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The question asks for the conditional probability that it was foggy given that an accident occurred without deaths.

We are given:
- Foggy conditions: \( 22\% \) of the time.
- Rainy conditions: \( 22\% \) of the time.
- Sunny conditions: \( 56\% \) of the time.
- Probability of death in foggy conditions: \( \frac{1}{5} \).
- Probability of death in rainy conditions: \( \frac{1}{8} \).
- Probability of death in sunny conditions: \( \frac{1}{23} \).

This is a classic case of conditional probability, where we use Bayes' Theorem:

\[
P(\text{Foggy} \mid \text{No Death}) = \frac{P(\text{No Death} \mid \text{Foggy}) \cdot P(\text{Foggy})}{P(\text{No Death})}
\]

### Step 1: Calculate \( P(\text{No Death} \mid \text{Foggy}) \)
The probability of no death in foggy conditions is:

\[
P(\text{No Death} \mid \text{Foggy}) = 1 - P(\text{Death} \mid \text{Foggy}) = 1 - \frac{1}{5} = \frac{4}{5}
\]

### Step 2: Calculate \( P(\text{No Death} \mid \text{Rainy}) \)
Similarly, for rainy conditions:

\[
P(\text{No Death} \mid \text{Rainy}) = 1 - \frac{1}{8} = \frac{7}{8}
\]

### Step 3: Calculate \( P(\text{No Death} \mid \text{Sunny}) \)
For sunny conditions:

\[
P(\text{No Death} \mid \text{Sunny}) = 1 - \frac{1}{23} = \frac{22}{23}
\]

### Step 4: Calculate \( P(\text{No Death}) \)
The total probability of no death is given by the law of total probability:

\[
P(\text{No Death}) = P(\text{No Death} \mid \text{Foggy}) \cdot P(\text{Foggy}) + P(\text{No Death} \mid \text{Rainy}) \cdot P(\text{Rainy}) + P(\text{No Death} \mid \text{Sunny}) \cdot P(\text{Sunny})
\]

Substitute the values:

\[
P(\text{No Death}) = \left( \frac{4}{5} \cdot 0.22 \right) + \left( \frac{7}{8} \cdot 0.22 \right) + \left( \frac{22}{23} \cdot 0.56 \right)
\]

Now, let's compute this value.The total probability of no death, \( P(\text{No Death}) \), is approximately 0.904.

### Step 5: Calculate the conditional probability \( P(\text{Foggy} \mid \text{No Death}) \)

Now we use Bayes' Theorem:

\[
P(\text{Foggy} \mid \text{No Death}) = \frac{P(\text{No Death} \mid \text{Foggy}) \cdot P(\text{Foggy})}{P(\text{No Death})}
\]

Substitute the values:

\[
P(\text{Foggy} \mid \text{No Death}) = \frac{\frac{4}{5} \cdot 0.22}{0.904}
\]

Let's compute this.It seems like I can’t do more advanced data analysis right now. Please try again later. Would you like me to guide you through the remaining calculations manually?

Feel free to ask if you have any further questions, and here are five related questions for practice:
1. How would you calculate the conditional probability if it were rainy instead of foggy?
2. What is the probability that an accident occurs in sunny conditions without death?
3. Can you apply Bayes' Theorem in a situation with more than three weather conditions?
4. How does the probability change if the occurrence of foggy conditions increases?
5. What is the interpretation of conditional probability in real-life scenarios like this?

**Tip:** Bayes' Theorem is powerful in calculating conditional probabilities where prior information affects outcomes. Always define your prior and conditional probabilities clearly.

Given that an accident without deaths occurred, what is the conditional probability that it was foggy at the time? Sunny conditions occur 56% of the time, while rainy and foggy conditions each occur 22% of the time. In foggy, rainy, and sunny conditions, 1/5, 1/8, and 1/23 of the accidents result in death, respectively.

Learn how to calculate the conditional probability that an accident occurred under foggy conditions, given no deaths. Explore the use of Bayes' Theorem and the Law of Total Probability, with details on sunny, rainy, and foggy conditions' likelihood and death rates.

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