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It rains in a city with a chance of 0.4. The weather forecast is not always accurate. When there will be a rain the next day, the forecast predicts the rain with probability 0.8; When there is no rain, the forecast falsely predicts a rain with probability 0.1. You take your umbrella every time rain is forecast, and you take your umbrella 25% of the times when rain is not forecast. Fnd the chance that it actually rains given that the forecast predicts rain. Given that it rains, what is the probability that you do not have your umbrella?

Let $R$ be the event about rain tomorrow, $FR$ be the event about weather forecast predicts will rain tomorrow and $U$ about I bring my umbrella.

For the first part, I have $$\mathbb{P}(R\,\vert\,FR)=\frac{\mathbb{P}(R\cap FR)}{\mathbb{P}(FR)}=\frac{\mathbb{P}(FR\,\vert\,R)\mathbb{P}(R)}{\mathbb{P}(FR\,\vert\,R)\mathbb{P}(R)+\mathbb{P}(FR\,\vert\,R^c)\mathbb{P}(R^c)}$$

For the second part, I have $$\mathbb{P}(U^c\,\vert\,R)=\frac{\mathbb{P}(R\,\vert\,U^c)\mathbb{P}(U^c)}{\mathbb{P}(R)}$$ I get stuck at this step, how to calculate $\mathbb{P}(R)$? And do I set up correctly on part two?


EDIT: I got a different answer after brainstorming, $$\mathbb{P}(U^c\,\vert\,R)=\mathbb{P}(FR\,\vert\,R)\mathbb{P}(U^c\,\vert\,FR\cap R)+\mathbb{P}(FR\,\vert\,R^c)\mathbb{P}(U^c\,\vert\,FN\cap R)$$ where $FN$ is the event about forecast predicts not rain tomorrow.

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    $\begingroup$ "It rains in a city with a chance of 0.4." $\endgroup$
    – user223391
    Oct 17, 2016 at 2:14
  • $\begingroup$ @ZacharySelk I miss that. Could you tell me how to find value of numerator? $\endgroup$
    – Simple
    Oct 17, 2016 at 2:57

1 Answer 1

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For the first part you got the formula and you can calculate the result in a straightforward way since you know that $\mathbb P(R)=0.4, \space \mathbb P(FR|R)=0.8, \space \mathbb P(FR|R^c)=0.1$

For the second part, I think it's simpler to look at it as: $$\mathbb P(U^c|R) = 1- \mathbb P(U|R) = 1- \frac{\mathbb P(U \cap R)}{\mathbb P(R)}$$

Then you only need to find $\mathbb P(U \cap R)$.

Expanding $R$ in two disjoint events we get: $R = (R\cap FR) \cup(R\cap FR^c)$ And because they are disjoint the probability becomes: $$\mathbb P(U \cap R) = \mathbb P(U \cap R\cap FR) + \mathbb P(U\cap R\cap FR^c)$$

The first term: $$\mathbb P(U \cap R\cap FR) = \mathbb P(U|FR)\cdot \mathbb P(FR|R)\cdot \mathbb P(R) = 1 \cdot 0.8 \cdot 0.4 = 0.32$$

The second term: $$\mathbb P(U \cap R\cap FR^c) = \mathbb P(U|FR^c)\cdot \mathbb P(FR^c|R)\cdot \mathbb P(R) = 0.25 \cdot 0.2 \cdot 0.4 = 0.02$$

So $\mathbb P(U \cap R) = 0.32 + 0.02 = 0.034$ and thus $\mathbb P(U^c|R) = 1- \frac{0.34}{0.4} = 0.15$

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