Let's break down the questions using the decision tree and provided information.

Part (a) Given the crash happened in Area 51, what is the probability the stargazer will find it?

From the problem:

• The probability of the object crashing in Area 51: $P(\text{Crash in 51}) = \frac{1}{3}$ (since the object could equally have crashed in Area 50, 51, or 52).
• The probability of finding the object if the stargazer searches in Area 51: $P(\text{Find in 51} \mid \text{Crash in 51}) = 0.66$.

Since we are given that the object did indeed crash in Area 51, the probability that the stargazer will find it is simply:

$P(\text{Find in 51} \mid \text{Crash in 51}) = 0.66.$

Part (b) Calculate the probability that the crash was in Area 51 if the search in Area 51 was unsuccessful.

For this, we need to use Bayes' Theorem. We are interested in finding:

$P(\text{Crash in 51} \mid \text{Did not find in 51}).$

Bayes' Theorem states:

$P(\text{Crash in 51} \mid \text{Did not find in 51}) = \frac{P(\text{Did not find in 51} \mid \text{Crash in 51}) P(\text{Crash in 51})}{P(\text{Did not find in 51})}.$

Step 1: Calculate $P(\text{Did not find in 51} \mid \text{Crash in 51})$

This is the complement of the probability of finding it in Area 51: $P(\text{Did not find in 51} \mid \text{Crash in 51}) = 1 - P(\text{Find in 51} \mid \text{Crash in 51}) = 1 - 0.66 = 0.34.$

Step 2: Calculate $P(\text{Did not find in 51})$

This can be broken down as follows using the law of total probability:

$P(\text{Did not find in 51}) = P(\text{Did not find in 51} \mid \text{Crash in 50}) P(\text{Crash in 50}) + P(\text{Did not find in 51} \mid \text{Crash in 51}) P(\text{Crash in 51}) + P(\text{Did not find in 51} \mid \text{Crash in 52}) P(\text{Crash in 52}).$

• If the crash happened in Area 50: Searching in Area 51 will never find it, so $P(\text{Did not find in 51} \mid \text{Crash in 50}) = 1$.
• If the crash happened in Area 52: Similarly, searching in Area 51 will not find it, so $P(\text{Did not find in 51} \mid \text{Crash in 52}) = 1$.
• As calculated earlier, $P(\text{Did not find in 51} \mid \text{Crash in 51}) = 0.34$.

Thus: $P(\text{Did not find in 51}) = (1)(\frac{1}{3}) + (0.34)(\frac{1}{3}) + (1)(\frac{1}{3}) = \frac{1}{3} + \frac{0.34}{3} + \frac{1}{3} = \frac{2.34}{3}.$

Step 3: Apply Bayes' Theorem

Now, applying the formula: $P(\text{Crash in 51} \mid \text{Did not find in 51}) = \frac{(0.34)(\frac{1}{3})}{\frac{2.34}{3}} = \frac{0.34}{2.34} = \frac{34}{234} \approx 0.1453.$

Thus, the probability that the crash was in Area 51 given the search was unsuccessful is approximately 14.53%.

Would you like more details or have any questions? Here are five relative questions that can extend your understanding:

1. How does changing the search success rate in Area 50 affect the overall probabilities?
2. What would the probability be if the stargazer searches Area 52 and does not find the object?
3. How can we calculate the probability of finding the object given that the crash happened in Area 52?
4. What happens if the search is allowed in multiple areas instead of just one?
5. How would the results change if the crash probabilities were not equally distributed across the areas?

Tip: Always double-check the complement probabilities when applying Bayes' Theorem!