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Let's analyze the given problem and work through the parts step by step.

### Problem Analysis:
The professor has **three chances** to hit the magpie with the following probabilities:
- For the **first shot**, the probability of hitting the magpie is **0.65**.
- If the magpie is **not hit on the first shot**, the probability of hitting it on the **second shot** decreases by **0.05**, making it **0.60** for the second shot.
- If the professor **misses again** on the second shot, the probability of hitting on the **third shot** further decreases by **another 0.05**, making it **0.55** for the third shot.

This defines a probability distribution for the random variable \( X \), which counts the number of shots the professor needs to hit the magpie. The professor has up to 3 shots.

#### Part (a) - Fill in the missing probabilities:

The table needs to be filled with the probabilities for:
- \( P(X = 2) \) (the professor hits on the **second** shot),
- \( P(X = 3) \) (the professor hits on the **third** shot).

Let's compute these step-by-step:

#### 1. Probability of hitting on the **first shot** (\( X = 1 \)):
This is directly given as \( P(X = 1) = 0.65 \).

#### 2. Probability of hitting on the **second shot** (\( X = 2 \)):
To hit on the second shot, the professor must **miss the first shot** and **hit on the second shot**.

- Probability of missing the first shot = \( 1 - 0.65 = 0.35 \),
- Probability of hitting on the second shot = \( 0.60 \).

Therefore:
\[
P(X = 2) = 0.35 \times 0.60 = 0.21.
\]

#### 3. Probability of hitting on the **third shot** (\( X = 3 \)):
To hit on the third shot, the professor must **miss the first two shots** and then **hit on the third shot**.

- Probability of missing the first shot = \( 0.35 \),
- Probability of missing the second shot = \( 1 - 0.60 = 0.40 \),
- Probability of hitting on the third shot = \( 0.55 \).

Therefore:
\[
P(X = 3) = 0.35 \times 0.40 \times 0.55 = 0.077.
\]

Now, the complete probability distribution table is:

| \( x \) | 1    | 2    | 3    |
|--------|------|------|------|
| \( P(X = x) \) | 0.65 | 0.21 | 0.077 |

#### Part (b) - Probability of taking **at least two shots** (\( P(X \geq 2) \)):

The probability that the professor needs at least two shots is the sum of \( P(X = 2) \) and \( P(X = 3) \).

\[
P(X \geq 2) = P(X = 2) + P(X = 3) = 0.21 + 0.077 = 0.287.
\]

### Final Answers:
- (a) The missing probabilities are:
  - \( P(X = 2) = 0.21 \),
  - \( P(X = 3) = 0.077 \).

- (b) \( P(X \geq 2) = 0.29 \) (rounded to two decimal places).

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Would you like further clarification on any of the steps? Here are some related questions to extend this concept:

1. What is the total probability of hitting the magpie in three shots or fewer?
2. How would the probabilities change if the initial probability of hitting the magpie was different, say 0.7 instead of 0.65?
3. What is the expected number of shots needed for the professor to hit the magpie?
4. Can this problem be modeled using geometric distribution, and why or why not?
5. How does the cumulative probability change for each subsequent shot?

**Tip:** When dealing with conditional probabilities (like missing one shot to hit on the next), always remember to account for both the conditions and the success probability at each stage.

A statistics professor has three chances to hit a magpie with a probability of 0.65 for the first shot, decreasing by 0.05 for each subsequent shot. Compute the missing probabilities and find the probability that at least two shots are needed.

In this problem, a statistics professor attempts to hit a magpie with a slingshot. Learn how to compute the missing probabilities and find the probability of needing at least two shots, given decreasing hit probabilities for each shot. This is a college-level statistics problem focusing on conditional probability and the multiplication rule for independent events.

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