# Carying an umbrella given rain forecast

We know it rains half of the time.

We also know that the weather forecaster is correct 2/3 of the time. I.e. the probability that it rains given they had predicted rain is the same as if it does not rain given they predicted it would not rain and is 2/3.

We know a man takes his umbrella when rain is in the forecast and takes his umbrella with probability 1/3 when rain is not in the forecast.

I have read other threads on here with this problem and I have two answers which i think are correct; however, the answer key in my book only agrees with one answer. The question asks us to find the probability that: a) the man has no umbrella given that it rains. Then b) the probability that he brings his umbrella given it does not rain.

This is how I attempted to solve this.

Let A be the event that the man has his umbrella and r be the event that it rains.

first we want the probability the man has no umbrella given that it rains

$$P(A^c|r)=\frac{P(A^c \cap r)}{P(r)}=\frac{\frac{1}{2}\frac{2}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$$

This answer matches the one in my answer key. The next one is the one that is off. My answer key says it should be $\frac{5}{12}$ which we will see is not what i get.

the probability that he brings his umbrella given it does not rain.

$$P(A|r^c)=\frac{P(A \cap r^c)}{P(r^c)}=\frac{\frac{1}{2}(1)\frac{1}{3}+\frac{1}{2}\frac{1}{3}\frac{2}{3}}{\frac{1}{2}} = \frac{5}{9}$$

Have I done something wrong? I am beginning to think the answer key is off.

• But the construction of your tree is not complete. How do you know that the probability of a rain forecast is 1/2? We are told is that the probability of rain is 1/2, not of the forecast. It's true, but it needs to be shown/derived. Sep 29, 2017 at 3:32
• @Mathemagical I Guess I did miss identify that part. I assume the probability come frrom $\frac{\frac{1}{3}}{\frac{2}{3}}$ but I do not see where this comes from. I assume the probability of rain given the forecast and the straight probability of rain. Sep 30, 2017 at 16:02
• I figured it out i think. The probability of a forecast for rain is $\frac{1}{2}$. I made a table of the probabilities that it rains and rain was forecast. Then i added the probabilities that rain was forecast divided by the total which comes out to $\frac{1}{2}$ Sep 30, 2017 at 16:13
• Great. I'm glad. Ha, it look me longer to typeset that answer for you. Do you still get your old answer? Sep 30, 2017 at 16:41

You have most ideas right. This is just a correction because the root of the tree has not been given (unconditional probability of a rain forecast). But posting it here because it's too large for the comments area. Let p be the unconditional probability of a forecast of rain(F). In 2/3 of those cases it rains(R). Denote the events no-rain and forecast of no-rain by R' and F' respectively. Since you know $P(R|F)=2/3$ and $P(R|F')=2/3$, you can solve for $p$ because $P(R)$ is given to be 1/2. $$P(R)=P(F)P(R|F) + P(F')P(R|F')=p\frac{2}{3}+(1-p)\frac{1}{3}=\frac{1}{2}$$ from which we know $p=\frac{1}{2}$.
From here, you can also calculate $P(F|R')=\frac{1}{3}$ and $P(F'|R')=\frac{2}{3}$.
$$P(U|R')=P(U|F')P(F'|R') + P(U|F)P(F|R')=\frac{1}{3}\times\frac{2}{3}+1 \times \frac{1}{3}=\frac{5}{9}$$