# The probability it's warm when it's sunny is 0.7. Is this a conditional probability?

The related question is the following: "Suppose the probability of snow is 0.2, of rain is 0.5 and of sun is 0.3. Suppose that it’s never warm when it snows, it’s warm with probability 0.3 when it rains and 0.7 when it is sunny. What’s the probability that it is warm? What’s the probability of it being warm given that it’s not sunny?"

I'm not sure whether I should interpret the question as saying P(Warm and Sunny)=0.7 or P(Warm|Sunny)=0.7 (and similarly for the other 'when' probabilities). I'm leaning towards the idea that it is conditional probability but how do I calculate the probabilities in question? I was looking at Bayes theorem, the law of total probability and also the formula $P(A|B)=P(A \cap B)/P(B)$ but doesn't look like these apply here so not sure what to use. It's probably something really basic I'm missing here but any help would be appreciated.

• Considering that $(\text{Warm}\cap \text{Sunny})\subseteq \text{Sunny}$ that would imply $Pr(\text{Warm}\cap \text{Sunny})\leq Pr(\text{Sunny})$. Now... does it make sense for $Pr(\text{Warm}\cap \text{Sunny})=0.7$ while we were told $Pr(\text{Sunny})=0.3$? – JMoravitz Mar 24 '17 at 21:35
• @JMoravitz Oh of course, thanks a lot! I completely overlooked that. So knowing it's a conditional probability, where do we go from there to find $P(\text{Warm})$? I was looking at using $P(\text{Warm and Sunny})=P(\text{Warm}|\text{Sunny})\times P(\text{Sunny})$ implying $P(\text{Warm and Sunny})=0.7\times 0.3-0.21$, but I don't think that actually helps. – Alex.F Mar 24 '17 at 21:46

## 1 Answer

Let snow, rain, sun, warm be represented by the events $I,R,S,W$ respectively (I chosen to represent ice instead to avoid overlap of letters). We further make the assumption that exactly one of the three precipitation conditions will be in effect at any given time... that is it is either snowing, it is raining, or it is sunny and there is no overlap between these three.

$0.7$ is indeed referring to the conditional probability $Pr(W\mid S)$

To continue, all of the mentioned tools in your post are indeed what you should be looking to use.

$Pr(W)=^\dagger Pr(W\cap(I\cup R\cup S))=^*Pr(W\cap I)+Pr(W\cap R)+Pr(W\cap S)$

The step labelled with a $\dagger$ is valid since $I\cup R\cup S$ forms the entire sample space and is guaranteed to happen so $W=W\cap \Omega = W\cap (I\cup R\cup S)$

The step labeled $*$ is valid because $W\cap (I\cup R\cup S) = (W\cap I)\cup (W\cap R)\cup (W\cap S)$ and each of $(W\cap I),(W\cap R),(W\cap S)$ are (assumed to be) pairwise mutually exclusive since it cannot simultaneously snow and rain just as it cannot simultaneously snow and be sunny, etc... regardless of whether or not it is warm at the same time.

Now... using the multiplication principle which is just a rearrangement of the definition for conditional probability: $Pr(A\cap B)=Pr(A)Pr(B\mid A)=Pr(B)Pr(A\mid B)$ we continue our manipulations giving:

$Pr(W\cap I)+Pr(W\cap R)+Pr(W\cap S) = Pr(I)Pr(W\mid I) + Pr(R)Pr(W\mid R) + Pr(S)Pr(W\mid S)$

Each of the numbers in the above expression on the right were given to you in the problem statement allowing you to just plug and chug at this point to get the answer for the probability that it is warm.

For the next part of the problem, note:

$Pr(A)=Pr(A\cap (B\cup B^c)) = Pr(A\cap B)+Pr(A\cap B^c)$, using the same observations as made at the beginning of the previous part.

This implies in particular that

$Pr(W)=Pr(W\cap S)+Pr(W\cap S^c) = Pr(S)Pr(W\mid S) + Pr(S^c)\underbrace{Pr(W\mid S^c)}_{\text{what we want to find}}$

We know all of the above numbers from the problem statement or the work in the previous step of the problem except for $Pr(W\mid S^c)$ which is what we are wishing to find. Rearranging terms appropriately using some highschool algebra and plugging in the necessary numbers will give us our answer.

• Aha! That makes perfect sense. Thank you kind sir. – Alex.F Mar 24 '17 at 22:31