Probability of Rain Question

Question

This is the question: Most mornings, Victor checks the weather report before deciding whether to carry an umbrella. If the forecast is “rain,” the probability of actually having rain that day is 80%. On the other hand, if the forecast is “no rain,” the probability of it actually raining is equal to 10%. During fall and winter the forecast is “rain” 70% of the time and during summer and spring it is 20%.

(a) One day, Victor missed the forecast and it rained. What is the probability that the forecast was “rain” if it was during the winter? What is the probability that the forecast was “rain” if it was during the summer?

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I've figured out the numerator for both winter and summer but I'm having issues coming up with the denominator. Any help is appreciated.

So far I have: Let A be the event that the forecast was rain. Let B be the event that it rained. Let p be the probability that the forecast says rain. So,

P(A|B) = P(B|A)P(A)

Answer 1

It becomes easier if you name the events "mnemonically" and use a simpler formula, eg

$R$ = event it Rains,
$F$ = event Forecast was rain

$P(R|F) = \dfrac{P(R\cap F)}{P(R\cap F)+P(R^c\cap F)}$

Answer 2

Let's do it for the winter. Using your notation: let $A$ be forecast rain, and $B$ is actual rain. During the winter $70\%$ of the forecasts are rain, thus $P(A) = 0.7$ and the probability that the forecast is not rain is $P(\bar A) = 1- 0.7 = 0.3$, hence using the Bayes rule \begin{align} P(A|B) &= \frac{P(B|A)P(A)}{P(B)}\\ & = \frac{ 0.8\times 0.7 }{ P(B|A)P(A) + P(B|\bar A) P( \bar A ) }\\ & = \frac{ 0.8\times 0.7 }{ 0.8\times 0.7 + 0.1\times 0.3 }, \end{align}

you can do the same for the summer by replacing $0.7$ by $0.2$ and $0.3$ by $0.8$.