Using odds instead of percentage makes this problem simple.
Our estimate of rain in Seattle is 1:9 (one rainy day for every nine sunny days).
Our friends tell us the truth with odds of 2:1 (two truthful reports for each false report).
So after calling one friend and hearing "yes it's raining" we multiply 1:9 by 2:1 and get odds of 2:9 as our new expectation of rain. That's 2/11, about 18%, so we are more confident of rain.
The second friend says "yes" and we multiply 2:9 by another 2:1 to get 4:9, or 4/13 which is about 31%.
The third friend says "yes" and we multiply 4:9 by 2:1 to get 8:9 which is 8/17, the final answer of about 47%.
I have found a way to force Bayes in, but don't know if it's correct. Replacing the A's and B's with more descriptive terms, Bayes' Theorem is
odds of rain, given a "yes" = odds of "yes," given rain × odds of rain / odds of "yes"
So, after hearing the first "yes" we have
odds of rain, given a "yes" = (2:1 the odds we will hear a "yes" on a rainy day) × (1:9 our prior estimate for rain) / (1, our certainty of hearing "yes" because we just talked with Albert and he definitely said "yes")
This gives us 2:9 divided by 1, which is the desired 2/11 result. That denominator seems forced and flaky, though, and I am not sure what to do if Albert says "no." Perhaps the odds of rain given a "no" are 1:2 × 1:9 / 1 again because we definitely heard "no" for a total of 1:18 or about 5% chance of rain.
Odds are good b_b will be able to straighten this out.