For the first part of this week’s assignment, there is a table of 90 events.

A1. Probability of Event A -> 30/90 -> 33.3%

A2. Probability of Event B -> 30/90 -> 33.3%

A3. Probability of Event A or B -> 50/90 -> 55.6%

A4. P(A or B) = P(A) + P(B) -> That equality wouldn’t be true because of the results that include both A and B. A and B have the same probability of occurring, but the total of their probabilities (66.7%) is higher than the probability of “A or B” (55.6%) because of the ten events where A and B are both true (which would be accounted for in the Addition Rule, where P(A or B) would be P(A) + P(B) – P(A and B)) – A and B are not mutually exclusive.

The second part concerns the probability that the weatherman is correct with regards to whether it will rain on a wedding day in Arizona.

B1. The answer given is almost true, I guess I’ll say. Technically false to three decimal places, but it’s an issue with the number crunching rather than the method presented being false, so it’s true at two decimal places.

B2. The formatting of the assignment looks like it leaves out a “/”, so for clarity I’ll write the formula to calculate P(A1|B) (using Bayes’ Theorem) on one line:

P(A1|B) = P(A1) * P(B|A1) / ( P(A1) * P(B|A1) + P(A2) * P(B|A2))

As the explanation in the assignment notes, that results in this equation:

P(A1|B) = 0.014 * 0.9 / (0.014 * 0.9 + 0.986 * 0.1)

By my calculations, rather than resulting in 0.111, the result of that calculation is 0.113.

Thus:

P(A1|B) = 0.113

And the probability of it raining on the wedding day is 11.3%.

I do agree that it feels like a weird result if one only looks at the probabilities for P(B|A1) and P(B|A2). Thinking about it further, though, the result we calculated comes because of the rarity of rainfall throughout the year. If the weatherman predicts rain on 10% of the days when it doesn’t rain, that means he predicts rain on 36 of the days when it doesn’t rain. Given that it only rains 5 days out of the year, the weatherman’s accuracy rate truly is terrible.