1. **State the problem:** We are given probabilities related to rain and flight delays at LaGuardia Airport. We want to find the probability that it is raining given that the flight has been delayed, i.e., $P(\text{Rain} \mid \text{Delayed})$. 2. **Given data:** - $P(\text{Rain}) = 0.17$ - $P(\text{Delayed}) = 0.11$ - $P(\text{No Rain and On Time}) = 0.74$ 3. **Important rules and formulas:** - The total probability must sum to 1: $$P(\text{Rain and Delayed}) + P(\text{Rain and On Time}) + P(\text{No Rain and Delayed}) + P(\text{No Rain and On Time}) = 1$$ - Conditional probability formula: $$P(\text{Rain} \mid \text{Delayed}) = \frac{P(\text{Rain and Delayed})}{P(\text{Delayed})}$$ 4. **Find $P(\text{No Rain})$:** $$P(\text{No Rain}) = 1 - P(\text{Rain}) = 1 - 0.17 = 0.83$$ 5. **Find $P(\text{No Rain and Delayed})$:** Since $P(\text{No Rain and On Time}) = 0.74$, and $P(\text{No Rain}) = 0.83$, then $$P(\text{No Rain and Delayed}) = P(\text{No Rain}) - P(\text{No Rain and On Time}) = 0.83 - 0.74 = 0.09$$ 6. **Find $P(\text{Rain and Delayed})$:** Using total probability: $$1 = P(\text{Rain and Delayed}) + P(\text{Rain and On Time}) + P(\text{No Rain and Delayed}) + P(\text{No Rain and On Time})$$ Also, $P(\text{Rain}) = P(\text{Rain and Delayed}) + P(\text{Rain and On Time}) = 0.17$ So, $$P(\text{Rain and On Time}) = 0.17 - P(\text{Rain and Delayed})$$ Substitute all known values: $$1 = P(\text{Rain and Delayed}) + (0.17 - P(\text{Rain and Delayed})) + 0.09 + 0.74$$ Simplify: $$1 = 0.17 + 0.09 + 0.74 = 1.0$$ This confirms the probabilities are consistent but does not directly give $P(\text{Rain and Delayed})$. 7. **Use $P(\text{Delayed})$ to find $P(\text{Rain and Delayed})$:** $$P(\text{Delayed}) = P(\text{Rain and Delayed}) + P(\text{No Rain and Delayed}) = 0.11$$ Substitute $P(\text{No Rain and Delayed}) = 0.09$: $$0.11 = P(\text{Rain and Delayed}) + 0.09$$ $$P(\text{Rain and Delayed}) = 0.11 - 0.09 = 0.02$$ 8. **Calculate the conditional probability:** $$P(\text{Rain} \mid \text{Delayed}) = \frac{P(\text{Rain and Delayed})}{P(\text{Delayed})} = \frac{0.02}{0.11} \approx 0.1818$$ 9. **Round to the nearest thousandth:** $$\boxed{0.182}$$