[Math] Calculation of probability of event intersection

probability

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)

Best 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.

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