Let's break down the problem based on the given diagram and probabilities.

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

We are told the stargazer is searching in Area 51 and the object has crashed in Area 51. While the success rate for finding the object in Area 51 is 66%, the question implies we should not take this at face value for the conditional probability.

From the tree diagram:

• The event of "crashing in Area 51 and being found in Area 51" has a corresponding probability of 1/3 (the probability of crashing in Area 51) multiplied by 0.66 (the chance of finding it in Area 51), which gives:

  $P(\text{Found in Area 51 | Crashed in Area 51}) = \frac{1}{3} \times 0.66 = 0.22$

Thus, the probability that the stargazer will find the object if it crashed in Area 51 is $66\%$ (as this is given directly and confirmed by the structure). However, you mention that the answer is not $0.66$. Therefore, we might need to confirm there isn't additional information that affects this.

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

This requires using Bayes' Theorem to calculate the posterior probability. We need to find:

$P(\text{Crashed in Area 51} | \text{Not Found in Area 51})$

To apply Bayes' Theorem, we need the following:

$P(\text{Crashed in Area 51} | \text{Not Found in Area 51}) = \frac{P(\text{Not Found in Area 51} | \text{Crashed in Area 51}) P(\text{Crashed in Area 51})}{P(\text{Not Found in Area 51})}$

Let's break this down:

• $P(\text{Crashed in Area 51}) = \frac{1}{3}$
• $P(\text{Not Found in Area 51} | \text{Crashed in Area 51}) = 1 - 0.66 = 0.34$
• $P(\text{Not Found in Area 51})$ is the total probability of not finding the object in any area, which can be calculated using the law of total probability:

$P(\text{Not Found in Area 51}) = P(\text{Crashed in 50}) P(\text{Not Found in Area 50}) + P(\text{Crashed in 51}) P(\text{Not Found in Area 51}) + P(\text{Crashed in 52}) P(\text{Not Found in Area 52})$

We are given:

• $P(\text{Crashed in Area 50}) = \frac{1}{3}$ and $P(\text{Not Found in Area 50}) = 1 - 0.59 = 0.41$
• $P(\text{Crashed in Area 51}) = \frac{1}{3}$ and $P(\text{Not Found in Area 51}) = 0.34$
• $P(\text{Crashed in Area 52}) = \frac{1}{3}$ and $P(\text{Not Found in Area 52}) = 1 - 0.56 = 0.44$

Thus:

$P(\text{Not Found in Area 51}) = \frac{1}{3} \times 0.41 + \frac{1}{3} \times 0.34 + \frac{1}{3} \times 0.44$ $P(\text{Not Found in Area 51}) = \frac{1}{3} \times (0.41 + 0.34 + 0.44) = \frac{1}{3} \times 1.19 = 0.3967$

Now, we can apply Bayes' Theorem:

$P(\text{Crashed in Area 51} | \text{Not Found in Area 51}) = \frac{\frac{1}{3} \times 0.34}{0.3967} = \frac{0.1133}{0.3967} \approx 0.2855$

Thus, the probability that the crash was in Area 51 given that the search in Area 51 was unsuccessful is approximately 28.55%.

Would you like further details or explanations on any part of the solution? Here are some related questions to expand on this information:

1. How would the probability change if the success rates for the other areas were different?
2. Can Bayes' Theorem be used to solve for the probability of crashes in other areas as well?
3. What if the stargazer could search two areas? How would the search strategy change?
4. How does the structure of the probability tree influence the final probabilities?
5. Could the stargazer improve their chances by allocating search time differently?

Tip: When applying Bayes' Theorem, ensure you calculate all relevant probabilities accurately by considering each event's impact on the overall situation.