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Every day before going out Alice looks at the weather forecast to decide whether to bring an umbrella or not. If the forecast says rain, the probability that it really rains that day is $80$%. If instead the forecast states that there will be no rain, the probability that that day will rain is $10$%. During the winter, the weather forecast announces rain with a probability of $70$%, while during the summer this probability is $20$%.

On a winter's day, Alice forgets to look at the weather forecast and finds herself in the rain. What is the probability that the forecast announced rain that day? And how much is the same probability worth if instead it had been a summer day?

Given $A=${the forecast says it will rain}, $B=${it rains that day}, $C=${it's winter}.

So

$P(B|A)=0.8$

$P(B|\overline{A})=0.1$

$P(A|C)=0.7$

$P(A|\overline{C})=0.2$

So the first probability that I have to calculate is

$P(A|(C∩B))=P(A∩B∩C)/P(C∩B)=$

Now I don't know how to calculate $P(A∩B∩C)$ and $P(C∩B)$.

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1 Answer 1

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You don't really need the probability of $C$, because it is given that it is winter. You can think of it this way: On that given winter day, there never was the possibility that it could be not winter. The probability that it is winter is $1$. Hence, you just need the probability $$P(A \vert B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B \vert A) \cdot P(A)}{P(B \vert A) \cdot P(A)+P(B \vert \overline{A}) \cdot P(\overline{A})}.$$ On a winter day, this equals $$P(A \vert B)=\frac{0.8\cdot 0.7}{0.8 \cdot 0.7+0.1 \cdot 0.3}=\frac{56}{59} \approx 94.915 \%.$$ On a summer day, the probability is $$P(A \vert B)=\frac{0.8\cdot 0.2}{0.8 \cdot 0.2+0.1 \cdot 0.8}=66.\bar{6} \%.$$

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  • $\begingroup$ why is P(B)=P(B|A)P(A) + P(B|notA)P(notA) ? $\endgroup$
    – user658422
    Apr 3, 2019 at 19:05
  • $\begingroup$ That's follows from the law of total probability: en.wikipedia.org/wiki/Law_of_total_probability. We either have $A$ or $\overline{A}$. Those two events are obviously disjoint. For both cases, we calculate the probability of $B$. Think of it like a basic probability tree: One branch looks like $-A-B$ and the other branch is $-\overline{A}-B$. $\endgroup$ Apr 3, 2019 at 19:11

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