1. **State the problem:** We want to find the probability that it is actually raining in Newcastle given that all three friends say it is raining. 2. **Define events:** Let $R$ be the event "it is raining" and $\neg R$ be "it is not raining." Each friend tells the truth with probability $\frac{2}{3}$ and lies with probability $\frac{1}{3}$. 3. **Assumptions:** Assume the prior probability of rain $P(R) = p$ and no rain $P(\neg R) = 1-p$. Since no prior is given, assume $p=\frac{1}{2}$ for simplicity. 4. **Calculate probability all three say "raining" given it is raining:** $$P(\text{all say rain}|R) = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$$ 5. **Calculate probability all three say "raining" given it is not raining:** Each friend lies with probability $\frac{1}{3}$, so all three lie: $$P(\text{all say rain}|\neg R) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$$ 6. **Use Bayes' theorem to find $P(R|\text{all say rain})$:** $$ P(R|\text{all say rain}) = \frac{P(\text{all say rain}|R)P(R)}{P(\text{all say rain}|R)P(R) + P(\text{all say rain}|\neg R)P(\neg R)} $$ Substitute values: $$ = \frac{\frac{8}{27} \times \frac{1}{2}}{\frac{8}{27} \times \frac{1}{2} + \frac{1}{27} \times \frac{1}{2}} = \frac{\frac{8}{54}}{\frac{8}{54} + \frac{1}{54}} = \frac{\frac{8}{54}}{\frac{9}{54}} = \frac{8}{9} $$ 7. **Final answer:** The probability it is actually raining given all three friends say it is raining is $$\boxed{\frac{8}{9} \approx 0.8889}$$ This means there is about an 88.89% chance it is raining in Newcastle based on your friends' reports.