Rain in Seattle

Solution & Explanation

To solve the problem of determining the probability that it is truly raining in Seattle given that all three friends say "Yes," we will use Bayes' Theorem. This theorem helps us update the probability of a hypothesis based on new evidence.

Step 1: Define the Events

• R: It is raining in Seattle.
• R': It is not raining in Seattle.
• Y: A friend says "Yes," it is raining.

Step 2: Define the Probabilities

• P(R): Prior probability that it is raining. Assume this is 0.5, indicating we have no prior information favoring rain or no rain.
• P(R'): Prior probability that it is not raining = 1 - P(R) = 0.5.
• P(Y|R): Probability a friend says "Yes" given it is raining = 2/3.
• P(Y|R'): Probability a friend says "Yes" given it is not raining = 1/3.

Step 3: Calculate the Probability of All Friends Saying "Yes"

• P(YYY|R): Probability that all three friends say "Yes" given it is raining = (2/3)^3 = 8/27.
• P(YYY|R'): Probability that all three friends say "Yes" given it is not raining = (1/3)^3 = 1/27.

Step 4: Apply Bayes' Theorem

Bayes' Theorem in this context is:

$P(R|YYY) = \frac{P(YYY|R) \cdot P(R)}{P(YYY)}$

Where:

$P(YYY) = P(YYY|R) \cdot P(R) + P(YYY|R') \cdot P(R')$

Substitute the known values:

$P(YYY) = \frac{8}{27} \cdot 0.5 + \frac{1}{27} \cdot 0.5 = \frac{4}{27} + \frac{0.5}{27} = \frac{4.5}{27} = \frac{1}{6}$

Now, substitute back into Bayes' Theorem:

$P(R|YYY) = \frac{\frac{8}{27} \cdot 0.5}{\frac{1}{6}} = \frac{4}{27} \div \frac{1}{6} = \frac{4 \times 6}{27} = \frac{24}{27} = \frac{8}{9}$

Thus, the probability that it is actually raining in Seattle, given that all three friends say "Yes," is $\frac{8}{9}$ or approximately 0.889, which is about 88.9%.

Conclusion

Given the information and assumptions, there is an 88.9% chance that it is truly raining in Seattle when all three friends independently affirm that it is raining. This high probability suggests that you should indeed pack an umbrella if you are traveling to Seattle.