# Calculation of probability of event intersection

This is a question from MIT 6.041 open courseware.

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?

(b) The probability of Victor missing the morning forecast is equal to 0.2 on any day in the year. If he misses the forecast, Victor will flip a fair coin to decide whether to carry an umbrella. On any day of a given season he sees the forecast, if it says “rain” he will always carry an umbrella, and if it says “no rain,” he will not carry an umbrella. Are the events “Victor is carrying an umbrella,” and “The forecast is no rain” independent? Does your answer depend on the season??

(c) Victor is carrying an umbrella and it is not raining. What is the probability that he saw the forecast? Does it depend on the season?

• Let C be the event Victor is carrying the umbrella
• Let R be the event It is not raining
• Let F be the event Victor saw the forecast

So to get the answer, we can use the formula: $$\mathbb{P}(F \mid C \cap R) = \frac{\mathbb{P}(F \cap C \cap R)}{\mathbb{P}(C \cap R)} = \frac{\mathbb{P}(F)\mathbb{P}(C \mid F)\mathbb{P}(R \mid C \cap F)}{\mathbb{P}(C \cap R)}$$

The numerator in the above equation is easy to calculate as the values are all known if we draw the sequential tree model (can be found here). It's equal to (0.8)p(0.2).

I am however not able to work out how to calculate $\mathbb{P}(C \cap R)$. The solution in the link above directly works out the value using the tree (which is simply sum of probability of carrying the umbrella and no rain for the initial condition on forecast being observed), but isn't there a more analytical approach? Can a solution be reached by using just the various formulas?

(edit: added the missing part of question)

• Perhaps I don't understand the question. Surely the fellow only picks up his umbrella if it is raining or if it is not raining but he heard a forecast saying that it will rain. As the first case is ruled out, we must be in the second...hence the answer is $100\%$ regardless of the season. Or have I misunderstood?
– lulu
Jun 19, 2016 at 21:13
• Just to emphasize: the problem says nothing about the circumstances which induce our man to pick up his umbrella. Perhaps he picks it up everyday, in which case we get no information from seeing him with it. I think my assumption (from the first comment) is the most logical reading, but one or way or another some assumption must get made.
– lulu
Jun 19, 2016 at 21:35
• @lulu sorry about the confusion, I did not include the question in its entirety. I have updated the question. Thanks. Jun 20, 2016 at 4:21
• Did you have your answer? Feb 2, 2020 at 6:54

## 1 Answer

The analytical approach is to carve up the event $C\cap R$ into two disjoint pieces, according to whether Victor checked the forecast: $$C\cap R = (F\cap C\cap R) \cup (F^c\cap C\cap R)$$ and therefore $$P(C\cap R) = P(F\cap C\cap R) +P (F^c\cap C\cap R).$$ Both of the probabilities on the RHS can be read off the sequential tree, which is just a tidy way to organize the events and their conditional probabilities. In fact you already computed the first of these in the numerator of your final expression; the second is computed in an analogous way.

• How do you explain where this comes from? P(actually rains | missed forecast) = (0.8)p + (0.1)(1 − p) = 0.1 + 0.7p Feb 2, 2020 at 8:43
• @J.Doe It is the unconditional probability that it rains. It is computed as P(rain) = P(rain | rain forecasted)P(rain forecasted) + P(rain | rain not forecasted)P(rain not forecasted). There's really no reason to condition on missed forecast, since the event of rain doesn't care (or know) whether Victor missed the forecast. Feb 3, 2020 at 18:15
• Thank you so much!!! Feb 4, 2020 at 5:04