I tried to teach myself some probability and I think that I still don't understand the difference between $\mathbb{P}(A | B)$ and $\mathbb{P}( A \cap B)$.
Here's a small problem I gave a try :
In London it rains on average 1 day over 2 hence the weather station forecast a rainy day half of the time. They're 2 chances over 3 that the forecasts are correct. When a rainy day is forecasted, Mr. P. takes his umbrella with a probability of 1 and when a dry day is forecasted Mr. P. takes is umbrella with a probability of $\frac{1}{3}$. Let $R$ be the event "it is a rainy day", let $W$ be the event "rain was forecasted" and let $U$ be the event "Mr. P. takes his umbrella".
My guess is that we know that $\mathbb{P}(W)= \frac{1}{2}=\mathbb{P}(\overline{W})$ but I am unsure if the sentence "the probability of a correct forecast is $\frac{2}{3}$" should be interpreted as $\mathbb{P}(W | R)= \frac{2}{3}$ or $\mathbb{P}( W \cap R)= \frac{2}{3}$. If I would make a probability tree, I would start by drawing a branch pointing to $W$ with probability $\frac{1}{2}$ but then what probability should I write over the second branch starting from $W$ and pointing towards $R$ ? Again, is this probability $\mathbb{P}(W | R)$ or $\mathbb{P}( W \cap R)$ ?
$\cdot$ $\overrightarrow{\frac{1}{2}}$ $W$ $\overrightarrow{?}$ $R$
I tried to figure out the probability that Mr. P. takes his umbrella using the law of total probabilities as follows :
$$ \mathbb{P}(U)=\mathbb{P}(U | W)\mathbb{P}(W) + \mathbb{P}(U | \overline{W})\mathbb{P}(\overline{W})=1 \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2}=\frac{2}{3}. $$
Is this correct ?
The second question is to evaluate $\mathbb{P}(\overline{U}|R)$. I first tried to use the definition of the conditional probability :
$$ \mathbb{P}(\overline{U}|R)=\frac{\mathbb{P}(\overline{U}\cap R)}{\mathbb{R}}. $$
But I had no clue on how to find out the value of $\mathbb{P}(\overline{U}\cap R)$ so I decided to try with Baye's theorem :
$$ \mathbb{P}(\overline{U}|R)=\frac{\mathbb{P}(R|\overline{U})\mathbb{P}(\overline{U})}{\mathbb{P}(R)}. $$
We know that $\mathbb{P}(R)=\frac{1}{2}$ and using the result above we get $\mathbb{P}(\overline{U})=1-\mathbb{P}(U)=1-\frac{2}{3}=\frac{1}{3}$. Now, $\mathbb{P}(R|\overline{U})$ hast to be $\frac{1}{3}$ because, if we know that Mr. P. didn't take his umbrella it's because a dry day was forecasted hence it means that the forecast was incorrect from which I deduce $\mathbb{P}(R|\overline{U})=1-\mathbb{P}(\text{correct forecast})=1-\frac{2}{3}=\frac{1}{3}$.
So I found :
$$ \mathbb{P}(\overline{U}|R)=\frac{\frac{1}{3} \cdot \frac{1}{3}}{\frac{1}{2}}=\frac{2}{9}. $$
Is it correct ? The part where I find the value of $\mathbb{P}(R|\overline{U})$ seems to be a bit "hand waving" I guess.
The last question I have is how to determine the value of $\mathbb{P}(\overline{R}|U)$.
Thank you for your help.
Best Answer
It’s vital to note the dependence of the events. $U$ depends only on $W$, which further depends only on $R$.
It should be interpreted as $P(W | R)=\frac 23$, or equivalently as $P(\overline W|\overline R)=\frac 23$.
Also, you write $P(W)=\frac 12 =P(\overline W)$ which is incorrect. $W$ should be replaced by $R$. So,
$$P(U) = P(W) \cdot 1 + P(\overline W) \cdot \frac 13 \\ $$
Now, $P(W) = P(R)\cdot P(W|R) + P(\overline R) \cdot P(W|\overline R) = \frac 12 \cdot \frac 23 + \frac 12\cdot \frac 13=\frac 12$
It just happens to also equal $\frac 12$ from which we deduce that $P(U)=\frac 23$.
Now,
$P(\overline U|R)=P(W|R)\cdot P(\overline U|W) + P(\overline W|R) \cdot P(\overline U|\overline W) \hspace{1 cm}\text{(using the law of total probability)}$
$=\frac 23\cdot 0 + \frac 13\cdot \frac 23=\frac 29$
Your method is correct too. Lastly,
$$P(\overline R|U) = P(U|\overline R) \cdot \frac{P(\overline R)}{P(U)} \\ =\left[P(W|\overline R)\cdot P(U|W) + P(\overline W|\overline R)\cdot P(U|\overline W)\right]\cdot\frac{\frac 12}{\frac 23} $$
Can you take it from here?