Interpretation of an ambiguous conditional probability

Question

This question and official solution come from MIT's 6.431 OCW.

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/assignments/

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I have trouble understanding the solution. Specifically, the solution mentions that $P($Forecast no rain$)=1-p$. This only seems true if we assume that when Victor misses the forecast, it no longer matters.

But how is this so? Surely just because he missed the forecast does not really mean the forecast somehow stops being broadcasted? Shouldn't we consider that no rain is forecast by adding branches on the top (following the one where Victor missed the forecast), to consider different forecast outcomes despite having missed the forecast?

Also, could anyone provide any suggestion on how to avoid these kinds of interpretation problems please? Ever since I have been following this course, I am having a lot of interpretation problems.

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Answer 1

There are four relevant events.

• $C$ carries an umbrella.
• $D$ forecast says "doesn't rain".
• $F$ forecast is seen.
• $H$ the toss shows heads.

The forecast is made whether it is seen or not ($D$ and $F$ are independent), but how the decision to carry an umbrella is made will depend on whether or not the forecast is seen. $$\begin{align}\mathsf P(D)&=1-p\\[2ex]\mathsf P(C)&=\mathsf P(F^\complement)\cdot \mathsf P(C\mid F^\complement)+\mathsf P(F)\cdot \mathsf P(C\mid F)\\[1ex]&=\mathsf P(F^\complement)\cdot \mathsf P(H)+\mathsf P(F)\cdot \mathsf P(D^\complement)\\[1ex]&=0.2\cdot 0.5+0.8\cdot p\\[2ex]\mathsf P(C\mid D)&=\mathsf P(F^\complement)\cdot\mathsf P(C\mid D\cap F^\complement)+\mathsf P(F)\cdot \mathsf P(C\mid D\cap F)\\[1ex]&=\mathsf P(F^\complement)\cdot \mathsf P(H)+\mathsf P(F)\cdot 0\\[1ex]&=0.2\cdot 0.5+0\end{align}$$

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I suppose you could use the following tree.

$$\begin{array}{c}&&&&\nearrow&\mathsf P(C\mid F^\complement,D^\complement)=0.5\\&&&\mathsf P(D^\complement\mid F^\complement)=p\\&&\nearrow&&\searrow&\mathsf P(C^\complement\mid F^\complement,D^\complement)=0.5\\&\mathsf P(F^\complement)=0.2\\&&\searrow&&\nearrow&\mathsf P(C\mid F^\complement,D)=0.5\\\nearrow&&&\mathsf P(D\mid F^\complement)=1-p\\&&&&\searrow&\mathsf P(C^\complement\mid F^\complement,D)=0.5\\\\&&&&\nearrow&\mathsf P(C\mid F,D^\complement)=1\\\searrow&&&\mathsf P(D^\complement\mid F)=p\\&&\nearrow&&\searrow&\mathsf P(C^\complement\mid F,D^\complement)=0\\&\mathsf P(F)=0.8\\&&\searrow&&\nearrow&\mathsf P(C\mid F,D)=0\\&&&\mathsf P(D\mid F)=1-p\\&&&&\searrow&\mathsf P(C^\complement\mid F,D)=1\end{array}$$

Answer 2

When Victor sees the forecast, there is one of two outcomes - either the forecast predicts rain, or it predicts no rain. So if we assume that the probability of the event

$$P(\text{Victor Carries An Umbrella}) = P(\text{missed forecast}).P(\text{umbrella}|\text{missed forecast}) + P(\text{saw forecast}).P(\text{umbrella}|\text{saw forecast}) $$

Now, the author assumes the probability of carrying an umbrella after seeing the forecast is $p$ (which is a seasonal parameter). Hence the probability of not carrying an umbrella after seeing the forecast is $1-p$ as the two events are mutually exclusive and complimentary, you cannot have any other outcomes after seeing the forecast.

He then goes on to show that the for the possible values of $p$, the events are not independent

Answer 3

First question asks about probability of Victor carrying an umbrella. Let us see when he carries one.

1. Half the times when he misses the forecast.
2. p fraction of the times when he sees the forecast because p fraction of the times depending on the seasons, it says it will rain. Let us see if these two can happen simultaneously. They can't because he can't see and not see the forecast. Did we miss any case? In our missed case he again has to see or not see the forecast and act based on it and we included all such cases. So we can just add these two probabilities to get the total probability by finite additivity. And each can be computed by using conditional probability easily.

The author assumes that p is the probability of forecast being "rain rain rain omg!" on that given day whether Victor sees it or not. The value of p can only be 0.2 or 0.7 depending on which season that particular day belonged to. So think about it more, you will get answer about the first part. To complete the picture imagine the hidden two branches with values p and 1-p of forecast being rain and not rain on branch of he missing the forecast. They are not shown because they are not relevant here. And if you want to that forecast of rain is still p, we can verify. It will be probability of seeing the forecast * p + probability of not seeing the forecast *p =p, which is obvious.

Now the question about how to avoid reasoning errors in such questions! The trick is becoming clear about what is the event we are calculating probability for and explicitly tracing only that event and ignoring the bigger problems because too many sentences, if you interpret all simultaneously, will clearly mess up your train of thought.