Let:
- \( R \) be the event that it is raining.
- \( D \) be the event that the flight is delayed.
From the problem, we know:
- \( P(R) = 0.06 \) (probability that it will rain)
- \( P(D) = 0.09 \) (probability that the flight will be delayed)
- \( P(R^c \cap D^c) = 0.86 \) (probability that it will not rain and the flight will leave on time)
We want to find \( P(R \cap D) \) (the probability that it is raining and the flight is delayed).
First, we can use the law of total probability to express \( P(D) \): \[ P(D) = P(D \mid R) P(R) + P(D \mid R^c) P(R^c) \]
We know \( P(R^c) = 1 - P(R) = 1 - 0.06 = 0.94 \).
Next, we can find \( P(D \mid R^c) \) using the information about \( P(R^c \cap D^c) \): Since \( P(R^c \cap D^c) = 0.86 \), we can find \( P(D^c \mid R^c) \) (the probability of the flight not being delayed given that it is not raining): \[ P(D^c \mid R^c) = \frac{P(R^c \cap D^c)}{P(R^c)} = \frac{0.86}{0.94} \] Calculating this gives: \[ P(D^c \mid R^c) \approx 0.9152 \] Now, we can find \( P(D \mid R^c) \):
\[ P(D \mid R^c) = 1 - P(D^c \mid R^c) \approx 1 - 0.9152 \approx 0.0848 \]
Now we have everything we need to use the law of total probability: \[ P(D) = P(D \mid R) P(R) + P(D \mid R^c) P(R^c) \] Substituting in known values: \[ 0.09 = P(D \mid R)(0.06) + (0.0848)(0.94) \]
Calculating \( 0.0848 \times 0.94 \): \[ 0.0848 \times 0.94 \approx 0.079792 \]
Now we substitute back into the equation: \[ 0.09 = P(D \mid R)(0.06) + 0.079792 \] Rearranging gives: \[ P(D \mid R)(0.06) = 0.09 - 0.079792 \approx 0.010208 \] Now, solve for \( P(D \mid R) \): \[ P(D \mid R) = \frac{0.010208}{0.06} \approx 0.17013333 \]
Finally, we can find \( P(R \cap D) \): \[ P(R \cap D) = P(D \mid R) P(R) \approx 0.17013333 \times 0.06 \approx 0.010208 \]
Rounding to the nearest thousandth gives: \[ \boxed{0.010} \]