Seattle Rain & Three Friends

ADVANCED

In Seattle, it rains on 60% of days. If it is raining, 80% of people you see will be carrying an umbrella. If it is not raining, 20% of people you see will still be carrying an umbrella (perhaps they are cautious or forgot to check the weather).

You look outside and see three of your friends, and all three of them are carrying umbrellas. Assuming their decisions to carry umbrellas are independent of each other given the weather, what is the probability that it is actually raining in Seattle?

Explanation: Seattle Rain & Three Friends

Let's break this down. We want to know the chance it's raining given that we see three friends all with umbrellas.

This is like updating our initial guess about rain based on new evidence (the umbrellas).

Initial Guess (Prior): We know it rains 60% of the time in Seattle. So, P(Rain) = 0.6.
Evidence: All three friends have umbrellas. Let's call this "E".
How likely is this evidence if it IS raining? If it's raining, one friend has an umbrella 80% of the time (0.8). For all three (assuming they decide independently based on the rain): 0.8 × 0.8 × 0.8 = 0.512. So, P(E | Rain) = 0.512.
How likely is this evidence if it's NOT raining? If it's not raining, one friend has an umbrella 20% of the time (0.2). For all three: 0.2 × 0.2 × 0.2 = 0.008. So, P(E | No Rain) = 0.008.
Bayes' Theorem helps us combine all this to find P(Rain | E), our updated belief. We also need the overall chance of seeing three umbrellas, P(E), which considers both rainy and non-rainy scenarios.

We will use Bayes' Theorem to solve this. Let the events be:

R: It is raining.
NR: It is not raining (complement of R).
U₁: Friend 1 has an umbrella.
U₂: Friend 2 has an umbrella.
U₃: Friend 3 has an umbrella.
E: All three friends have umbrellas (i.e., U₁ and U₂ and U₃).

We want to find P(R | E).

1. Identify Given Probabilities (Priors and Likelihoods for one person)

Prior probability of rain: P(R) = 0.60
Prior probability of no rain: P(NR) = 1 - P(R) = 1 - 0.60 = 0.40
Likelihood of one person carrying an umbrella if it rains: P(Umbrella | R) = 0.80
Likelihood of one person carrying an umbrella if it does not rain: P(Umbrella | NR) = 0.20

2. Calculate Likelihoods for All Three Friends (P(E|R) and P(E|NR))

Since the friends' decisions to carry umbrellas are independent given the weather:

Likelihood of all three friends having umbrellas given it is raining:
P(E | R) = P(U₁|R) × P(U₂|R) × P(U₃|R) = (0.80) × (0.80) × (0.80) = 0.8³ = 0.512
Likelihood of all three friends having umbrellas given it is not raining:
P(E | NR) = P(U₁|NR) × P(U₂|NR) × P(U₃|NR) = (0.20) × (0.20) × (0.20) = 0.2³ = 0.008

3. Calculate the Total Probability of the Evidence (P(E))

We use the Law of Total Probability:
P(E) = P(E | R) × P(R) + P(E | NR) × P(NR)

P(E) = (0.512 × 0.60) + (0.008 × 0.40)
P(E) = 0.3072 + 0.0032
P(E) = 0.3104

This is the overall probability of seeing three friends with umbrellas, regardless of whether it's raining or not.

4. Apply Bayes' Theorem to Find P(R | E)

Bayes' Theorem states:
P(R | E) = [P(E | R) × P(R)] / P(E)

P(R | E) = (0.512 × 0.60) / 0.3104
P(R | E) = 0.3072 / 0.3104
P(R | E) ≈ 0.9896907...

Final Result

The probability that it is raining, given that you see three friends all carrying umbrellas, is:

P(Rain | All 3 umbrellas) ≈ 0.990

Or about 99.0%.

Interpretation: Our initial belief (prior) that it's raining was 60%. After observing strong evidence (all three friends with umbrellas, which is much more likely if it's raining than if it's not), our updated belief (posterior) that it's raining has increased significantly to approximately 99%. This demonstrates the power of Bayes' Theorem in updating probabilities based on new evidence.

Seattle Rain & Three Friends

Updating Beliefs: Advanced Bayes' Theorem

Bayes' Theorem with Multiple Observations

Calculating Combined Likelihoods