \right)}& = &{p\left( {s{\text{ and }}S} \right) + p\left( {s{\text{ and }}R} \right)} \\ {}& = &{p\left( {s|S} \right)p\left( S \right) + p\left( {s|R} \right)p\left( R \right)} \\ {}& = &{\tfrac{2}{3} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot \tfrac{1}{2}} \\ {}& = &{\tfrac{1}{2}} \end{array}$$
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 $\frac{2}{9}$.